4 Marks - Page 5 - Maths STD 12 Science Questions - Vidyadip

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4 Marks

Question 2014 Marks

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line $\frac{\text{x}-3}{1}=\frac{\text{y}-6}{5}=\frac{\text{z}-4}{4}$

Answer

Required equation of plane is passing through the point (3, 2, 0)
$\therefore$ a(x - 3) + b(y - 2) + c(z - 0) = 0
⇒ a(x - 3) + b(y - 2) + cz = 0 ....(i)
Required equation of plane also contains the line $\frac{\text{x}-3}{1}=\frac{\text{y}-6}{5}=\frac{\text{z}-4}{4},$ so it passes through the point (3, 2, 0)
⇒ a(3 - 3) + b(6 - 2) + c(4) = 0
⇒ 4b + 4c = 0 ...(ii)
Also plane will be parallel to,
a(1) + b(5) + c(4) = 0
a + 5b + 4c = 0 ....(iii)
Solving (ii) and (iii) by cross multiplication,
$\frac{\text{a}}{16-20}=\frac{\text{b}}{4-0}=\frac{\text{c}}{0-4}=\lambda(\text{say})$
$-\frac{\text{a}}{4}=\frac{\text{b}}{4}=-\frac{\text{c}}{4}=\lambda(\text{say})$
$\Rightarrow\text{a}=-\lambda,\text{ b}=\lambda,\text{ c}=-\lambda$
Put $\text{a}=-\lambda,\text{ b}=\lambda,\text{ c}=-\lambda$ in equation (i) we get
$(-\lambda)(\text{x}-3)+(\lambda)(\text{y}-2)+(-\lambda)\text{z}=0$
$\Rightarrow-\text{x}+3+\text{y}-2-\text{z}=0$
$\Rightarrow\text{x}-\text{y}+\text{z}-1=0$

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Question 2024 Marks

Find the coordinates of the foot of the perependicular drawn from the origin to the plane 2x - 3y + 4z - 6 = 0.

Answer

Let M be the foot of the perpendicular of the origin P(0, 0, 0) in the plane 2x - 3y + 4z - 6 =0.
Then, PM is normal to the plane. So, the direction rations of PM are proportional to 2, -3, 4.
Since PM passes through P(0, 0, 0) and has direction ratios proportional to 2, -3, 4 the equation ot PQ is
$\frac{\text{x}-0}{2}=\frac{\text{y}-0}{-3}=\frac{\text{z}-0}{4}=\text{r (say)}$
Let the coordinted of M be (2r, -3r, 4r)
Since M lies in the plane 2x - 3y + 4z - 6 = 0,
2(2r) - 3(-3r) + 4(4r) - 6 = 0
⇒ 4r + 9r + 16r - 6 = 0
⇒ 29r - 6 = 0
$\Rightarrow\ \text{r}=\frac{6}{29}$
Substituting the value of r in the coordinated of Ml we get
$\text{M}=(2\text{r},-3\text{r},4\text{r})=\bigg(2\Big(\frac{6}{29}\Big),-3\Big(\frac{6}{29}\Big),4\Big(\frac{6}{29}\Big)\bigg)$
$=\Big(\frac{12}{29},\frac{-18}{29},\frac{24}{29}\Big)$

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Question 2034 Marks

Show that the vectors $\overrightarrow{\text{a}},\overrightarrow{\text{b}} \text{and}{\overrightarrow{\text{c}}}$ are coplanar $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}}+\overrightarrow{\text{c}}\text{and} \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar.

Answer

Given that $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar
$\therefore[\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}] = 0$
$\text{i.e} \overrightarrow{\text{(a}} +\overrightarrow{\text{b})}, \big\{\overrightarrow{\text{(b}} + \overrightarrow{\text{c})} \times\overrightarrow{\text{(c}} + \overrightarrow{\text{a})}\big\} = 0$
$\overrightarrow{\text{(a}} +\overrightarrow{\text{b})}. \big\{\overrightarrow{\text{(b}} \times \overrightarrow{\text{c}}+ \overrightarrow{\text{b}} \times\overrightarrow{\text{a}}\times\overrightarrow{\text{c}} \times \overrightarrow{\text{a})}\big\} = 0$
$\Rightarrow\overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{a}}(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})}+\overrightarrow{\text{b}}. (\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})} = 0$
$\Rightarrow 2 [\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}] = \text{0 or} [\overrightarrow{\text{a}},\overrightarrow{\text{b}}, \overrightarrow{\text{c] }} = 0$
$\Rightarrow\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}\text{are coplanar}.$

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Question 2044 Marks

Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}$ and$\frac{\text{x}}{-3}=\frac{\text{y}}{2}=\frac{\text{z}}{5}.$

Answer

Let theD.R’s of the required line be a,b , c

$\therefore\ $a + 2b + 3c = 0

and –3a + 2b + 5c = 0

$\Rightarrow\ \frac{\text{a}}{4}=\frac{\text{b}}{-14}=\frac{\text{c}}{8}\ \therefore\ \text{DRS are} \ 2,-7,4$

$\therefore\ $ Equations of line are $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{-7}=\frac{\text{z-3}}{4}$

which, in vector form is,$\overrightarrow{\text{r}}=(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})+\lambda(2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}})$

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Question 2054 Marks

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0

Answer

The direction ratios of normal to the plane, L₁: a₁x + b₁y + c₁z = 0,
are a₁, b₁, c₁ and L₂: a₂x + b₂y + c₂z = 0 are a₂, b₂, c₂
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between L₁ and L₂ is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the given planes are 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0
Here, a₁ = 2, b₁ = -2, c₁ = 4 and a₂ = 3, b₂ = -3, c₂ = 6
a₁a₂ +b₁b₂ + c₁c₂ = 2 × 3 + (-2) × (-3) + 4 × 6 = 6 + 6 + 24
$=36\neq0$
Thus, the given planes are not perpendicular to each other.
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-3}=\frac{2}{3}\text{ and } \frac{\text{c}_1}{\text{c}_2}=\frac{4}{6}=\frac{2}{3}$
$\therefore\ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Thus, the given planes are parallel to each other.

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Question 2064 Marks

Find the vector equation of the line through the origin which is perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=3$

Answer

Required line is perpendicular to plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=3,$ so line is parallel to the normal vector $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ of plane.
And it is passing through point $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
We know that equation of a passing through $\vec{\text{a}}$ and parallel to vector $\vec{\text{b}}$ is,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\ ...(\text{i})$
Here, $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{b}}=\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
So, $\vec{\text{r}}=(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
Hence, equation required line is
$\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$

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Question 2074 Marks

Find the angle between line $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{1}$ and the plane 2x + y - z = 4.

Answer

We know that the angle $\theta$ between the line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and plane a₂x + b₂y + c₂z + d₂= 0 is given by
$\sin\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}\ ...(\text{i})$
Given, equation of line is
$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{1}$
So, a₁ = 1, b₁ = -1, c₁ = 1
Given equation of plane is 2x + y - z - 4 = 0
So, a₂ = 2, b₂ = 1, c₂ = -1
Put these value in equation (i),
$\sin\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}$
$\sin\theta=\frac{(1)(2)+(-1)(1)+(1)(-1)}{\sqrt{(1)^2+(-1)^2+(1)^2}\sqrt{(2)^2+(1)^2+(-1)^2}}$
$\sin\theta=\frac{2-1-1}{\sqrt{1+1+1}\sqrt{4+1+1}}$
$\sin\theta=\frac{0}{\sqrt{3}\sqrt{6}}$
$\sin\theta=0$
$\theta=0^{\circ}$
angle between plane and line $=0^{\circ}$

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Question 2084 Marks

Find the angle between the lines whose direction cosines are given by the equations:
2l + 2m - n = 0, mn + ln + lm = 0

Answer

The given equation are,
2l + 2m - n = 0 .....(1)
mn + ln + lm = 0 .....(2)
From (1), We get n = 2l + 2m.
Putting n = 2l + 2m in (2), We get
m(2l + 2m) + l(2l + 2m) + lm = 0
2lm + 2m² + 2l² + 2ml + lm = 0
2ml² + 5lm + 2l² = 0
2m² + 4lm + lm + 2l² = 0
(2m + l) (m + 2l) = 0
$\Rightarrow\text{m}=-\frac{1}{2}$ or $\text{m}=-2\text{l}$
By putting $\text{m}=-\frac{\text{l}}{2}$ in (1) we get n = l
By putting m = 2l in (i) we get n = -2l
So, vector parallel to these lines are
$\vec{\text{a}}=\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ respectively.
If $\theta$ is the angle between the lines, then $\theta$ is also the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$
then,
$\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{1+1-2}{\sqrt{1+\frac{1}{4}+1}\sqrt{1+4+9}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$.

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Question 2094 Marks

Find a vector of magnitude 26 units normal to the plane 12x - 3y + 4z = 1

Answer

Given, equation of plane is,
12x - 3y + 4z = 1
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(12\hat{\text{i}}-3\hat{\text{j}}-4\hat{\text{k}})=1$
$\vec{\text{r}}\cdot\vec{\text{n}}=1$
So, normal to the plane is
$\vec{\text{n}}=12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{(12)^2+(-3)^2+(4)^2}$
$=\sqrt{144+9+16}$
$=\sqrt{144+25}$
$=\sqrt{169}$
$=13$
$\text{unit vector}\hat{\text{ n}}=\frac{12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}}{13}$
$=\frac{12\hat{\text{i}}}{13}-\frac{3}{13}\hat{\text{j}}+4\hat{\text{k}}$
A vector to the plane with magnitude
$26=26\hat{\text{n}}$
$=26\Big(\frac{12\hat{\text{i}}}{13}-3\hat{\text{j}}+4\hat{\text{k}}\Big)$
Required vector $=24\hat{\text{i}}-6\hat{\text{j}}+8\hat{\text{k}}$

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Question 2104 Marks

Show that the lines $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{4}$ and $\frac{\text{x}-4}{5}=\frac{\text{y}-1}{2}=\text{z}$ intersect. Also, find their point of intersection.

Answer

We have lines
$\text{L}_1:\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{4}=\lambda$
and $\text{L}_2:\frac{\text{x}-4}{5}=\frac{\text{y}-1}{2}=\text{z}=\mu$
Any point on the line L₁is $(2\lambda+1,3\lambda+2,4\lambda+3)$
Any point on the line L₂is $(5\mu+4,2\mu+1,\mu)$
$(2\lambda+1,3\lambda+2,4\lambda+3)=(5\mu+4,2\mu+1,\mu)$
$\Rightarrow2\lambda+1=5\mu+4,3\lambda+2=2\mu+1$ and $4\lambda+3=\mu$
Solving first two equations we get $\lambda=-1,\mu=-1$
These values of $\lambda=-1,\mu=-1$ also satisfy the third equation.
Thus lines intersect.
Also the point of intersection is (-1, -1, -1).

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Question 2114 Marks

If the line drawn from (4, -1, 2) meets a plane at right at the point (-10, 5, 4) find the equation of the plane.

Answer

The normal is passing through the point A(4, -1, 2) and B(-10, 5, 4) So,
$\overrightarrow{\text{n}}=\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(-10\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})-(4\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$
$=-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}}$
Since the plane passes through (-10, 5, 4), $\vec{\text{a}}=-10\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is
$\vec{\text{r}}\cdot(-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}})=(-10\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})(-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}})=140+30+8$
$\Rightarrow\vec{\text{r}}\cdot\big(-2(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})\big)=178$
$\Rightarrow\vec{\text{r}}\cdot(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})=-89$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})=-89$
$\Rightarrow7\text{x}-3\text{y}-\text{z}=-89$
$\Rightarrow7\text{x}-3\text{y}-\text{z}+89=0$

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Question 2124 Marks

Find the shortest distance between the lines
$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$ $\text{and}\ \vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big).$

Answer

$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$

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Question 2134 Marks

Find the shortest distance between the following two lines:

$\vec{\text{r}}=\text{(1 +}\lambda)\hat{\text{i}}+\text{(2 -}\lambda)\hat{\text{j}}+(\lambda+\text{1)}\hat{\text{k}};$

$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})+\mu(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}).$

Answer

Here $\vec{\text{a}_{1}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k},}\vec{\text{ b}_{1}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$

$\vec{\text{a}_{2}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k},}\vec{\text{ b}_{2}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{2k}}$

$\vec{\text{a}_{2}}-\vec{\text{a}_{1}}=\hat{\text{i}}-3{\text{j}}-2{\text{k}}$

$\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}=-3\hat{\text{i}}+3{\text{k}}$

Shortest distance (d) = $\Bigg|\frac{\Big(\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}\Big)\cdot\Big(\vec{\text{a}_{2}}-\vec{\text{a}_{1}}\Big)}{\Big|\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}\Big|}\Bigg|$

=$\Bigg|\frac{\Big(-3\hat{\text{i}}+3\hat{\text{k}}\Big)\cdot\Big(\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\Big)}{\sqrt{9 + 9}}\Bigg|$

$=\frac{9}{3\sqrt{2}}\text{ OR }\frac{3\sqrt{2}}{2}$.

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Question 2144 Marks

Find the length and the foot of the perpendicular drawn from the point (2, –1, 5) to the line $\frac{\text{x - 11}}{10}=\frac{\text{y + 2}}{-4}=\frac{\text{z + 8}}{-11}.$

Answer

Any point P on the line is given by

x = 10$\lambda$ + 11, y = -4$\lambda$ - 2, z = - 11$\lambda$ - 8

The given point is Q (2, -1, 5)

Direction Ratio's of PQ are 10$\lambda$ + 9, - 4$\lambda$ - 1, - 11$\lambda$ - 13

PQ $\perp$ to the given line

$\therefore$ 10 (10$\lambda$ + 9) - 4 (-4$\lambda$ - 1) - 11 (-11$\lambda$ - 13) = 0

100 $\lambda$ + 90 + 16$\lambda$ + 4 + 121$\lambda$ + 143 = 0

237$\lambda$ + 237 = 0 $\Rightarrow$ $\lambda$ = - 1

$\therefore$ The point P is (11 - 10, 4 - 2, 11 - 8)

OR (1, 2, 3)

$\therefore$ PQ² = (2 - 1)² + (-1 - 2)² + (5 - 3)²

= 1 + 9 + 4 = 14

$\Rightarrow$ PQ = $\sqrt{14}$.

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Question 2154 Marks

Show that the points A, B, C with position vectors $2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \hat{\text{i}} - 3\hat{\text{j}} - 5\hat{\text{k}} \text{ and } 3\hat{\text{i}} - 4\hat{\text{j}} - 4\hat{\text{k}}$ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.

Answer

$\vec{\text{AB}} = - \hat{\text{i}} - 2\hat{\text{j}} - 6\hat{\text{k}}, \vec{\text{BC}} = 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \vec{\text{CA}} = -\hat{\text{i}} + 3\hat{\text{j}} + 5\hat{\text{k}}$
Since $\vec{\text{AB}}, \vec{\text{BC}}, \vec{\text{CA}},$ are not parallel vectors, and $\vec{\text{AB}} + \vec{\text{BC}} + \vec{\text{CA}} = \vec{0} \therefore \text{A, B, C}$ form a triangle
$\text{Also} \vec{\text{ BC}}. \vec{\text{CA}} = 0 \text{ }\text{ }\text{ }\text{ }\text{ } \therefore\text{A, B, C}$ form a right triangle
$\text{Area of} \Delta = \frac{1}{2} | \vec{\text{AB}} \times \vec{\text{BC}}| = \frac{1}{2} \sqrt{210}$

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Question 2164 Marks

Find the position vector of the food of perpendicular and the perpendicular distance from the point P with position vector $2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-26=0.$ Also find image or P in the plane.

Answer

Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) in the plane
$\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-26=0$ or 2x + y + 3z - 26 = 0
Then, PM is the normal to the plane. So, the direction rations of PM are proportional to 2, 1, 3.
Since PM passes throught P(2, 3, 4) and has direction rations proportional to 2, 1, 3 so the equation or PM is
$\frac{\text{x}-2}{2}=\frac{\text{y}-3}{1}=\frac{\text{z}-4}{3}=\text{r (say)}$
Let the coordinates or M be (2r + 2, r + 3, 3r + 4). Since M lies in the plane 2x + y + 3z - 26 = 0, So
2(2r + 2) + r + 3 + 3(3r + 4) - 26 = 0
⇒ 4r + 4 + r + 3 + 9r + 12 - 26 = 0
⇒ 14r - 7 = 0
$\Rightarrow\text{r}=\frac{1}{2}$
Therefore, the coordinates of M are
$(2\text{r}+2,\text{r}+3,3\text{r}+4)\\=\Big(2\times\frac{1}{2}+2,\frac{1}{2}+3,3\times\frac{1}{2}+4\Big)\\=\Big(3,\frac{7}{2},\frac{11}{2}\Big)$
Thus, the position vector of the foot of perpendicular are $3\hat{\text{i}}+\frac{7}{2}\hat{\text{j}}+\frac{11}{2}\hat{\text{k}}.$
Now,
Length of the perpendicular from P on to the given plane
$=\Big|\frac{2\times2+1\times3+3\times4-26}{\sqrt{4+1+9}}\Big|$
$=\frac{7}{\sqrt{14}}$
$=\sqrt{\frac{7}{2}}\text{ units}$
Let Q(x₁, y₁, z₁) be the image of point P in the given plane.
Then, the coordinates of M are $\Big(\frac{\text{x}_1+2}{2},\frac{\text{y}_1+3}{2},\frac{\text{z}_1+4}{2}\Big)$
But, the coordinate or M are $\Big(3,\frac{7}{2},\frac{11}{2}\Big)$
$\therefore\Big(\frac{\text{x}_1+2}{2},\frac{\text{y}_1+3}{2},\frac{\text{z}_1+4}{2}\Big)=\Big(3,\frac{7}{2},\frac{11}{2}\Big)$
$\Rightarrow \frac{\text{x}_1+2}{2}=3,\frac{\text{y}_1+3}{2}=\frac{7}{2},\frac{\text{z}_1+4}{2}=\frac{11}{2}$
$\Rightarrow\text{x}_1=4,\text{y}_1=4,\text{z}_1=7$
Thus, the coordinates of the image of the point P in the given plane are (4, 4, 7).

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Question 2174 Marks

Find the distance of the point (-1, -5, -10) from the point of intersection of the line $\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$ and the plane $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5.$

Answer

The given equation of the line is
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}=(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}$
The coordinates of any point on this line are of the form
$(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}$
or $(2+3\lambda,-1+4\lambda,2+2\lambda)$
Since this point lies on the plane $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5,$
$\Big[(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}\Big]\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$
$\Rightarrow2+3\lambda+1-4\lambda+2+2\lambda-5=0$
$\Rightarrow\lambda=0$
So, the coordinates of the point are
$(2+3\lambda,-1+4\lambda,2+2\lambda)$
$=(2+0,-1+0,2+0)$
$=(2,-1,2)$
Distance between (2, -1, 2) and (-1, -5, -10)
$=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}$
$=\sqrt{9+16+144}$
$=13\text{ units}$

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Question 2184 Marks

If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane $\vec{\text{r}}\Big(3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}\Big)+13=0,$ then find the value of p.

Answer

Equation of the given plane is $\vec{\text{r}}\Big(3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}\Big)+13=0$
$\Rightarrow\ \ \Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big).\Big(3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}\Big)+13=0$
$\Big[\because\ \vec{\text{r}}=\text{Position vector of any point (x, y, z) on the plane }\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big]$
⇒ 3x + 4y - 12z + 13 = 0 ....(i)
Also, the point (1, 1, p) and (-3, 0, 1) are equidistance from plane (i)
⇒ (Perpendicular) distance of point (1, 1, p) from plane (i) = Distance of point (-3, 0, 1) from plane (i)
$\Rightarrow\ \ \ \frac{|3(1)+4(1)-12(\text{p})+13|}{\sqrt{9+16+144}}=\frac{|3(-3)+4(0)-12(1)+13|}{\sqrt{9+16+144}}$
$\Rightarrow\ \ \ \frac{|3+4-12\text{p}+13|}{13}=\frac{|-9-12+13|}{13}$
⇒ |20 - 12p| = |-8|
$\Rightarrow\ \ 20-12\text{p}=\pm8\ \ \ [\because\ \text{If }|\text{x}|=\text{a, a}\geq0,\ \text{then x}=\pm\text{a}]$

Taking positive sign,	20 - 12p = 8	⇒	-12p = -12	⇒	p = 1
Taking negative sign,	20 - 12p = -8	⇒	-12p = -28	⇒	$\text{p}=\frac{-28}{-12}=\frac{7}{3}$

Hence, the values of p are $1\ \text{or}\ \frac{7}{3}.$

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Question 2194 Marks

Find the equation of the plane through (2, 3, -4) and (1, -1, 3) and parallel to x-axis.

Answer

The equation of the plane through (2, 3, -4) is
a(x - 2) + b(y - 3) + c(z + 4) = 0 ....(i)
This plane passes through (1, -1, 3). So,
a(1 - 2) + b(-1 - 3) + c(3 + 4) = 0
⇒ -a - 4b + 7c = 0 ....(ii)
Again plane (i) is parallel to x-axis. It means that plane (i) is perpendicular to the yz-plane whose equation is x = 0 or 1x + 0y + 0z = 0
⇒ a(1) + b(0) + c(0) = 0 ....(iii) (Because a₁a₂ + b₁b₂ + c₁c₂= 0)
Solving (i), (ii) and (iii), we get
$\begin{vmatrix}\text{x}-3&\text{y}-3&\text{z}+4\\-1&-4&7\\1&0&0\end{vmatrix}=0$
$\Rightarrow0(\text{x}-3)+7(\text{y}-3)+4(\text{z}+4)=0$
$\Rightarrow7\text{y}+4\text{z}-5=0$

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Question 2204 Marks

Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+\mu(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$

Answer

Given, equation of plane,
$\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+\mu(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
We know that, $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represent a plane passing through a point having position vector $\vec{\text{a}}$ and parallel to vectors $\vec{\text{b}}$ and $\vec{\text{c}}.$
Here, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
The given plane is perpendicular to a vector
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&-1\\-1&1&-2\end{vmatrix}$
$=\hat{\text{i}}(-4+1)-\hat{\text{j}}(-2-1)+\hat{\text{k}}(1+2)$
$=-3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
We know that, the equation of plane in scalar product form is given by,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})=(\hat{\text{i}}+\hat{\text{j}})(-3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})$
$\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})=(-1)(-3)+(1)(3)+(0)+(3)$
$\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})=-3+3$
$\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})=0$
Dividing by 3, we get
$\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
Equation in required form is,
$\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$

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Question 2214 Marks

Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).

Answer

Let P(1, 1, -1), Q(6, 4, -5) and R(-4, -2, 3) be three points on a plane having position vectors $\vec{\text{p}},\vec{\text{q}}$ and $\vec{\text{s}}$ respectively. Then the vectors $\overrightarrow{\text{PQ}}$ and $\overrightarrow{\text{PR}}$ are in the same plane.
Therefore, $\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$ is a vector perpendicular to the plane.
Let $\vec{\text{n}}=\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$
$\overrightarrow{\text{PQ}}=(6-1)\hat{\text{i}}+(4-1)\hat{\text{j}}+(-5-(-1))\hat{\text{k}}$
$\Rightarrow\overrightarrow{\text{PQ}}=5\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$
Similarly,
$\Rightarrow\overrightarrow{\text{PR}}=(-4-1)\hat{\text{i}}+(-2-1)\hat{\text{j}}+(3-(-1))\hat{\text{k}}$
$\Rightarrow\overrightarrow{\text{PR}}=-5\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
Thus,
Here, $\overrightarrow{\text{PQ}}=-\overrightarrow{\text{PR}}$
Therefore, the given points are collinear.
Thus, $\vec{\text{n}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ where, 5a + 3b - 4c = 0
The plane passes through the point P with position vector $\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Thus, its vector equation is,
$\big\{\vec{\text{r}}-(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})\big\}\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=0$ Where, 5x + 3b - 4c = 0

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Question 2224 Marks

$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$

Let c₁ = 1 and c₂ = 2, find c₃ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
If c₂ = –1 and c₃ = 1, show that no value of c₁ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$

Answer

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$

$\text{c}_{1} = 1, \text{ c}_{2} = 2$

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$

$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$

$\text{c}_{2} = -1, \text{c}_{3} = 1$

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$

$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$

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Question 2234 Marks

Determine whether the following pair of lines intersect or not:
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$

Answer

Given equation of lines are
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
If these lines intersect each other, there must be some common point, So, we must have $\lambda$ and $\mu$ such that
$\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$(1+2\lambda)\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}=(2+\mu)\hat{\text{i}}+(-1+\mu)\hat{\text{j}}-\mu\hat{\text{k}}$
Equation the cofficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$
$1+2\lambda=2+\mu\Rightarrow2\lambda-\mu=1\dots(1)$
$-1=-1+\mu\Rightarrow\mu=0\dots(2)$
$\lambda=-\mu\Rightarrow\lambda=0\dots(3)$
Put value of $\lambda$ and $\mu$ in equation (1),
$2\lambda=\mu=1$
$2(0)-(0)=1$
$0=1$
$\text{LHS}\neq\text{RHS}$
Since, the values of $\lambda$ and $\mu$ form equation (2) and (3) dose not satisfy equation (1),
Hence, given lines do not intersect each other.

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Question 2244 Marks

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, -3), B(-2, -3, 5) nad C(5, 3, -3).

Answer

The given points are A(2, 5, -3), B(-2, -3, -3). The equation of the plane ABC is given by
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{x}_3-\text{x}_1&\text{y}_3-\text{y}_1&\text{z}_3-\text{z}_1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-5&\text{z}-(-3)\\-2-2&-3-5&5-(-3)\\5-2&3-5&-3-(-3)\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-5&\text{z}+3\\-4&-8&8\\3&-2&0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-5&\text{z}+3\\1&2&-2\\3&-2&0\end{vmatrix}=0$
$\Rightarrow-4(\text{x}-2)-6(\text{y}-5)-8(\text{z}+3)=0$
$\Rightarrow2(\text{x}-2)+3(\text{y}-5)+4(\text{z}+3)=0$
$\Rightarrow2\text{x}+3\text{y}+4\text{z}-7=0$
Distance between the point (7, 2, 4) and the plane 2x + 3y + 4z - 7 = 0
Distance between the point (7, 2, 4) to the plane 2x + 3y + 4z - 7 = 0
$=\bigg|\frac{2\times7+3\times2+4\times4-7}{\sqrt{2^2+3^2+4^2}}\bigg|$
$=\bigg|\frac{14+16-16-7}{\sqrt{4+9+16}}\bigg|$
$=\Big|\frac{29}{\sqrt{29}}\Big|$
$=\sqrt{29}\text{ units}$
Thus, the required distance between the given point is $\sqrt{29}\text{ units}.$

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Question 2254 Marks

Find the equation of the plane that contains the line of intersection of the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})-4=0$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})+5=0$ and which is perpendicular to the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})+8=0.$

Answer

The equation of the plane passing through the line of intersection of the given planes is
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})-4+\lambda\Big[\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})+5\Big]=0$
$\vec{\text{r}}\cdot\Big[(1+2\lambda)\hat{\text{i}}+(2+\lambda)\hat{\text{j}}+(3-\lambda)\hat{\text{k}}\Big]-4+5\lambda=0\ ...(\text{i})$
The plane is perpendicular to $\vec{\text{r}}(5\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})+8=0$ So,
$5(1+2\lambda)+3(2+\lambda)-6(3-\lambda)=0$ (Because a₁a₂ + b₁b₂ + c₁c₂= 0)
$\Rightarrow5+10\lambda+6+3\lambda-18+6\lambda=0$
$\Rightarrow19\lambda-7=0$
$\Rightarrow\lambda=\frac{7}{19}$
Substituting this in (i) we get
$\vec{\text{r}}\cdot\Big[\Big(1+2\Big(\frac{7}{19}\Big)\Big)\hat{\text{i}}+\Big(2+\frac{7}{19}\Big)\hat{\text{j}}+\Big(3-\frac{7}{19}\Big)\hat{\text{k}}\Big]-4+5\Big(\frac{7}{19}\Big)=0$
$\Rightarrow\vec{\text{r}}(33\hat{\text{i}}+45\hat{\text{j}}+50\hat{\text{k}})-41=0$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(33\hat{\text{i}}+45\hat{\text{j}}+50\hat{\text{k}})-41=0$
$\Rightarrow33\text{x}+45\text{y}+50\text{z}-41=0$

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Question 2264 Marks

Find the shortest distance between the lines given by $\vec{\text{r}}=(8+3\lambda)\hat{\text{i}}-(9-16\lambda)\hat{\text{j}}+(10+7\lambda)\hat{\text{k}}$ and $\vec{\text{r}}=15\hat{\text{i}}+29\hat{\text{j}}+5\hat{\text{k}}+\mu(3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}).$

Answer

We have $\vec{\text{r}}=(8+3\lambda)\hat{\text{i}}-(9-16\lambda)\hat{\text{j}}+(10+7\lambda)\hat{\text{k}}$
$=8\hat{\text{i}}-9\hat{\text{j}}+10\hat{\text{k}}+\lambda(3\hat{\text{i}}-16\hat{\text{j}}+7\hat{\text{k}})$
$\Rightarrow\vec{\text{a}}_1=8\hat{\text{i}}-9\hat{\text{j}}+10\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-16\hat{\text{j}}+7\hat{\text{k}}\ ......(\text{i})$
Also $\vec{\text{r}}=15\hat{\text{i}}+29\hat{\text{j}}+5\hat{\text{k}}+\mu(3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}})$
$\Rightarrow\vec{\text{a}}_2=15\hat{\text{i}}+29\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\text{b}}_2=3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}\ .....(\text{ii})$
Now, shortest distance between two lines is given by $\Bigg|\frac{(\vec{\text{b}}_1\times\vec{\text{b}_2})\cdot(\vec{\text{a}}_1-\vec{\text{a}_2})}{(\vec{\text{b}}_1\times\vec{\text{b}_2})}\Bigg| $
$\therefore\ \vec{\text{b}}_1\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\3&-16&7\\3&8&-5 \end{vmatrix}$
$=\hat{\text{i}}(80-56)-\hat{\text{j}}(-15-21)+\hat{\text{k}}(24+48)$
$=24\hat{\text{i}}+36\hat{\text{j}}+72\hat{\text{k}}$
$\Rightarrow|\vec{\text{b}_1}\times\vec{\text{b}_2}|=\sqrt{24^2+36^2+72^2}$
$=12\sqrt{2^2+3^2+72^2}=84$
Now $(\vec{\text{a}}_2-\vec{\text{a}}_1)=(15-8)\hat{\text{i}}+(29+9)\hat{\text{j}}+(5-10)\hat{\text{k}}$
$=7\hat{\text{i}}+38\hat{\text{j}}-5\hat{\text{k}}$
$\therefore$ Shortest distance $=\Bigg|\frac{(24\hat{\text{i}}+36\hat{\text{j}}+72\vec{\text{k}})\cdot(7\hat{\text{i}}+38\hat{\text{j}}-5\vec{\text{k}})}{84}\Bigg|$
$=\Bigg|\frac{(24\hat{\text{i}}+36\hat{\text{j}}+72\vec{\text{k}})\cdot(7\hat{\text{i}}+38\hat{\text{j}}-5\vec{\text{k}})}{7}\Bigg|$
$\Big|\frac{14+114-30}{7}\Big|=14$

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Question 2274 Marks

Show that the line $\frac{\text{x}+3}{-3}=\frac{\text{y}-1}{1}=\frac{\text{z}-5}{5}$ and $\frac{\text{x}+1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-5}{5}$ are coplanar. Hence, find the equation of the plane containing these lines.

Answer

The lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
The given lines are $\frac{\text{x}+3}{-3}=\frac{\text{y}-1}{1}=\frac{\text{z}-5}{5}$ and $\frac{\text{x}+1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-5}{5}$
Now, $\begin{vmatrix}-1-(-3)&2-1&5-5\\-3&1&5\\-1&2&5\end{vmatrix}=\begin{vmatrix}2&1&0\\-3&1&5\\-1&2&5\end{vmatrix}$
$=2(5-10)-1(-15+15)+0=-10+10+0=0$
So, the given lines are coplanar.
The equation of the containing the given lines is
$\begin{vmatrix}\text{x}-(-3)&\text{y}-1&\text{z}-5\\-3&1&5\\-1&2&5\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}+3&\text{y}-1&\text{z}-5\\-3&1&5\\-1&2&5\end{vmatrix}=0$
$\Rightarrow(\text{x}+3)(5-10)-(\text{y}-1)(-15+5)+(\text{z}-5)(-6+1)=0$
$\Rightarrow-5(\text{x}+3)+10(\text{y}-1)-5(\text{z}-5)=0$
$\Rightarrow\text{x}-2\text{y}+\text{z}=0$
Thus, the equation of the containing the given lines is x - 2y + z = 0.

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Question 2284 Marks

If O is the origin and the coordinates of A are (a, b, c) Find the direction cosines of OA and the equation of the plane through A at right angles to OA.

Answer

It is given that O is the origin and the coordinates of A are (a, b, c)
The direction of OA are proportional to
a - 0, b - 0, c - 0 or a, b, c
$\therefore$ Direction cosines of OA are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
The normal vector to the required plane is $(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})$
The vector equation of the plane through A(a, b, c) and perpendicular to OA is
$\big[\vec{\text{r}}-(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})\big]\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=0$
$\Rightarrow\vec{\text{r}}-(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}-(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=\text{a}^2+\text{b}^2+\text{c}^2$
The cartesian equation of this plane is
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=\text{a}^2+\text{b}^2+\text{c}^2$
Or $\text{ax}+\text{by}+\text{cz}=\text{a}^2+\text{b}^2+\text{c}^2$

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Question 2294 Marks

Find the equation of the plane which contains two parallel lines $\frac{\text{x}-4}{1}=\frac{\text{y}-3}{-4}=\frac{\text{z}-2}{5}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-4}=\frac{\text{z}}{5}.$

Answer

We know that the equation of the plane containing two parallel lines $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$ and $\frac{\text{x}-\text{x}_2}{\text{a}}=\frac{\text{y}-\text{y}_2}{\text{b}}=\frac{\text{z}-\text{z}_2}{\text{c}}$ is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}&\text{b}&\text{c} \end{vmatrix}=0$
Here, $\text{x}_1=4,\text{ y}_1=3,\text{ z}_1=2,\text{ x}_2=3,\text{ y}_2=-2,\text{ z}_2=0$
$\text{l}_1=1,\text{ m}_1=-4,\text{ n}_1=5,\text{ l}_2=1,\text{ m}+2=-4,\text{ n}_2=5$
Now, $\begin{vmatrix}\text{x}-4&\text{y}-3&\text{z}-2\\3-4&-2-3&0-2\\1&-4&5 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-4&\text{y}-3&\text{z}-2\\3-4&-2-3&0-2\\1&-4&5 \end{vmatrix}=0$
$\Rightarrow-33(\text{x}-4)+3(\text{y}-3)+9(\text{z}-2)=0$
$\Rightarrow11(\text{x}-4)-(\text{y}-3)-3(\text{z}-2)=0$
$\Rightarrow11\text{x}-\text{y}-3\text{z}=35$

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Question 2304 Marks

Show that the points (1, 1, 1) and (-3, 0, 1) are equidistant from the plane 3x + 4y - 12z + 13 = 0.

Answer

We know that the distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is given by
$\text{D}=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\ ...(\text{i})$
Let D₁ be the distance of the point (1, 1, 1) from plane 3x + 4y - 12z + 13 = 0,
So, using (i) we get
$\text{D}_1=\Bigg|\frac{(3)(1)+(4)(1)-12(1)+(13)}{\sqrt{(3)^2+(4)^2+(-12)^2}}\Bigg|$
$=\Big|\frac{3+4-12+13}{\sqrt{9+16+144}}\Big|$
$=\Big|\frac{8}{\sqrt{169}}\Big|$
$\text{D}_1=\frac{8}{13}\text{ units}\ ...(\text{ii})$
Let D₁ be the distance of the point (-3, 0, 1) from plane 3x + 4y - 12z + 13 = 0,
So, using equation (i)
$\text{D}_2=\Bigg|\frac{(3)(-3)+(4)(0)-12(1)+(13)}{\sqrt{(3)^2+(4)^2+(-12)^2}}\Bigg|$
$=\Big|\frac{-9+0-12+13}{\sqrt{9+4+144}}\Big|$
$=\Big|-\frac{8}{\sqrt{169}}\Big|$
$\text{D}_2=\frac{8}{13}\text{ units}\ ...(\text{iii})$
Hence, from equation (ii) and (iii)
$\text{D}_1=\text{D}_2$

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Question 2314 Marks

Find the direction cosines of the lines, connected by the relations: l + m + n = 0 and $\frac{2}{\text{m}}+\frac{2}{\text{n}}-\text{mn}=0$.

Answer

Given:
l + m + n = 0 ......(1)
2lm + 2ln + nm = 0 ......(2)
From (1), we get
l = m - n
Substituting l = -m - n in (2), we get
2(-m - n) m + 2(-m - n)n - mn = 0
⇒ -2m² - 2mn - 2mn - 2n² - mn = 0
⇒ 2m² + 2n² + 5mn = 0
⇒ (m + 2n) (2m + n) = 0
$\Rightarrow\text{m}=-2\text{n},-\frac{\text{n}}{2}$
If m = -2n, then from (1), we get l = n.
If $\text{m}=-\frac{\text{n}}{2},$ then from (1), we get $\text{l}=-\frac{\text{n}}{2}.$
Thus, the direction ratios of the two lines are proportional to n, - 2n, n and$-\frac{\text{n}}{2},-\frac{\text{n}}{2},\text{n},\text{i}.\text{e}.1,-2, 1$and $-\frac{1}{2},-\frac{1}{2},1$
Hence, their direction cosines are
$\pm\frac{1}{\sqrt{6}},\pm\frac{-2}{\sqrt{6}},\pm\frac{1}{\sqrt{6}}$
$\pm\frac{-1}{\sqrt{6}},\pm\frac{-1}{\sqrt{6}},\pm\frac{2}{\sqrt{6}}$.

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Question 2324 Marks

Show that the lines $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ are coplanar. Also, find the equation of the plane containing them.

Answer

We know that lines $\frac{\text{x}-\text{x}_1}{\text{l}_1}=\frac{\text{y}-\text{y}_1}{\text{m}_1}=\frac{\text{z}-\text{z}_1}{\text{n}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{l}_2}=\frac{\text{y}-\text{y}_2}{\text{m}_2}=\frac{\text{z}-\text{z}_2}{\text{n}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\end{vmatrix}=0$
And equation of plane containing them is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\end{vmatrix}=0$
Here, equation of lines are
$\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$
So, $\text{x}_1=-1,\text{ y}_1=3,\text{ z}_1=-2,\text{ l}_1=-3,\text{ m}_1=2,\text{ n}_1=1$
$\text{x}_2=0,\text{ y}_2=7,\text{ z}_2=-7,\text{ l}_2=1,\text{ m}_1=-3,\text{ n}_1=2$
So, $\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2 \end{vmatrix}=0$
$=\begin{vmatrix}0+1&7-3&-7+2\\-3&2&1\\1&-3&2\end{vmatrix}$
$=\begin{vmatrix}1&4&-5\\-3&2&1\\1&-3&2\end{vmatrix}$
$=1(4+3)-4(6-1)-5(9-2)$
$=7+28-35$
$=0$
So, lines are coplanar.
Equation of plane containing line is
$\begin{vmatrix}\text{x}+1&\text{y}-3&\text{z}+2\\-3&2&1\\1&-3&2\end{vmatrix}=0$
$(\text{x}+1)(4+3)-(\text{y}-3)(-6-1)+(\text{z}+2)(9-2)=0$
$7\text{x}+7+7\text{y}-21+7\text{z}+14=0$
$7\text{x}+7\text{y}+7\text{z}=0$

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Question 2334 Marks

Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{p}^2}.$

Answer

We know that equation of plane making intercepts a, b, c (on the axes) is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}-1=0$
Given: Perpendicular distance of the origin (0, 0, 0) from plane = p
$\therefore\ \frac{|\text{a}\text{x}_1+\text{b}\text{y}_1+\text{c}\text{z}_1+\text{d}|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}=\frac{\Big|\frac{0}{\text{a}}+\frac{0}{\text{b}}+\frac{0}{\text{c}}-1\Big|}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}=\text{p}$
$\Rightarrow\frac{|-1|}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}=\text{p}$
Squaring both sides, $\frac{1}{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}=\text{p}^2$
$\Rightarrow\ \text{p}^2\Big({\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}\Big)=1$
$\Rightarrow\ \Big({\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}\Big)=\frac{1}{\text{p}^2}$

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Question 2344 Marks

Find the vector equation of the line passing through the point (1, -1, 2) and perpendicular to the plane 2x - y + 3z - 5 = 0.

Answer

Let a, b, c be the direction ratios of the given line.
Since the line passes through the point (1, -1, 2) is
$\frac{\text{x}-1}{\text{a}}=\frac{\text{y}+1}{\text{b}}=\frac{\text{z}-2}{\text{c}}\ ...(\text{i})$
$\text{a}=2\lambda,\text{ b}=-\lambda,\text{ c}=3\lambda$
Substituting these values in (i) we get
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-2}{3},$ which is the cartesian from of the line.
Vector form
The given line passes through a point whose position vector is $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
So, its equation in vector form is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(\hat{2\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$

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Question 2354 Marks

Find the distance of a point (2, 4, –1) from the line $\frac{\text{x}+5}{1}+\frac{\text{y}+3}{4}+\frac{\text{z}-6}{-9}.$

Answer

Find the distance of a point (2, 4, -1) from the line $\frac{\text{x}+5}{1}+\frac{\text{y}+3}{4}+\frac{\text{z}-6}{-9}=\lambda$
$\Rightarrow\text{x}=\lambda-5,\text{y}=4\lambda-3,\text{z}=6-9\lambda$
Let the coordinates of L be $(\lambda-5,4\lambda-3,6-9\lambda),$ then Dr’s of PL are $(\lambda-7,4\lambda-7,7-9\lambda).$
Also, the direction ratios of given line are proportional to 1, 4, -9.
Since, P L is perpendicular to the given line.
$\therefore(\lambda-7)\cdot1+(4\lambda-7)\cdot4+(7-9\lambda)\cdot(-9)=0$
$\Rightarrow\lambda-7+16\lambda-28+81\lambda-63=0$
$\Rightarrow98\lambda=98$
$\Rightarrow\lambda=1$
So, the coordinates of L are (-4, 1, -3).
$\therefore$ Requires distance, $\text{PL}=\sqrt{(-4-2)^2+(1-4)^2+(-3+1)^2}$
$=\sqrt{36+9+4}$
$=7\text{ units}$

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Question 2364 Marks

Find the equations of the two lines through the origin which intersect the line $\frac{\text{x}-3}{2}-\frac{\text{y}-3}{1}=\frac{\text{z}}{1}$ at angles of $\frac{\pi}{3}$ each.

Answer

Given Equation of the line is, $\frac{\text{x}-3}{2}-\frac{\text{y}-3}{1}=\frac{\text{z}}{1}=\lambda$
So, direction ratios of the line are (2, 1, 1) = (a₁, b₁, c₁)
Any point on the given line is $\text{P}(2\lambda+3,\lambda+3,\lambda)$
So, direction ratios of OP are:
$(2'\lambda+3,\lambda+3,\lambda)=(\text{a}_2,\text{b}_2,\text{c}_2)$
Now, angle between given line and OP is $\frac{\pi}{3}.$
$\because\cos\frac{\pi}{3}=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_1+\text{c}_1\text{c}_1}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}}$
$\therefore\cos\frac{\pi}{3}=\frac{(4\lambda+6)+(\lambda+3)+(\lambda)}{\sqrt{6}\sqrt{(2\lambda+3)^2+(\lambda+3)^2+\lambda^2}}$
$\Rightarrow\frac{1}{2}=\frac{6\lambda+9}{\sqrt{6}\sqrt{(4\lambda^2+9+12\lambda+\lambda^2+9+6\lambda+\lambda^2)}}$
$\Rightarrow\frac{\sqrt{6}}{2}=\frac{6\lambda+9}{\sqrt{6\lambda^2+18\lambda+18}}$
$\Rightarrow6\sqrt{(\lambda^2+3\lambda+3)}=2(6\lambda+9)$
$\Rightarrow36(\lambda^2+3\lambda+3)=36(4\lambda^2+9+12\lambda)$
$\Rightarrow\lambda^2+3\lambda+3=4\lambda^2+9+12\lambda$
$\Rightarrow3\lambda^2+9\lambda+6=0$
$\Rightarrow\lambda^2+3\lambda+2=0$
$\Rightarrow\lambda(\lambda+2)+1(\lambda+2)=0$
$\Rightarrow(\lambda+1)(\lambda+2)=0$
$\Rightarrow\lambda=-1-2$
So, the directions cosines are 1, 2, -1 and -1, 1, -2.
Also, both the required lines pass through origin.
So, the equations of required lines are $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{-1}$ and $\frac{\text{x}}{-1}=\frac{\text{y}}{1}=\frac{\text{z}}{-2}.$

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Question 2374 Marks

$\overrightarrow{\text{AB}}=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{CD}}=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$ are two vectors. The position vectors of the points A and C are $6\hat{\text{i}}+7\hat{\text{j}}+4\hat{\text{k}}$ and $-9\hat{\text{j}}+2\hat{\text{k}},$ respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that $\overrightarrow{\text{PQ}}$ is perpendicular to $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{CD}}$ both.

Answer

We have $\overrightarrow{\text{AB}}=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{CD}}=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
Also, the position vectors of A and C are $6\hat{\text{i}}+7\hat{\text{j}}+4\hat{\text{k}}$ and $-9\hat{\text{j}}+2\hat{\text{k}},$ respectively.
Since, $\overrightarrow{\text{PQ}}$ is perpendicular to both $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{CD}}.$
So, P and Q will be foot of perpendicular to both the liens through A abd C.
Now, equation of the through C and parallel to the vector $\overrightarrow{\text{CD}}$ is given by
$\vec{\text{r}}=(6\hat{\text{i}}+7\hat{\text{j}}+4\hat{\text{k}})+\lambda(3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})\ ....(\text{i})$
and the line throught c and parallel to the vector $\overrightarrow{\text{CD}}$ is given by
$\vec{\text{r}}=-9\hat{\text{j}}+2\hat{\text{k}}+\mu(-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}})\ ......(\text{ii})$
Let $\text{P}(6+3\lambda,7-\lambda,4+\lambda)$ is any point on the first line and Q be any point on second line is given by $(-3\mu,-9+2\mu,2+4\mu).$
$\therefore\overrightarrow{\text{PQ}}=(-3\mu-6-3\lambda)\hat{\text{i}}-(2\mu+\lambda-16)\hat{\text{j}}+(4\mu-\lambda-2)\hat{\text{k}}$
If $\overrightarrow{\text{PQ}}$ is perpendiculas to the first line, then
$3(-3\mu-6-3\lambda)-(2\mu+\lambda-16)+(4\mu-\lambda-2)=0$
$\Rightarrow-7\mu-11\lambda-4=0\ .....(\text{iii})$
If $\overrightarrow{\text{PQ}}$ is perpendiculas to the second line, then
$-3(-3\mu-6-3\lambda)+2(2\mu+\lambda-16)+4(4\mu-\lambda-2)=0$
$\Rightarrow29\mu+7\lambda-22=0$
On solving Eqs. (iii) and (iv), we get
$\mu=1$ and $\lambda=-1$
$\therefore\overrightarrow{\text{OP}}=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ [from (i)]
and $\overrightarrow{\text{OP}}=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ [from (ii)]

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Question 2384 Marks

Find the vector equation of the plane passing through the point (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x - 5y - 15 = 0. Also, show that the plane thus obtaines contains the line

Answer

Let the equation of the plane be $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{i})$
Plane is passing through (3, 4, 2) and (7, 0, 6)
$\frac{3}{\text{a}}+\frac{4}{\text{b}}+\frac{2}{\text{c}}=1$
$\frac{7}{\text{a}}+\frac{0}{\text{b}}+\frac{6}{\text{c}}=1$
Required plane is perpendicular to 2x - 5y - 15 = 0
$\frac{2}{\text{a}}+\frac{-5}{\text{b}}+\frac{0}{\text{c}}=0$
$\Rightarrow2\text{b}=5\text{a}$
$\therefore\text{ b}=2.5\text{a}$
$\frac{3}{\text{a}}+\frac{4}{\text{2.5a}}+\frac{2}{\text{c}}=1$
$\frac{7}{\text{a}}+\frac{6}{\text{b}}=1$
Solving the above 2 equations,
$\text{a}=3.4=\frac{17}{5},\text{ b}=8.5=\frac{17}{2}$ and $\text{c}=\frac{-34}{6}=-\frac{17}{3}$
Substituting the values in (i)
$\frac{\text{x}}{\frac{17}{5}}+\frac{\text{y}}{\frac{17}{2}}+\frac{\text{z}}{-\frac{17}{3}}=1$
$\Rightarrow\frac{5\text{x}}{17}+\frac{2\text{y}}{17}-\frac{3\text{z}}{17}=1$
$\Rightarrow2\text{x}+2\text{y}-3\text{z}=17$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
Vector equation of the plane is $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
The line passes through B(1, 3, -2)
5(1) + 2(3) - 3(-2) = 17
The point B lies on the plane.
$\therefore$ The line $\vec{\text{r}}=\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}+\lambda(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})$ lies on the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$

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Question 2394 Marks

Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}-3}{4}$ and $\frac{\text{x}-2}{3}=\frac{\text{y}-3}{4}=\frac{\text{z}-5}{5}$

Answer

The equation of the given are
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}-3}{4}\dots(1)$
$\frac{\text{x}-2}{3}=\frac{\text{y}-3}{4}=\frac{\text{z}-5}{5}\dots(2)$
Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$
Here,
$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_1=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.
Its vector equation is
$\vec{\text{a}}_2-\vec{\text{a}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&4\\3&4&5\end{vmatrix}$
$=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\Big|=\sqrt{(-1)^2+2^2+(-1)^2}$
$=\sqrt{1+4+1}$
$=\sqrt{6}$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=-1+2-2$
$=-1$
Now,
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{-1}{\sqrt{6}}\Big|$
$=\frac{1}{\sqrt{6}}$

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Question 2404 Marks

By computing the shortest distance determine whether the following pairs of lines intersect or not:
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)$ and $\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$

Answer

Given equations of lines are,
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big),\vec{\text{b}}_1=\big(3\hat{\text{i}}-\hat{\text{j}}\big)$
and, $\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_2=\big(4\hat{\text{i}}-\hat{\text{k}}\big),\vec{\text{b}}_2=\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$
We know that, shortest distance between lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(4\hat{\text{i}}-\hat{\text{k}}\big)-\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=4\hat{\text{i}}-\hat{\text{k}}-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$=3\hat{\text{i}}-\hat{\text{j}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&0\\2&0&3 \end{vmatrix}$
$=\hat{\text{i}}(-3-0)-\hat{\text{j}}(9-0)+\hat{\text{k}}(0+2)$
$=-3\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-3)^2+(-9)^2+(2)^2}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{9+81+4}$
$=\sqrt{94}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(3\hat{\text{i}}-\hat{\text{j}}\big)\big(-3\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(3)(-3)+(-1)(-9)+(0)(2)$
$=-9+9+0$
$=0$
Using $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$ and $\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|$ in equation (1) to get shortest distance between given lines, so
$\text{S.D.}=\Big|\frac{0}{\sqrt{94}}\Big|$
$\text{S.D.}=0$
Since, shortest distance between the given lines is not zero, so lines are intersecting.

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Question 2414 Marks

Find the length and the foot of perpendicular from the poin $\Big(1,\frac{3}{2},2\Big)$ to the plane 2x - 2y + 4z + 5 = 0.

Answer

Equation of the given plane is 2x - 2y + 4z + 5 = 0 .....(i)
Thus, normal to the plane is $\vec{\text{n}}=2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
So, the equation of line through $\Big(1,\frac{3}{2},2\Big)$ and parallel to n is given by
$\frac{\text{x}-1}{2}=\frac{\text{y}-\frac{3}{2}}{-2}=\frac{\text{z}-2}{4}=\lambda$
Thus any point on thus line is $\Big(\text{x}=2\lambda+1,\text{y}=-2\lambda+\frac{3}{2},\text{z}=4\lambda+2\Big)$
If this point lies on the given plane, then
$2(2\lambda+1)-2\Big(-2\lambda+\frac{3}{2}\Big)+4(4\lambda+2)+5=0$ [Using Eq. (i)]
$\Rightarrow4\lambda+2+4\lambda-3+16\lambda+8+5=0$
$\Rightarrow24\lambda=-12$
$\Rightarrow\lambda=\frac{-1}{2}$
$\therefore$ Required foot of perpendicular
$=\Big[2\times\Big(\frac{-1}{2}\Big)+1,-2\times\Big(\frac{-1}{2}\Big)+\frac{3}{2},4\times\Big(\frac{-1}{2}\Big)+2\Big]$ i.e., $\Big(0,\frac{5}{2},0\Big)$
$\therefore$ Required length of perpendicular
$=\sqrt{(1-0)^2+\Big(\frac{3}{2}-\frac{5}{2}\Big)^2+(2-0)^2}$
$=\sqrt{1+1+1}$
$=\sqrt{6}\text{ units}$

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Question 2424 Marks

Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}\big)$

Answer

$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}\big)$
Comparing the given equations with the equations $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ we get
$\vec{\text{a}}_1=\hat{\text{i}}+\hat{\text{j}}$
$\vec{\text{a}}_2=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{b}}_1=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=\hat{\text{i}}-\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-1&1\\3&-5&2\end{vmatrix}$
$=3\hat{\text{i}}-\hat{\text{j}}-7\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{3^2+(-1)^2+(-7)^2}{}$
$=\sqrt{9+1+49}$
$=\sqrt{59}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(\hat{\text{i}}-\hat{\text{k}}\big).\big(3\hat{\text{i}}-\hat{\text{j}}-7\hat{\text{k}}\big)$
$=3+7$
$=10$
The shaortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{10}{\sqrt{59}}\Big|$
$=\frac{10}{\sqrt{59}}$

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Question 2434 Marks

Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.

Answer

Given, equation of plane is,
2x + 2y + 2z = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=3$
$\vec{\text{r}}\cdot(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=3$
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{d}}$
Normal to the plane $\vec{\text{n}}=2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Direction ratio of $\vec{\text{n}}=2,2,2$
Direction cosine of $\vec{\text{n}}=\frac{2}{|\vec{\text{n}}|},\frac{2}{|\vec{\text{n}}|},\frac{2}{|\vec{\text{n}}|}$
$|\vec{\text{n}}|=\sqrt{(2)^2+(2)^2+(2)^2}$
$=\sqrt{4+4+4}$
$=\sqrt{12}$
$|\vec{\text{n}}|=2\sqrt{3}$
Direction cosine of $|\vec{\text{n}}|=\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}},\frac{ 2}{ 2\sqrt{3}}$
$=\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
So, $\text{l}=\frac{1}{\sqrt{3}},\text{ m}=\frac{1}{\sqrt{3}},\text{ n}=\frac{1}{\sqrt{3}}$
Let $\alpha,\beta,\gamma$ be the angle that normal $\vec{\text{n}}$ makes with the coordinate axes respectively.
$\text{l}=\cos\alpha=\frac{1}{\sqrt{3}}$
$\alpha=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\ ...(\text{i})$
$\text{m}=\cos\beta=\frac{1}{\sqrt{3}}$
$\beta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
$\text{n}=\cos\gamma=\frac{1}{\sqrt{3}}$
$\gamma=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\ ...(\text{ii})$
From equation (i), (ii), (iii),
$\alpha=\beta=\gamma$
So, normal to the plane, $\vec{\text{n}}$ is equally inclined with the coordinate axes.

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Question 2444 Marks

If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} \text{and} \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then find the value of k and hence find the equation of the plane containing these lines.

Answer

Any point on line $\frac{\text{x} - 1}{2} = \frac{\text{y} + 1}{3} = \frac{\text{z} - 1}{4} \text{is} ( 2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$
$\therefore \frac{2\lambda + 1 - 3}{1} = \frac{3\lambda - 1 - \text{k}}{2} = \frac{4\lambda + 1}{1} \Rightarrow \lambda = -\frac{3}{2}, \text{hence k} = \frac{9}{2}$
Eqn. of plane containing three lines is
$\begin{vmatrix} \text{x - 1} & \text{y + 1} & \text{z - 1} \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0 $
$\Rightarrow \text{-5 ( x - 1) + 2 (y + 1) + 1 (z - 1) = 0}$
$\text{i.e 5x - 2y - z - 6 = 0}$

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Question 2454 Marks

Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector $2\hat{i} + 3\hat{j} + 4\hat{k}$ to the plane $\overrightarrow{\text{r}}. (2\hat{i} + \hat{j} + 3\hat{k}) - 26 = 0.$ Also find image of P in the plane.

Answer

Line through ‘P’ and perpendicular to plane is:
$\overrightarrow{\text{r}} = (2\hat{\text{i}} + 3\hat{\text{j}} + 4\hat{\text{k}}) + \lambda(2\hat{\text{i}} + \hat{\text{j}} + 3\hat{\text{k}})$
General point on line is: $\overrightarrow{\text{r}} = (2 + 2\lambda) \hat{\text{i}} + (3 + \lambda) \hat{\text{j}} + (4 + 3\lambda) \hat{\text{k}}$
For some $\lambda \in \text{R}, \overrightarrow{\text{r}}$ is the foot of perpendicular, say Q, from P to the plane, since it lies on plane
$\therefore[(2 + 2\lambda)\text{i} + (3 +\lambda) \hat{\text{j}} + (4 + 3\lambda) \hat{\text{k}}]. (2\hat{\text{i}} + \hat{\text{j}} + 3\hat{\text{k}}) - 26 = 0$
$\Rightarrow 4 + 4\lambda + 3 + \lambda + 12 + 9\lambda - 26 = 0 \Rightarrow \lambda = \frac{1}{2}$
$\therefore$ Foot of perpendicular is Q $\bigg(3\hat{\text{i}} + \frac{7}{2}\hat{\text{j}} + \frac{11}{2}\hat{\text{k}}\bigg)$
let P' $(\text{a}\hat{\text{i}} + \text{b}\hat{\text{j}} + \text{c}\hat{\text{k}})$ be the image of P in the plane Q is mid point of PP'
$\therefore \text{Q}\bigg(\frac{\text{a} + 2}{2}\hat{\text{i}} + \frac{\text{b + 3}}{2}\hat{\text{j}} + \frac{\text{c + 4}}{2}\hat{\text{k}}\bigg) = \text{Q}\bigg(3\hat{\text{i}} + \frac{7}{2} \hat{\text{j}} + \frac{11}{2}\hat{\text{k}}\bigg) $
$\Rightarrow\frac{\text{a + 2}}{2} = 3, \frac{\text{b + 3}}{2} = \frac{7}{2}, \frac{\text{c + 4}}{2} = \frac{11}{2}\Rightarrow\text{a = 4, b = 4, c = 7}\therefore\text{P'} (4\hat{\text{i}} + 4\hat{\text{j}} + 7\hat{\text{k}})$
Perpendicular distance of P from plane = $\text{PQ} =\sqrt{(2 - 3)^{2} + \bigg(3 - \frac{7}{2}\bigg)^{2} + \bigg(4 - \frac{11}{2}\bigg)^{2}} = \sqrt\frac{7}{2}$

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Question 2464 Marks

Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=(\lambda-1)\hat{\text{i}}+(\lambda+1)\hat{\text{j}}-(1+\lambda)\hat{\text{k}}$ and $\vec{\text{r}}=(1-\mu)\hat{\text{i}}+(2\mu-1)\hat{\text{j}}+(\mu+2)\hat{\text{k}}$

Answer

$\vec{\text{r}}=(\lambda-1)\hat{\text{i}}+(\lambda+1)\hat{\text{j}}-(1+\lambda)\hat{\text{k}}$ and $\vec{\text{r}}=(1-\mu)\hat{\text{i}}+(2\mu-1)\hat{\text{j}}+(\mu+2)\hat{\text{k}}$
The vector equation of the given lines can be re-written as
$\vec{\text{r}}=-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}+\lambda\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$ and $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+\mu\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$
Comparing the given equation with the equations $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ we get
$\vec{\text{a}}_1=-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{b}}_2=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{J}}&\hat{\text{k}}\\1&1&-1\\-1&2&1 \end{vmatrix}$
$=3\hat{\text{i}}+3\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{3^2+3^2}$
$=\sqrt{9+9}$
$=3\sqrt{2}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big).\big(3\hat{\text{i}}+3\hat{\text{k}}\big)$
$=6+9$
$=15$
The shoetest distance between the line $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{15}{3\sqrt{2}}\Big|$
$=\frac{5}{\sqrt{2}}$

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Question 2474 Marks

Find the equation of the plane through the line of intersection of the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+3\hat{\text{j}})+6=0$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})=0,$ which is at a unit distance from the origin.

Answer

The equation of the plane passing through the line intersection of the given planes is,
$\vec{\text{r}}\cdot(\hat{\text{i}}+3\hat{\text{j}})+6+\lambda\big(\vec{\text{r}}\cdot(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})\big)$
$\vec{\text{r}}\cdot\Big[(1+3\lambda)\hat{\text{i}}+(3-\lambda)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]+6=0\ ...(\text{i})$
$\vec{\text{r}}\cdot\Big[(1+3\lambda)\hat{\text{i}}+(3-\lambda)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]=-6$
$\vec{\text{r}}\cdot\Big[(-1-3\lambda)\hat{\text{i}}+(\lambda-3)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]=6$
Dividing both sides by $\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2},$ we get
$\vec{\text{r}}\cdot\frac{\Big[-1-3\lambda)\hat{\text{i}}+(\lambda-3)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}=\frac{6}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}$
Which is the normal form of plane (i), where
The perpendicular distance of plane (i) from the origin
$=\frac{6}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}$
$\Rightarrow1=\frac{6}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}\text{ (Given})$
$\Rightarrow\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}=6$
$\Rightarrow1+9\lambda^2+6\lambda+\lambda^2+9-6\lambda+16\lambda^2=36$
$\Rightarrow26\lambda^2-26=0$
$\Rightarrow\lambda^2=1$
$\Rightarrow\lambda=1,-1$
Case 1: Substituting $\lambda=1$ in (i) we get
$\vec{\text{r}}\cdot\Big[4\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\Big]+6=0$
Case 2: Substituting $\lambda=-1$ in (i) we get
$\vec{\text{r}}\cdot\Big[-2\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}\Big]+6=0$

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Question 2484 Marks

By computing the shortest distance determine whether the following pairs of lines intersect or not:3
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{-5}=\frac{\text{z}+3}{-5}$ and $\frac{\text{x}-8}{7}=\frac{\text{y}-7}{1}=\frac{\text{z}-5}{3}$

Answer

Given lines are ,
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{-5}=\frac{\text{z}+3}{-5}=\lambda$ (say)
$\Rightarrow\text{x}=4\lambda+5,\text{y}=-5\lambda+7,\text{z}=-5\lambda-3$
$\Rightarrow\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$=(4\lambda+5)\hat{\text{i}}+(-5\lambda+7)\hat{\text{j}}+(-5\lambda-3)\hat{\text{k}}$
$\vec{\text{r}}=\big(5\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(4\hat{\text{i}}-5\hat{\text{j}}-5\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(5\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big),\vec{\text{b}}_1=\big(4\hat{\text{i}}-5\hat{\text{j}}-5\hat{\text{k}}\big)$
and, $\frac{\text{x}-8}{7}=\frac{\text{y}-7}{1}=\frac{\text{z}-5}{3}=\mu$ (say)
$\Rightarrow\text{x}=7\mu+8,\text{y}=\mu+7,3\mu+5$
$\Rightarrow\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$=(7\mu+8)\hat{\text{i}}+(\mu+7)\hat{\text{j}}+(3\mu+5)\hat{\text{k}}$
$\vec{\text{r}}=\big(8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big)+\mu\big(7\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}=\big(8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big),\vec{\text{b}}_2=\big(7\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
we know that, shortest distance between lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big)-\big(5\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big)$
$=8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}-5\hat{\text{i}}-7\hat{\text{j}}+3\hat{\text{k}}$
$=3\hat{\text{i}}+8\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-5&-5\\7&1&3 \end{vmatrix}$
$=\hat{\text{i}}(-15+5)-\hat{\text{j}}(12+35)+\hat{\text{k}}(4+35)$
$=-10\hat{\text{i}}-47\hat{\text{j}}+39\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(3\hat{\text{i}}+8\hat{\text{k}}\big)\big(-10\hat{\text{i}}-47\hat{\text{j}}+39\hat{\text{k}}\big)$
$=(3)(-10)+(0)(-4)+(8)(39)$
$=-30+312$
$=282$
Using equation (1) to get the shortest distance between the given lines, so
$\text{S.D.}=\Bigg|\frac{282}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$\text{S.D.}\neq0$
Since, the shortest distance between given lines is not equal to zero, so Given lines are not intersecting.

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Question 2494 Marks

Find the equatoion of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

Answer

We know that, equation of plane passing through the point (x₁, y₁, z₁) is given by,
a(x - x₁) + b(y - y₁) + c(z - z₁) = 0
Here, the plane is pasing through (2, 2, 1)
a(x - 2) + b(y - 2) + c(z - 1) = 0 ....(i)
It is also passing through (9, 3, 6), so it must satisfy the equation (i),
a(9 - 2) + b(3 - 2) + c(6 - 1) = 0
7a + b + 5c = 0 ....(ii)
We know that, plane a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are perpendicular if
a₁a₂ + b₁b₂ + c₁c₂= 0 ....(iii)
Given that, plane (i) is perpendicular to plane
2x + 6y + 6z = 1 ....(iv)
Using plane (i), (iv) in equation (iii),
a₁a₂ + b₁b₂ + c₁c₂= 0
(a)(2) + (b)(6) + (c)(6) = 0
2a + 6b + 6c = 0 ....(v)
Solving (ii) and (v) by cross-multiplication,
$\frac{\text{a}}{(1)(6)-(5)(6)}=\frac{\text{b}}{(2)(5)-(7)(6)}=\frac{\text{c}}{(7)(6)-(2)(1)}$
$\frac{\text{a}}{6-30}=\frac{\text{b}}{10-42}=\frac{\text{c}}{42-2}$
$\frac{\text{a}}{-24}=\frac{\text{b}}{-32}=\frac{\text{c}}{40}=\lambda(\text{say})$
$\text{a}=-24\lambda,\text{b}=-32\lambda,\text{c}=40\lambda$
Put a, b, c in equation (i),
$\text{a}(\text{x}-2)+\text{b}(\text{y}-2)+\text{c}(\text{z}-1)=0$
$(-24\lambda)(\text{x}-2)+(-32\lambda)(\text{y}-2)+(40\lambda)(\text{z}-1)=0$
$-24\lambda\text{x}+48\lambda-32\lambda\text{y}+64\lambda+40\lambda\text{z}-40\lambda=0$
$-24\lambda\text{x}-32\lambda\text{y}+40\lambda\text{z}+72\lambda=0$
Dividing by $(-8\lambda),$
3x + 4y - 5z - 9 = 0
Equation of required plane is,
3x + 4y - 5z = 9

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Question 2504 Marks

Determine the equation of the line passing through the points (1, 2, -4) and perpendicular to the lines $\frac{\text{x}-8}{8}=\frac{\text{y}+9}{-16}=\frac{\text{z}-10}{7}$ and $\frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}.$

Answer

We have
$\frac{\text{x}-8}{8}=\frac{\text{y}+9}{-16}=\frac{\text{z}-10}{7}$
$\frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}$
Let:
$\vec{\text{b}}_1=8\hat{\text{i}}-16\hat{\text{j}}+7\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}$
Since the required line is perpendicular to the lines parallel to the vectors
$\vec{\text{b}}_1=8\hat{\text{i}}-16\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{b}}_2=3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}$
it is parallel to the vector $\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
Now,
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\8&-16&7\\3&8&-5\end{vmatrix}$
$=24\hat{\text{i}}+61\hat{\text{j}}+112\hat{\text{k}}$
The direction of the required line are proportional to 24, 81, 112.
The equation of the required line passing through the point (1, 2, -4) and having direction ratios proportional to 24, 61, 112 is $\frac{\text{x}-1}{24}=\frac{\text{y}-2}{61}=\frac{\text{z}+4}{112}$

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