Question 2015 Marks
Find the coordinates of the foot of the perependicular drawn from the origin to the plane 2x - 3y + 4z - 6 = 0.
AnswerLet 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
$\text{M}=(2\text{r},-3\text{r},4\text{r})=\bigg(2\Big(\frac{6}{29}\Big),-3\Big(\frac{6}{29}\Big),4\Big(\frac{6}{29}\Big)\bigg)$
$=\Big(\frac{12}{29},\frac{-18}{29},\frac{24}{29}\Big)$
View full question & answer→Question 2025 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.
AnswerGiven that $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar
$\therefore[\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}] = 0$
$\text{i.e} \overrightarrow{\text{(a}} +\overrightarrow{\text{b})}, \big\{\overrightarrow{\text{(b}} + \overrightarrow{\text{c})} \times\overrightarrow{\text{(c}} + \overrightarrow{\text{a})}\big\} = 0$
$\overrightarrow{\text{(a}} +\overrightarrow{\text{b})}. \big\{\overrightarrow{\text{(b}} \times \overrightarrow{\text{c}}+ \overrightarrow{\text{b}} \times\overrightarrow{\text{a}}\times\overrightarrow{\text{c}} \times \overrightarrow{\text{a})}\big\} = 0$
$\Rightarrow\overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{a}}(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})}+\overrightarrow{\text{b}}. (\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})} = 0$
$\Rightarrow 2 [\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}] = \text{0 or} [\overrightarrow{\text{a}},\overrightarrow{\text{b}}, \overrightarrow{\text{c] }} = 0$
$\Rightarrow\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}\text{are coplanar}.$
View full question & answer→Question 2035 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$\frac{\text{x}}{-3}=\frac{\text{y}}{2}=\frac{\text{z}}{5}.$
AnswerLet 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,$\overrightarrow{\text{r}}=(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})+\lambda(2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}})$
View full question & answer→Question 2045 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 $3x - 3y + 6z - 1 = 0$
AnswerThe direction ratios of normal to the plane, $L_1: a_1x + b_1y + c_1z = 0,$
are $a_1, b_1, c_1$ and $L_2: a_2x + b_2y + c_2z = 0$ are $a_2, b_2, c_2$
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between $L_1$ and $L_2$ 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, $a_1 = 2, b_1 = -2, c_1 = 4$ and $a_2 = 3, b_2 = -3, c_2 = 6$
$a_1a_2 +b_1b_2 + c_1c_2 = 2 \times 3 + (-2) \times (-3) + 4 \times 6 = 6 + 6 + 24$
$=36\neq0$
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.
View full question & answer→Question 2055 Marks
Find the vector equation of the line through the origin which is perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=3$
AnswerRequired 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
$\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
View full question & answer→Question 2065 Marks
Find the angle between line $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{1}$ and the plane $2x + y - z = 4.$
AnswerWe 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 $a_2x + b_2y + c_2z + d_2= 0$ 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, $a_1 = 1, b_1 = -1, c_1 = 1$
Given equation of plane is $2x + y - z - 4 = 0$
So, $a_2 = 2, b_2 = 1, c_2 = -1$
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 $=0^{\circ}$
View full question & answer→Question 2075 Marks
Find the angle between the lines whose direction cosines are given by the equations:
$2l + 2m - n = 0, mn + ln + lm = 0$
AnswerThe 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
$m(2l + 2m) + l(2l + 2m) + lm = 0$
$2lm + 2m^2 + 2l^2 + 2ml + lm = 0$
$2ml^2 + 5lm + 2l^2 = 0$
$2m^2 + 4lm + lm + 2l^2 = 0$
$(2m + l) (m + 2l) = 0$
$\Rightarrow\text{m}=-\frac{1}{2}$ 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,
$\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{1+1-2}{\sqrt{1+\frac{1}{4}+1}\sqrt{1+4+9}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$.
View full question & answer→Question 2085 Marks
Find a vector of magnitude 26 units normal to the plane 12x - 3y + 4z = 1
AnswerGiven, 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 $=24\hat{\text{i}}-6\hat{\text{j}}+8\hat{\text{k}}$
View full question & answer→Question 2095 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.
AnswerWe 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 $L_1$ is $(2\lambda+1,3\lambda+2,4\lambda+3)$
Any point on the line $L_2$ is $(5\mu+4,2\mu+1,\mu)$
$(2\lambda+1,3\lambda+2,4\lambda+3)=(5\mu+4,2\mu+1,\mu)$
$\Rightarrow2\lambda+1=5\mu+4,3\lambda+2=2\mu+1$ 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 $(-1, -1, -1).$
View full question & answer→Question 2105 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.
AnswerThe 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
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})=-89$
$\Rightarrow7\text{x}-3\text{y}-\text{z}=-89$
$\Rightarrow7\text{x}-3\text{y}-\text{z}+89=0$
View full question & answer→Question 2115 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$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$
View full question & answer→Question 2125 Marks
Find the shortest distance between the following two lines:$\vec{\text{r}}=\text{(1 +}\lambda)\hat{\text{i}}+\text{(2 -}\lambda)\hat{\text{j}}+(\lambda+\text{1)}\hat{\text{k}};$
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})+\mu(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}).$
AnswerHere $\vec{\text{a}_{1}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k},}\vec{\text{ b}_{1}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ $\vec{\text{a}_{2}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k},}\vec{\text{ b}_{2}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{2k}}$ $\vec{\text{a}_{2}}-\vec{\text{a}_{1}}=\hat{\text{i}}-3{\text{j}}-2{\text{k}}$ $\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}=-3\hat{\text{i}}+3{\text{k}}$ Shortest distance (d) = $\Bigg|\frac{\Big(\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}\Big)\cdot\Big(\vec{\text{a}_{2}}-\vec{\text{a}_{1}}\Big)}{\Big|\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}\Big|}\Bigg|$=$\Bigg|\frac{\Big(-3\hat{\text{i}}+3\hat{\text{k}}\Big)\cdot\Big(\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\Big)}{\sqrt{9 + 9}}\Bigg|$
$=\frac{9}{3\sqrt{2}}\text{ OR }\frac{3\sqrt{2}}{2}$.
View full question & answer→Question 2135 Marks
Find the length and the foot of the perpendicular drawn from the point $(2, –1, 5)$ to the line $\frac{\text{x - 11}}{10}=\frac{\text{y + 2}}{-4}=\frac{\text{z + 8}}{-11}.$
AnswerAny 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 $(1, 2, 3)$
$\therefore PQ^2 = (2 - 1)^2 + (-1 - 2)^2 + (5 - 3)^2= 1 + 9 + 4 = 14$
$\Rightarrow PQ = \sqrt{14}.$
View full question & answer→Question 2145 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
$\text{Area of} \Delta = \frac{1}{2} | \vec{\text{AB}} \times \vec{\text{BC}}| = \frac{1}{2} \sqrt{210}$
View full question & answer→Question 2155 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.
AnswerLet $M$ 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
$2(2r + 2) + r + 3 + 3(3r + 4) - 26 = 0$
$\Rightarrow 4r + 4 + r + 3 + 9r + 12 - 26 = 0$
$\Rightarrow 14r - 7 = 0$
$\Rightarrow\text{r}=\frac{1}{2}$
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 $Q(x_1, y_1, z_1)$ 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 $(4, 4, 7).$
View full question & answer→Question 2165 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 $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5.$
AnswerThe 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)
$=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}$
$=\sqrt{9+16+144}$
$=13\text{ units}$
View full question & answer→Question 2175 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.
AnswerEquation 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 $1\ \text{or}\ \frac{7}{3}.$ View full question & answer→Question 2185 Marks
Find the equation of the plane through $(2, 3, -4)$ and $(1, -1, 3)$ and parallel to $x-$axis.
AnswerThe 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,
$a(1 - 2) + b(-1 - 3) + c(3 + 4) = 0$
$\Rightarrow -a - 4b + 7c = 0 ....(ii)$
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 $1x + 0y + 0z = 0$
$\Rightarrow a(1) + b(0) + c(0) = 0 ....(iii) ($Because $a_1a_2 + b_1b_2 + c_1c_{2 }= 0)$
Solving $(i), (ii)$ and $(iii),$ we get
$\begin{vmatrix}\text{x}-3&\text{y}-3&\text{z}+4\\-1&-4&7\\1&0&0\end{vmatrix}=0$
$\Rightarrow0(\text{x}-3)+7(\text{y}-3)+4(\text{z}+4)=0$
$\Rightarrow7\text{y}+4\text{z}-5=0$
View full question & answer→Question 2195 Marks
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+\mu(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
AnswerGiven, 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,
$\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
View full question & answer→Question 2205 Marks
Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).
AnswerLet 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
View full question & answer→Question 2215 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 $c_1 = 1$ and $c_2 = 2,$ find $c_3$ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
- If $c_2 = –1$ and $c_3 = 1,$ show that no value of $c_1 $ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$
Answer$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$
- $\text{c}_{1} = 1, \text{ c}_{2} = 2$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$
$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$
- $\text{c}_{2} = -1, \text{c}_{3} = 1$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$
$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$ View full question & answer→Question 2225 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 $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
AnswerGiven 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.
View full question & answer→Question 2235 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).
AnswerThe 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 $\sqrt{29}\text{ units}.$
View full question & answer→Question 2245 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 $\vec{\text{r}}\cdot(5\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})+8=0.$
AnswerThe 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 $a_1a_2 + b_1b_2 + c_1c_{2 }= 0)$
$\Rightarrow5+10\lambda+6+3\lambda-18+6\lambda=0$
$\Rightarrow19\lambda-7=0$
$\Rightarrow\lambda=\frac{7}{19}$
Substituting this in $(i)$ we get
$\vec{\text{r}}\cdot\Big[\Big(1+2\Big(\frac{7}{19}\Big)\Big)\hat{\text{i}}+\Big(2+\frac{7}{19}\Big)\hat{\text{j}}+\Big(3-\frac{7}{19}\Big)\hat{\text{k}}\Big]-4+5\Big(\frac{7}{19}\Big)=0$
$\Rightarrow\vec{\text{r}}(33\hat{\text{i}}+45\hat{\text{j}}+50\hat{\text{k}})-41=0$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(33\hat{\text{i}}+45\hat{\text{j}}+50\hat{\text{k}})-41=0$
$\Rightarrow33\text{x}+45\text{y}+50\text{z}-41=0$
View full question & answer→Question 2255 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 $\vec{\text{r}}=15\hat{\text{i}}+29\hat{\text{j}}+5\hat{\text{k}}+\mu(3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}).$
AnswerWe 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 $=\Bigg|\frac{(24\hat{\text{i}}+36\hat{\text{j}}+72\vec{\text{k}})\cdot(7\hat{\text{i}}+38\hat{\text{j}}-5\vec{\text{k}})}{84}\Bigg|$
$=\Bigg|\frac{(24\hat{\text{i}}+36\hat{\text{j}}+72\vec{\text{k}})\cdot(7\hat{\text{i}}+38\hat{\text{j}}-5\vec{\text{k}})}{7}\Bigg|$
$\Big|\frac{14+114-30}{7}\Big|=14$
View full question & answer→Question 2265 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.
AnswerThe 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.
View full question & answer→Question 2275 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.
AnswerIt 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 $\text{ax}+\text{by}+\text{cz}=\text{a}^2+\text{b}^2+\text{c}^2$
View full question & answer→Question 2285 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 $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-4}=\frac{\text{z}}{5}.$
AnswerWe 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, $\begin{vmatrix}\text{x}-4&\text{y}-3&\text{z}-2\\3-4&-2-3&0-2\\1&-4&5 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-4&\text{y}-3&\text{z}-2\\3-4&-2-3&0-2\\1&-4&5 \end{vmatrix}=0$
$\Rightarrow-33(\text{x}-4)+3(\text{y}-3)+9(\text{z}-2)=0$
$\Rightarrow11(\text{x}-4)-(\text{y}-3)-3(\text{z}-2)=0$
$\Rightarrow11\text{x}-\text{y}-3\text{z}=35$
View full question & answer→Question 2295 Marks
Show that the points $(1, 1, 1)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0.$
AnswerWe know that the distance of the point $(x_1, y_1, z_1)$ 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 $D_1$_ 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 $D_1$_ 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 $(iii)$
$\text{D}_1=\text{D}_2$
View full question & answer→Question 2305 Marks
Find the direction cosines of the lines, connected by the relations: $l + m + n = 0$ and $\frac{2}{\text{m}}+\frac{2}{\text{n}}-\text{mn}=0$.
AnswerGiven:
$l + m + n = 0 ......(1)$
$2lm + 2ln + nm = 0 ......(2)$
From $(1),$ we get
$l = m - n$
Substituting $l = -m - n$ in $(2),$ we get
$2(-m - n) m + 2(-m - n)n - mn = 0$
$\Rightarrow -2m^2 - 2mn - 2mn - 2n^2 - mn = 0$
$\Rightarrow 2m^2 + 2n^2 + 5mn = 0$
$\Rightarrow (m + 2n) (2m + n) = 0$
$\Rightarrow\text{m}=-2\text{n},-\frac{\text{n}}{2}$
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
$\pm\frac{1}{\sqrt{6}},\pm\frac{-2}{\sqrt{6}},\pm\frac{1}{\sqrt{6}}$
$\pm\frac{-1}{\sqrt{6}},\pm\frac{-1}{\sqrt{6}},\pm\frac{2}{\sqrt{6}}$.
View full question & answer→Question 2315 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.
AnswerWe 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
$\begin{vmatrix}\text{x}+1&\text{y}-3&\text{z}+2\\-3&2&1\\1&-3&2\end{vmatrix}=0$
$(\text{x}+1)(4+3)-(\text{y}-3)(-6-1)+(\text{z}+2)(9-2)=0$
$7\text{x}+7+7\text{y}-21+7\text{z}+14=0$
$7\text{x}+7\text{y}+7\text{z}=0$
View full question & answer→Question 2325 Marks
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{p}^2}.$
AnswerWe 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, $\frac{1}{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}=\text{p}^2$
$\Rightarrow\ \text{p}^2\Big({\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}\Big)=1$
$\Rightarrow\ \Big({\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}\Big)=\frac{1}{\text{p}^2}$
View full question & answer→Question 2335 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.
AnswerLet 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
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(\hat{2\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$
View full question & answer→Question 2345 Marks
Find the distance of a point (2, 4, –1) from the line $\frac{\text{x}+5}{1}+\frac{\text{y}+3}{4}+\frac{\text{z}-6}{-9}.$
AnswerFind 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, $\text{PL}=\sqrt{(-4-2)^2+(1-4)^2+(-3+1)^2}$
$=\sqrt{36+9+4}$
$=7\text{ units}$
View full question & answer→Question 2355 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.
AnswerGiven 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 $(2, 1, 1) = (a_1, b_1, c_1)$
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 $\frac{\text{x}}{-1}=\frac{\text{y}}{1}=\frac{\text{z}}{-2}.$
View full question & answer→Question 2365 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.
AnswerWe 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)]
View full question & answer→Question 2375 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
AnswerLet 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 $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
View full question & answer→Question 2385 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 $\frac{\text{x}-2}{3}=\frac{\text{y}-3}{4}=\frac{\text{z}-5}{5}$
AnswerThe 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
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{-1}{\sqrt{6}}\Big|$
$=\frac{1}{\sqrt{6}}$
View full question & answer→Question 2395 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 $\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$
AnswerGiven 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.
View full question & answer→Question 2405 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.
AnswerEquation 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
$=\sqrt{(1-0)^2+\Big(\frac{3}{2}-\frac{5}{2}\Big)^2+(2-0)^2}$
$=\sqrt{1+1+1}$
$=\sqrt{6}\text{ units}$
View full question & answer→Question 2415 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
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{10}{\sqrt{59}}\Big|$
$=\frac{10}{\sqrt{59}}$
View full question & answer→Question 2425 Marks
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
AnswerGiven, 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.
View full question & answer→Question 2435 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.
AnswerAny 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 $
$\Rightarrow \text{-5 ( x - 1) + 2 (y + 1) + 1 (z - 1) = 0}$
$\text{i.e 5x - 2y - z - 6 = 0}$
View full question & answer→Question 2445 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.
AnswerLine 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 = $\text{PQ} =\sqrt{(2 - 3)^{2} + \bigg(3 - \frac{7}{2}\bigg)^{2} + \bigg(4 - \frac{11}{2}\bigg)^{2}} = \sqrt\frac{7}{2}$
View full question & answer→Question 2455 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
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{15}{3\sqrt{2}}\Big|$
$=\frac{5}{\sqrt{2}}$
View full question & answer→Question 2465 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.
AnswerThe 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
$\vec{\text{r}}\cdot\Big[-2\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}\Big]+6=0$
View full question & answer→Question 2475 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 $\frac{\text{x}-8}{7}=\frac{\text{y}-7}{1}=\frac{\text{z}-5}{3}$
AnswerGiven 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.
View full question & answer→Question 2485 Marks
Find the equatoion of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z = 1.$
AnswerWe know that, equation of plane passing through the point $(x_1, y_1, z_1)$ is given by,
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
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 $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Given that, plane $(i)$ is perpendicular to plane
$2x + 6y + 6z = 1 ....(iv)$
Using plane $(i), (iv)$ in equation $(iii),$
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a)(2) + (b)(6) + (c)(6) = 0$
$2a + 6b + 6c = 0 ....(v)$
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,
$3x + 4y - 5z = 9$
View full question & answer→Question 2495 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 $\frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}.$
AnswerWe 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 $\frac{\text{x}-1}{24}=\frac{\text{y}-2}{61}=\frac{\text{z}+4}{112}$
View full question & answer→Question 2505 Marks
Find the equation of the plane passing through the line of intersection of the planes $2x - y = 0$ and $3z - y= 0$ and perpendicular to the plane $4x + 5y - 3z = 8.$
AnswerWe know that, equation of a plane passing through the line of intersection of $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by,
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
So, equation of plane passing through the line of intersection of plane $2x - y = 0$ and $3z - y = 0$ is,
$(2\text{x}-\text{y})+\lambda(3\text{z}-\text{y})=0$
$2\text{x}-\text{y}+3\lambda\text{z}-\lambda\text{y}=0$
$\text{x}(2)+\text{y}(-1-\lambda)+\text{z}(3\lambda)=0\ ...(\text{i})$
We know that, two planes are perpendicular if,
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0\ ...(\text{ii})$
Given, plane $(i)$ is parpendicular to plane
$4\text{x}+5\text{y}-3\text{z}=8\ ...(\text{iii})$
Using $(i)$ and $(iii)$ in equation $(ii),$
$(2)(4)+(-1-\lambda)(5)+(3\lambda)(-3)=0$
$8-5-5\lambda-9\lambda=0$
$3-14\lambda=0$
$\lambda=\frac{3}{14}$
Put the value of $\lambda$ in equation $(i),$
$2\text{x}+\text{y}(-1-\lambda)+\text{z}(3\lambda)=0$
$2\text{x}+\text{y}\Big(-1-\frac{3}{14}\Big)+3\text{z}\Big(\frac{3}{14}\Big)=0$
$2\text{x}+\text{y}\Big(\frac{-14-3}{14}\Big)+\frac{9\text{z}}{14}=0$
$2\text{x}+\text{y}\Big(-\frac{17}{14}\Big)+\frac{9\text{z}}{14}=0$
Multiplying with $14,$ we get
$28x - 17y + 9z = 0$
Equation of required plane is,
$28x - 17y + 9z = 0$
View full question & answer→