4 Marks - Page 3 - Maths STD 12 Science Questions - Vidyadip

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4 Marks

Question 1014 Marks

Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.

Answer

The equation of the plane passing through the line of intersection of the given planes is,
$\text{x}+\text{y}+\text{z}-1+\lambda(2\text{x}+3\text{y}+4\text{z}-5)=0$
$(1+2\lambda)\text{x}+(1+3\lambda)\text{y}+(1+4\lambda)\text{z}-1-5\lambda=0\ ...(\text{i})$
This plane is perpendicular to x - y + z = 0. So,
$1+2\lambda-1(1+3\lambda)+1+4\lambda=0$ (Because a₁a₂ + b₁b₂ + c₁c₂= 0)
$\Rightarrow1+2\lambda-1-3\lambda+1+4\lambda=0$
$\Rightarrow3\lambda+1=0$
$\Rightarrow\lambda=\frac{-1}{3}$
Substituting this in (i), we get
$\Big(1+2\Big(\frac{-1}{3}\Big)\Big)\text{x}+\Big(1+3\Big(\frac{-1}{3}\Big)\Big)\text{y}+\Big(1+4\Big(\frac{-1}{3}\Big)\text{z}-1-5\Big(\frac{-1}{3}\Big)\Big)=0$
$\Rightarrow\text{x}-\text{z}+2=0$

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Question 1024 Marks

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, -4, 3). Find the equation of the plane.

Answer

The given equation of the line are
$\frac{\text{x}+3}{3}=\frac{\text{y}}{-2}=\frac{\text{z}-7}{6}\ ....(\text{i})$
$\frac{\text{x}+6}{1}=\frac{\text{y}+5}{-3}=\frac{\text{z}-1}{2}\ ....(\text{ii})$
Let the directions of the plane be proportional to a, b, c.
Since the plane contains line (i), it should pass through (-3, 0, 7)
And is parallel to the line (i)
Equation of the plane through (i) is,
a(x + 3) + b(y) + c(z - 7) = 0 ....(iii)
Where 3a - 2b + 6c = 0 ...(iv)
Since the plane contains line (ii), the plane is parallel to line (ii) also
⇒ a - 3b + 2c = 0 ....(v)
Solving (iv) and (v) using cross-multiplication, we get
$\frac{\text{a}}{14}=\frac{\text{b}}{0}=\frac{\text{c}}{-7}$
Substituting a, b and c in (iii) we get
14(x + 3) + 0(y) - 7(z - 7) = 0
⇒ 2(x + 3) + 0(y) - 1(z - 7) = 0
⇒ 2x - z + 13 = 0

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Question 1034 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 1044 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 1054 Marks

Find the equation of the plane through (3, 4, -1) which is parallel to the plane $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+2=0$

Answer

Let the equation of a plane parallel to the given plane be
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=\text{k}\ ...(\text{i})$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=\text{k}$
This passes through (3, 4, -1). so,
$(3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=\text{k}$
$\text{k}=6-12-5=-11$
Substituting this in (i), we get
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=-11$
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+11=0,$ which is the equation of the required plane.

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Question 1064 Marks

Find the distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z.

Answer

The equation of line parallel to the line x = y = z and passing through the point (1, -5, 9) is
$\frac{\text{x}-1}{1}=\frac{\text{y}+5}{1}=\frac{\text{z}-9}{1}\ ...(\text{i})$
Any point on this line is of the form (k + 1, k - 5, k + 9)
If (k + 1, k - 5, k + 9) be the point of intersection of line (i) and the given plane, then
(k + 1) - (k - 5) + (k + 9) = 5
⇒ k = -10
So, the point of intersection of line (i) and the given plane is (-10 + 1, -10 - 5, -10 + 9) i.e., (-9, -15, -1).
$\therefore$ Required distance = Distance between (1, -5. 9) and (-9, -15, -1)
$=\sqrt{(1+9)^2+(-5+15)^2+(9+1)^2}$
$=\sqrt{3\times10^2}$
$=10\sqrt{3}\text{ units}$

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Question 1074 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 1084 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 1094 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 1104 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 1114 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.

Answer

We know that, equation of a plane passing through the line of intersection of a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 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

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Question 1124 Marks

Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z - 4 = 0, 2x + y - z + 5 = 0.

Answer

We know that, equation of plane passing through the line of intersection of planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 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 given two planes is x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0 is given by,
$(\text{x}+2\text{y}+3\text{z}-4)+\lambda(2\text{x}+\text{y}-\text{z}+5)=0$
$\text{x}(1+2\lambda)+\text{y}(2+\lambda)+\text{z}(3-\lambda)+4+5\lambda=0\ ...(\text{i})$
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 that plane (i) is perpendicular to plane,
$5\text{x}+3\text{y}+6\text{z}+8=0\ ...(\text{iii})$
Using (i) and (iii) in equation (ii),
$(5)(1+2\lambda)+(3)(2+\lambda)+(6)(3-\lambda)=0$
$5+10\lambda+6+3\lambda+18-6\lambda=0$
$29+7\lambda=0$
$7\lambda=-29$
$\lambda=-\frac{29}{7}$
Put the value of $\lambda$ in equation (i),
$\text{x}(1+2\lambda)+\text{y}(2+\lambda)+\text{z}(3-\lambda)-4+5\lambda=0$
$\text{x}\Big(1-\frac{58}{7}\Big)+\text{y}\Big(2-\frac{29}{7}\Big)+\text{z}\Big(3+\frac{29}{7}\Big)-4-\frac{145}{7}=0$
$\text{x}\Big(\frac{7-58}{7}\Big)+\text{y}\Big(\frac{14-29}{7}\Big)+\text{z}\Big(\frac{21+29}{7}\Big)\frac{-28-145}{7}=0$
$\text{x}\Big(-\frac{51}{7}\Big)+\text{y}\Big(-\frac{15}{7}\Big)+\text{z}\Big(\frac{50}{7}\Big)-\frac{173}{7}=0$

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Question 1134 Marks

Find the image of the point with position vector $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ in the plane $\vec{\text{r}}. (2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4.$ Also, find the position vectors of the foot of the prependicular and the equation of the perpendicular line through $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$

Answer

Let Q be the image the point $\text{P}(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$ in the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4$
Since PQ passes through P and is normal to the given plane, it is parallel to the normal vector $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$. So, the equation of PQ is
$\vec{\text{r}}=\big(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
As Q lies on PQ, let the position vector of Q be $(3+2\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}+(2+\lambda)\hat{\text{k}}.$
$=\frac{\Big[(3+2\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}+(2+\lambda)\hat{\text{k}}\Big]+\Big[3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\Big]}{2}$
$=\frac{(6+2\lambda)\hat{\text{i}}+(2-\lambda)\hat{\text{j}}+(4+\lambda)\hat{\text{k}}}{2}$
$=(3+\lambda)\hat{\text{i}}+\Big(1-\frac{\lambda}{2}\Big)\hat{\text{j}}+\Big(2+\frac{\lambda}{2}\Big)\hat{\text{k}}$
Since R lies in the plane $\vec{\text{r}}.\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)=4$
$=\Big[(3+\lambda)\hat{\text{i}}+\Big(1-\frac{\lambda}{2}\Big)\hat{\text{j}}+\Big(2+\frac{\lambda}{2}\Big)\hat{\text{k}}\Big].\Big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\Big)=4$
$\Rightarrow 6+2\lambda-1+\frac{\lambda}{2}+2+\frac{\lambda}{2}=4$
$\Rightarrow 7+2\lambda+\frac{\lambda}{2}+\frac{\lambda}{2}=4$
$\Rightarrow 14+6\lambda=8$
$\Rightarrow 6\lambda=8-14$
$\Rightarrow \lambda=-1$
Putting $\lambda=-1$ in Q, we get
$\text{Q}=(3+2(-1))\hat{\text{i}}+(1-(-1))\hat{\text{j}}+(2+(-1))\hat{\text{k}}$
$=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{h}}$ or (1, 2, 1)
Therefore, by putting $\lambda=-1$ in R, we get
$\text{R}=(3+(-1))\hat{\text{i}}+\Big(1-\frac{(-1)}{2}\Big)\hat{\text{j}}+\Big(2+\frac{(-1)}{2}\Big)\hat{\text{k}}$
$=2\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+\frac{3}{2}\hat{\text{k}}$

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Question 1144 Marks

Find the equation of the plane passing through the line of intersection of the planes 2x - 7y + 4z = 0, 3x - 5y + 4z + 11 = 0 and the point (-2, 1, 3).

Answer

We know that, equation of a plane passing through the line of intersection of two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
Given, equation of plane is,
2x - 7y + 4z - 3 = 0 and
3x - 5y + 4z + 11 = 0
So, equation of plane passing through the line of intersection of given two planes is,
$(-2)(2+3\lambda)+(1)(-7-5\lambda)+(3)(4+4\lambda)-3+11\lambda=0$
$-4-6\lambda-7-5\lambda+12+12\lambda-3+11\lambda=0$
$-2+12\lambda=0$
$12\lambda=2$
$\lambda=\frac{2}{12}$
$\lambda=\frac{1}{6}$
Put $\lambda$ in equation (i),
$\text{x}(2+3\lambda)+\text{y}(-7+5\lambda)+\text{z}(4+4\lambda)-3+11\lambda=0$
$\text{x}\Big(2+\frac{3}{6}\Big)+\text{y}\Big(-7-\frac{5}{6}\Big)+\text{z}\Big(4+\frac{4}{6}\Big)-3+\frac{11}{6}=0$
$\text{x}\Big(\frac{12+3}{6}\Big)+\text{y}\Big(\frac{-42-5}{6}\Big)+\text{z}\Big(\frac{24+4}{6}\Big)-\frac{18+11}{6}=0$
$\frac{15}{6}\text{x}-\frac{47}{6}\text{y}+\frac{28}{6}\text{z}-\frac{7}{6}=0$
Multiplying by 6, we get
15x - 47y + 28z - 7 = 0
Therefore, equation of required plane is,
15x - 47y + 28z = 7

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Question 1154 Marks

Find the equation of the containing the line $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and the point (0, 7, -7) and show that the line $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ also lies in the same plane.

Answer

We know that plane through (x₁, y₁, z₁) is given by
a(x - x₁) + b(y - y₁) + c(z - z₁) = 0 ...(i)
Required plane is passing through (0, 7, -7), so
a(x - 0) + b(y - 7) + c(z + 7) = 0
ax + b(y - 7) + c(z + 7) = 0 ....(ii)
Plane (ii) also contains line $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ so, it passes through point (-1, 3, -2),
a(-1) + b(3 - 7) + c(-2 + 7) = 0
-a - 4b + 5c = 0 ....(iii)
Also, plane (iii) will be parallel to line
So, a₁a₂ + b₁b₂ + c₁c₂ = 0
(a)(-3) + (b)(2) + (c)(1) = 0
-3a + 2b + c = 0 .....(iv)
Solving (iii) and (iv) by cross-multiplication,
$\frac{\text{a}}{(-4)(1)-(5)(2)}=\frac{\text{b}}{(-3)(5)-(-1)(1)}=\frac{\text{c}}{(-1)(2)-(-4)(-3)}$
$\frac{\text{a}}{-4-10}=\frac{\text{b}}{-15+1}=\frac{\text{c}}{-2-12}$
$\frac{\text{a}}{-14}=\frac{\text{b}}{-14}=\frac{\text{c}}{14}=\lambda(\text{say})$
$\text{a}=-14\lambda,\text{b}=-14\lambda,\text{c}=-14\lambda$
Put a, b, c in equation (ii),
$\text{ax}+\text{b}(\text{y}-7)+\text{c}(\text{z}+7)=0$
$(-14\lambda)\text{x}+(-14\lambda)(\text{y}-7)+(-14\lambda)(\text{z}+7)=0$
Dividing by $(-14\lambda),$ we get
$\text{x}+\text{y}-7+\text{z}+7=0$
$\text{x}+\text{y}+\text{z}=0$
So, equation of plane containing the given point and line is x + y + z = 0
The other line is $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$
So, a₁a₂ + b₁b₂ + c₁c₂ = 0
(1)(1) + (1)(-3) + (1)(2) = 0
1 - 3 + 2 = 0
0 = 0
LHS = RHS
So, $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ lie on plane x + y + z = 0

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Question 1164 Marks

Find the shortest distance between the lines $\frac{\text{x}-2}{-1}=\frac{\text{y}-5}{2}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}-0}{2}=\frac{\text{y}+5}{-1}=\frac{\text{z}-1}{2}.$

Answer

Consider,
$\text{l}_1:\frac{\text{x}-2}{-1}=\frac{\text{y}-5}{2}=\frac{\text{z}-0}{3}$
$\text{l}_2:\frac{\text{x}-0}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-1}{2}$
Clearly line l₁ passes through the point P(2, 5, 0)
The equation of a plane containing line l₂ is
a(x - 0) + b(y + 1) + c(z - 1) = 0 .....(i)
Where 2a - b + 2c = 0
If it is paralle to line l₁ then
-a + 2b + 3c = 0
Therefore
$\frac{\text{a}}{-7}=\frac{\text{b}}{-8}=\frac{\text{c}}{3}$
Substituting values of a, b, c in the equation (i) we obtain
a(x - 0) + b(y + 1) + c(z - 1) = 0
-7(x - 0) - 8(y + 1) + 3(z - 1) = 0
-7x - 8y - 8 + 3z - 3 = 0
7x + 8y - 3z + 11 = 0 .....(ii)
This is the equation of the plane contaning line 22 and paralle to line l1
Shortest distance between l₁ and l₂ = Distance between point P(2, 5, 0) and plane (ii)
$=\bigg|\frac{14+40+11}{\sqrt{7^2+8^2+(-3)^2}}\bigg|=\frac{65}{\sqrt{122}}$

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Question 1174 Marks

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

Answer

The equation of any plane passing through (-1, 3, 2) is
Given (x + 1) + b(y - 3) + c(z - 2) = 0 ...(i)
Given that, plane (i) is perpendicular to the planes
x + 2y + 3z = 5
and
3x + 3y + z = 0
Therefore, we have,
a + 2b + 3c = 0 ...(ii)
Solving equations (ii) and (iii) by cross multiplication, we have
$\frac{\text{a}}{2\times1-3\times3}=\frac{\text{b}}{3\times3-1\times1}=\frac{\text{c}}{1\times3-3\times2}=\lambda(\text{say})$
$\Rightarrow\frac{\text{a}}{2-9}=\frac{\text{b}}{9-1}=\frac{\text{c}}{3-6}=\lambda$
$\Rightarrow\frac{\text{a}}{-7}=\frac{\text{b}}{8}=\frac{\text{c}}{-3}=\lambda$
Thus, we have
$\text{a}=-7\lambda,\text{b}=8\lambda$ and $\text{c}=-3\lambda$
Substituting the above values in equation (i), we get
$7\lambda(\text{x}+1)+8\lambda(\text{y}-3)-3\lambda(\text{z}-2)=0$
Since $\lambda\neq0,$ we get
-7(x + 1) + 8(y - 3) - 3(z - 2) = 0
⇒ -7x - 7 + 8y - 24 - 3z + 6 = 0
⇒ -7x + 8y - 3z - 25 = 0
⇒ 7x - 8y + 3z + 25 = 0
Thus the required of the plane is 7x - 8y + 3z + 25 = 0

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Question 1184 Marks

Find the equation of the plane passing through the point (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.

Answer

The equation of any plane passing through (2, 1, -1) is,
a(x - 2) + b(y - 1) + c(z + 1) = 0 ....(i)
It is given that (i) is passing through (-1, 3, 4). So,
a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0
⇒ -3a + 2b + 5c ....(ii)
It is given that (i) is perpendicular to the plane x - 2y + 4z = 10. So,
a + 2b + 4c = 0 ....(iii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}+1\\-3&2&5\\1&-2&4\end{vmatrix}=0$
⇒ 18(x - 2) + 17(y - 1) + 4(z + 1) = 0
⇒ 18x + 17y + 4z - 49 = 0

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Question 1194 Marks

Find the equation of the plane passing through the intersection of the planes 2x + 3y - z + 1 = 0 and x + y - 2z + 3 = 0 and perpendicular to the plane 3x - y - 2z - 4 = 0.

Answer

We know that equation of a plane passing through the line of intersection of two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 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 planes 2x + 3y - z + 1 = 0 and x + y - 2z + 3 = 0 is,
$(2\text{x}+3\text{y}-\text{z}+1)+\lambda(\text{x}+\text{y}-2\text{z}+3)=0$
$\text{x}(2+\lambda)\text{y}(3+\lambda)+\text{z}(-1-2\lambda)+1+3\lambda=0\ ...(\text{i})$
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 perpendicular to the plane,
$3\text{x}-\text{y}-2\text{z}-4=0\ ...(\text{iii})$
Using (i) and (iii) in equation (ii),
$(3)(2+\lambda)+(-1)(3+\lambda)+(-2)(-1-2\lambda)=0$
$6+3\lambda-3-\lambda+2+4\lambda=0$
$6\lambda+5=0$
$6\lambda=-5$
$\lambda=-\frac{5}{6}$
Put the value of $\lambda$ in equation (i),
$\text{x}(2+\lambda)+\text{y}(3+\lambda)+\text{z}(-1-2\lambda)+1+3\lambda=0$
$\text{x}\Big(2-\frac{5}{6}\Big)+\text{y}\Big(3-\frac{5}{6}\Big)+\text{z}\Big(1+\frac{10}{6}\Big)+1-\frac{15}{6}=0$
$\text{x}\Big(\frac{12-5}{6}\Big)+\text{y}\Big(\frac{18-5}{6}\Big)+\text{z}\Big(\frac{-6+10}{6}\Big)+\frac{6-15}{6}=0$
$\frac{7\text{x}}{6}+\frac{13\text{y}}{6}+\frac{4\text{z}}{6}-\frac{9}{6}=0$

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Question 1204 Marks

Find the equation of the plane passing through the intersection of the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})=7,\vec{\text{r}}\cdot(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})=9$ and the point (2, 1, 3).

Answer

The equation of the plane passing through the line of intersection of the given planes is,
$\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-7+\lambda\big(\vec{\text{r}}\cdot(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})-9\big)=0$
$\vec{\text{r}}\cdot\Big[(2-2\lambda)\hat{\text{i}}+(1+5\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}\Big]-7-9\lambda=0\ ...(\text{i})$
This passes through $(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})$
$\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)\Big[(2-2\lambda)\hat{\text{i}}+(1+5\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}\Big]-7-9\lambda=0$
$\Rightarrow4+4\lambda+1+5\lambda+9+9\lambda-7-9\lambda=0$
$\Rightarrow9+\lambda+7=0$
$\Rightarrow\lambda=\frac{-7}{9}$
Substituting this in (i) we get
$\vec{\text{r}}\cdot\Big[\Big(2+2\Big(\frac{-7}{9}\Big)\Big)\hat{\text{i}}+\Big(1+5\Big(\frac{-7}{9}\Big)\Big)\hat{\text{j}}+\Big(3+3\Big(\frac{-7}{9}\Big)\hat{\text{k}}\Big]-7-9\Big(\frac{-7}{9}\Big)=0$
$\Rightarrow\vec{\text{r}}\cdot(4\hat{\text{i}}-26\hat{\text{j}}+6\hat{\text{k}})=0$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-13\hat{\text{j}}+3\hat{\text{k}})=0$

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Question 1214 Marks

Prove that the line of section of the planes 5x + 2y - 4z + 2 = 0 and 2x + 8y + 2z - 1 = 0 is parallel to the plane 4x - 2y - 5z - 2 = 0.

Answer

Firstly we have to find that line of section of planes 5x + 2y - 4z + 2 = 0 and 2x + 8y + 2z - 1 = 0
Let a₁, b₁, c₁ be the direction ratios of the line 5x + 2y - 4z + 2 = 0 and 2x + 8y + 2z - 1 = 0
Since, line lies in both the planes, so it is perpendicular to both planes, so
5a₁ + 2b₁- 4c₁ = 0 ...(i)
2a₁ + 8b₁ + 2c₁ = 0 ....(ii)
Solving equation (i) and (ii), by cross-multiplication
$\frac{\text{a}_1}{(2)(2)-(-4)(8)}=\frac{\text{b}_1}{(2)(-4)-(5)(2)}=\frac{\text{c}_1}{(5)(8)-(2)(2)}$
$\frac{\text{a}_1}{4+32}=\frac{\text{b}_1}{-8-10}=\frac{\text{c}_1}{40-4}$
$\frac{\text{a}_1}{36}=\frac{\text{b}_1}{-18}=\frac{\text{c}_1}{36}$
$\frac{\text{a}_1}{2}=\frac{\text{b}_1}{-1}=\frac{\text{c}_1}{2}=\lambda(\text{say})$
$\Rightarrow\text{a}_1=2\lambda,\text{b}_1=-\lambda,\text{c}_1=2\lambda$
We know that, line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{{\text{y}}-{\text{y}_1}}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ is parallel to plane a₂x + b₂y + c₂z + d₂= 0 if
a₁a₂ + b₁b₂ + c₁c₂= 0 ....(iii)
Here line with direction ratio a₁, b₁, c₁ is parallel to plane 4x - 2y - 5z - 2 = 0,
a₁a₂ + b₁b₂ + c₁c₂
= (2)(4) + (-1)(-2) + (2)(-5)
= 8 + 2 - 10
= 0
Therefore, line of section is parallel to the plane.

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Question 1224 Marks

Find the vector equation of the plane passing through three point with position vectors $\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}},2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}.$ Also, find coordinates of the point of intersection of this plane and the line $\vec{\text{r}}=3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\lambda(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).$

Answer

Let A(1, 1, -2), B(2, -1, 1) and C(1, 2, 1) be the point represented by the given position vectors.
The required planes through the point A(1, 1, -1) whose position vector is
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
Clearly, $\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
$=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{AC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$
$=(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
$=0\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&3\\0&1&3\end{vmatrix}$
$=-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$
The vector equation of the required plane is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})=(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
To find the point of intersection of the plane
The given equation of the line is
$\vec{\text{r}}=(3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})+\lambda(2\hat{\text{i}}-2\text{j}+\hat{\text{k}})$
$\vec{\text{r}}=(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}$
The coordinates of any point on this line are in the form of
$(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}$
Since this point lies on the plane $\vec{\text{r}}\cdot(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14,$
$\Big[(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}\Big]\cdot\big(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)=14$
$\Rightarrow27+18\lambda-3-6\lambda+1-\lambda=14$
$\Rightarrow11\lambda=-11$
$\Rightarrow\lambda=-1$
So, the coordinates of the point are
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})=-9-3-2$
$\Rightarrow\vec{\text{r}}\cdot\Big[-\big(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)\Big]=-14$
Or $(3+2\lambda,-1-2\lambda,-1+\lambda)$
$(3+2\lambda,-1-2\lambda,-1+\lambda)$
$=(3-2,-1+2,-1-1)$
$=(1,1,-2)$

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Question 1234 Marks

Find the equation of the plane passing through the points whose coordinates are (-1, 1, 1) and (1, -1, 1) and perpendicular to the plane x + 2y + 2z = 5.

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 required plane is pasing through (-1, 1, 1)
a(x + 1) + b(y - 1) + c(z - 1) = 0 ....(i)
It is also passing through (-1, 1, 1), so it must satisfy the equation (i),
a(1 + 1) + b(1 - 1) + c(1 - 1) = 0
2a - 2b = 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
x + 2y + 2z = 5 ....(iv)
Using plane (i), (iv) in equation (iii),
a₁a₂ + b₁b₂ + c₁c₂= 0
(a)(1) + (b)(2) + (c)(2) = 0
a + 2b + 2c = 0 ....(v)
Solving (ii) and (v) by cross-multiplication,
$\frac{\text{a}}{(-2)(2)-(2)(0)}=\frac{\text{b}}{(1)(0)-(2)(2)}=\frac{\text{c}}{(2)(2)-(1)(-2)}$
$\frac{\text{a}}{-4-0}=\frac{\text{b}}{0-4}=\frac{\text{c}}{4+2}$
$\frac{\text{a}}{-4}=\frac{\text{b}}{-4}=\frac{\text{c}}{6}=\lambda(\text{say})$
$\text{a}=-4\lambda,\text{b}=-4\lambda,\text{c}=6\lambda$
Put a, b, c in equation (i),
$\text{a}(\text{x}+1)+\text{b}(\text{y}-1)+\text{c}(\text{z}-1)=0$
$(-4\lambda)(\text{x}+1)+(-4\lambda)(\text{y}-1)+(6\lambda)(\text{z}-1)=0$
$-4\lambda\text{x}+4\lambda-4\lambda\text{y}+4\lambda+6\lambda\text{z}-6\lambda=0$
$-4\lambda\text{x}-4\lambda\text{y}+6\lambda\text{z}-6\lambda=0$
Dividing by $(-2\lambda),$ we get
2x + 2y - 3z + 3 = 0
The equation of required plane is,
2x + 2y - 3z + 3 = 0

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Question 1244 Marks

Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x - y - z = 7. Also, find the length of the perpendicular.

Answer

Let Q be the foot of the perpendicular.
Here, Direction ratios of normal to plane is 3, -1, -1
⇒ Line PQ is parallel to normal to plane
⇒ Direction ratios of PQ are proportional to 3, -1, -1 and PQ is passing through P(2, 3, 7).
So,
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\frac{\text{x}-2}{3}=\frac{\text{y}-3}{-1}=\frac{\text{z}-7}{-1}=\lambda\text{ (say)}$
General ponit on line PQ
$=(3\lambda+2,-\lambda+3,-\lambda+7)$
Coordinates of Q be $(3\lambda+2,-\lambda+3,-\lambda+7)$
Point Q lies on the plane 3x - y - z = 7.
So,
$3(3\lambda+2)-(-\lambda+3)-(-\lambda+7)=7$
$9\lambda+6+\lambda-3+\lambda-7=7$
$11\lambda=7+4$
$11\lambda=11$
$\lambda=\frac{11}{11}$
$\lambda=1$
$\therefore$ Coordinate of Q $=(3\lambda+2,-\lambda+3,-\lambda+7)$
$=\big(3(1)+2,-(1)+3,-(1)+7\big)$
$=(5,2,6)$
Length of the perpendicular PQ
$=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{(2-5)^2+(3-2)^2+(7-6)^2}$
$=\sqrt{9+1+1}$
$=\sqrt{11}$

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Question 1254 Marks

Find the equation of the plane through the points (3, 4, 1) and (0, 1, 0) and parallel to the line $\frac{\text{x}+3}{2}=\frac{\text{y}-3}{7}=\frac{\text{z}-2}{5}.$

Answer

We know that equation of a line passing through (x₁, y₁, z₁) is given by
a(x - x₁) + b(y - y₁) + c(z - z₁) ....(i)
Given that, requaired equation of plane is passing through (3, 4, 1), so
a(x - 3) + b(y - 4) + c(z - 1) = 0 ....(ii)
Plane (ii) is also passing through (0, 1, 0), so
a(0 - 3) + b(1 - 4) + c(0 - 1) = 0
-3a - 3b - c = 0
3a + 3b + c = 0 ....(iii)
We know that, plane a₁x + b₁y + c₁z + d₁ = 0 and line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ are parallel if a₁a₂ + b₁b₂ + c₁c₂= 0
Here, line $\frac{\text{x}+3}{2}=\frac{\text{y}-3}{7}=\frac{\text{z}-2}{5}$ is parallel to plane (ii), so
$(2)(\text{a})+(7)(\text{b})+(\text{c})(5)=0$
$2\text{a}+7\text{b}+5\text{c}=0\ ...(\text{iv})$
Solving (iii) and (iv) by croaa-multiplication,
$\frac{\text{a}}{(3)(5)-(7)(1)}=\frac{\text{b}}{(2)(1)-(3)(5)}=\frac{\text{c}}{(3)(7)-(2)(3)}$
$=\frac{\text{a}}{15-7}=\frac{\text{b}}{2-15}=\frac{\text{c}}{21-6}$
$\frac{\text{a}}{8}=\frac{\text{b}}{-13}=\frac{\text{c}}{15}=\lambda(\text{ say})$
$\text{a}=8\lambda,\text{d}=-13\lambda,\text{c}=15\lambda$
Put a, b, c in equation (ii),
$\text{a}(\text{x}-3)+\text{b}(\text{y}-4)+\text{c}(\text{z}-1)=0$
$8\lambda(\text{x}-3)+(-13\lambda)(\text{y}-4)+(15\lambda)(\text{z}-1)=0$
$8\lambda\text{x}-24\lambda-13\lambda\text{y}+52\lambda+15\lambda\text{z}-15\lambda=0$
$8\lambda\text{x}-13\lambda\text{y}+15\lambda\text{z}+13\lambda=0$
Dividing by $\lambda,$ equation of required plane is,
$8\text{x}-13\text{y}+15\text{z}+13=0$

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Question 1264 Marks

Find the equation of the plane mid-parallel to the planes 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0

Answer

We know that the distance between two planes ax + by + cz = d₁ and ax + by + cz = d₂ is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
The equation of plane thatb is mid-parallel to the planes
2x - 2y + z + 3 = 0 ...(i)
2x - 2y + z + 9 = 0 ....(ii)
is of the form 2x - 2y + z + k = 0 ...(iii)
It meance that the distance between (i) and (iii) = distance between (i) and (ii)
$\Rightarrow\frac{|\text{k}-3|}{\sqrt{4+4+1}}=\frac{|\text{k}-9|}{\sqrt{4+4+1}}$
⇒ |k - 3| = |k - 9|
⇒ k - 3 = k - 9 or k - 3 = -(k - 9)
⇒ 3 = 9 (false); k - 3 = -k + 9
⇒ 2k = 12
⇒ k = 6
Substituting this in (iii) we get 2x - 2y + z + 6 = 0, which is the required equation of the plane.

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Question 1274 Marks

Find the equation of the plane which contains planes is the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0 and whose x-intercept is twice its z-intercept.

Answer

Hence write the vector equation of a plane passing through the point (2, 3, -1) and parallel to the plane obtained above.

Solution:

Given, eqcation of planes are

$\text{x}+2\text{y}+3\text{z}-4=0$ . . .(i)

and $2\text{x}+\text{y}+\text{z}+5=0$ . . . (ii)

Clearly, the equation of the plan which contains the line of intersection of planes (i)and (ii), is

$(\text{x}+2\text{y}+3\text{z}-4)+\lambda(2\text{x}+\text{y}+\text{z}+5)=0$

or $(1+2\lambda)\text{x}+(2+\lambda)\text{y}+(3-\lambda)\text{z}+5\lambda-4=0$ . . .(iii)

This equation can be written in intercept form as

$\frac{\text{x}}{\frac{4-5\lambda}{1+2\lambda}}+\frac{\text{y}}{\frac{4-5\lambda}{2+\lambda}}+\frac{\text{z}}{\frac{4-5\lambda}{3+\lambda}}=1$

Since, it is given that the x-intercept of plane (iii) is twice its z-intercept.

$\therefore \frac{4-5\lambda}{1+2\lambda}=2\bigg(\frac{4-5\lambda}{3-\lambda}\bigg) $

or $3-5\lambda=2+4\lambda$

or $5\lambda=1 $or $ \lambda=\frac{1}{5}$

So,the required equation of plane is

$\bigg(1+\frac{2}{5}\bigg)\text{x}+\bigg(2+\frac{1}{5}\bigg)\text{y}+\bigg(3-\frac{1}{5}\bigg)\text{z}=4-5.\frac{1}{5} $

or $\frac{7}{5}\text{x}+\frac{11}{5}\text{y}+\frac{14}{5}\text{z}=\frac{15}{5}$

or $7\text{x}+11\text{y}+14\text{z}=15$ ...(iii)

Clearly the Dr's of normal to the plane , which is parallel to the plane (iv), are 7, 11 and 14.

$\therefore$ The vector equation of a plane passing through the (2, 3, -1) and parallel to the plane (iv), is

$\big[\overrightarrow{\text{r}}-(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\big].(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=0$

or $\overrightarrow{\text{r}}.(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}).(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})$

or $\overrightarrow{\text{r}}(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=14+33-14$

or $\overrightarrow{\text{r}}(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=33$\

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Question 1284 Marks

Find the vector equation of the following planes in non-parametric form.
$\vec{\text{r}}=(2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\mu(5\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}})$

Answer

We know that equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}} $ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to t.

Here, $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\hat{\text{c}}=5\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$

Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$

$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\5&-2&7\end{vmatrix}$

$=20\hat{\text{i}}+8\hat{\text{j}}-12\hat{\text{k}}$

The vector equation of the plane in scalar product form is,

$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$

$\Rightarrow\vec{\text{r}}\cdot(20\hat{\text{i}}+8\hat{\text{j}}-12\hat{\text{k}})=(2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})(20\hat{\text{i}}+8\hat{\text{j}}-12\hat{\text{k}})$

$\Rightarrow\vec{\text{r}}\cdot\big(4(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\big)=40+16+12$

$\Rightarrow\vec{\text{r}}\cdot\big(4(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\big)=68$

$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$

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