3 Marks - Maths STD 12 Science Questions - Vidyadip

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Question 13 Marks

A plane meets the coordinate axes at A, B and C, respectively, such that the centriod of triangle ABC is (1, -2, 3). Find the equation of the plane.

Answer

Let a, b and c be the intercepts of the given plane on the coordinate axes.
Then the plane meets the coordinate axes at
A(a, 0, 0), B(0, b, 0) and C(0, 0, c)
Given that the centroid of the triangle is (1, -2, 3)
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(1,-2,3)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(1,-2,3)$
$\Rightarrow\frac{\text{a}}{3}=1,\frac{\text{b}}{3}=-2,\frac{\text{c}}{3}=3$
$\Rightarrow\text{a}=3,\text{b}=-6,\text{c}=9\ ...(\text{i})$
Equation of the plane whose intercepts on the coordinate axes a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{-6}+\frac{\text{z}}{9}=1$ [From (i)]
$\Rightarrow6\text{x}-3\text{y}+2\text{x}=18$

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Question 23 Marks

Find the vector and cartesian equations of a plane passing through the point (1, -1, 1) and normal to the line joining the points (1, 2, 5) and (-1, 3, 1)

Answer

Since the given plane passes through the point (1, -1, 1) and is normal to the line joining A(1, 2, 5) and B(-1, 3, 1)
$\overrightarrow{\text{n}}=\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(-\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}})$
$=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
We know that the vector equation of the passing through a point $\overrightarrow{\text{a}}$ and normal to $\overrightarrow{\text{n}}$ is,
$\overrightarrow{\text{r}}\cdot\overrightarrow{\text{n}}=\overrightarrow{\text{a}}\cdot\overrightarrow{\text{n}}$
Substituting $\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{n}}=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ we get
$\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=-2-1-4$
$\Rightarrow\overrightarrow{\text{r}}\cdot\big[-(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})\big]=-7$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=7$
For cartesian form, we need to substitute $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation.
Then, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})=7$
$\Rightarrow2\text{x}-\text{y}+4\text{z}=7$

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Question 33 Marks

Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
4x + 3y - 6z - 12 = 0

Answer

Equation of the given plane is,
4x + 3y - 6z - 12 = 0
⇒ 4x + 3y - 6z = 12
Dividng both sides by 12, we get
$\frac{4\text{x}}{12}+\frac{3\text{y}}{12}+\frac{(6\text{z})}{12}=\frac{12}{12}$
$\Rightarrow\frac{4\text{x}}{12}+\frac{3\text{y}}{12}-\frac{6\text{z}}{12}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{4}+\frac{\text{z}}{-2}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=3;\text{ b}=4;\text{ c}=-2$

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Question 43 Marks

Find the distance of the point P(-1, -5, -10) from the point of intersection of the line joining the points A(2, -1, 2) and B(5, 3, 4) with the plane x - y + z = 5.

Answer

Equation of the line through the points A(2, -1, 2) and B(5, 3, 4) is $\frac{\text{x}-2}{5-2}=\frac{\text{y}+1}{3+1}=\frac{\text{z}-2}{4-2}=\text{r}$
$\Rightarrow\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-2}{2}=\text{r}$
$\Rightarrow\text{x}=3\text{r}+2,\text{ y}=4\text{r}-1,\text{ z}=2\text{r}+2$
Substituting these in the plane equation we get
$(3\text{r}+2)-(4\text{r}-1)+(2\text{r}+2)=5$
$\Rightarrow\text{r}=0$
$\Rightarrow\text{x}=2,\text{ y}=-1,\text{ z}=2$
Distance of (2, -1, 2) from (-1, -5, -10) is
$=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{3^2+4^2+12^2}$
$=\sqrt{169}=13$

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Question 53 Marks

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
zx-plane

Answer

Direction ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the zx-plane y = 0
$\Rightarrow-3\text{r}+1=0$
$\Rightarrow\text{r}=\frac{1}{3}$
$\Rightarrow\text{x}=2\Big(\frac{1}{3}\Big)+5=\frac{17}{3}$
$\Rightarrow\text{z}=5\Big(\frac{1}{3}\Big)+6=\frac{23}{3}$
Hence, the corrdinates of the point are $\Big(\frac{17}{3},0,\frac{23}{3}\Big)$

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Question 63 Marks

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
yz-plane

Answer

Direction ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the yz-plane x = 0
$\Rightarrow2\text{r}+5=0$
$\Rightarrow\text{r}=-\frac{5}{2}$
$\Rightarrow\text{y}=-3\Big(-\frac{5}{2}\Big)+1=\frac{17}{2}$
$\Rightarrow\text{z}=5\Big(-\frac{5}{2}\Big)+6=-\frac{13}{2}$
Hence, the corrdinates of the point are $\Big(0,\frac{17}{2},-\frac{13}{2}\Big)$

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Question 73 Marks

Write the vector equation of the line passing through the point (1, -2, -3) and normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5.$

Answer

The required line is normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5$ and it is parallel to the normal vector of the plane.
So, the required line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
It is given that the line passes through the point (1, -2, -3) whose position vector is given by $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
We know that the equation of the line passing through the point whose position vector is $\vec{\text{a}}$ and parrallel to the vector $\vec{\text{b}}$ is given by
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$

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Question 83 Marks

Find the angle between the plane:
2x + y - 2z = 5 and 3x - 6y - 2z = 7

Answer

We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x + y - 2z = 5 and 3x - 6y - 2z = 7 is given by
$\cos\theta=\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\sqrt{2^2+1^2+(-2)^2}\sqrt{3^2+(-6)^2+(-2)^2}}$
$=\frac{6-6+4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{4}{(3)(7)}=\frac{4}{21}$
$\theta=\cos^{-1}\Big(\frac{4}{21}\Big)$

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Question 93 Marks

Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=6$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})=9$

Answer

We know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{n}}_2=3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}}\big|}$
$=\frac{6-6-4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{-4}{(3)(7)}$
$=\frac{-4}{21}$
$\theta=\cos^{-1}\Big(\frac{-4}{21}\Big)$

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Question 103 Marks

Find the distance of the plane 2x- 3y + 4z - 6 = 0 from the origin.

Answer

The given equation of the plane is,
2x - 3y + 4z = 6 .....(i)
Now, $\sqrt{2^2+(-3)^2+4^2}$
$=\sqrt{4+9+16}$
$=\sqrt{29}$
Dividing (i) by $\sqrt{29},$ we get
$\frac{2}{\sqrt{29}}\text{x}-\frac{3}{\sqrt{29}}\text{y}+\frac{4}{\sqrt{29}}\text{z}=\frac{6}{\sqrt{29}},$ which is the normal form of plane (i)
So, the length of the perpendicular from the origin to the plane $=\frac{6}{\sqrt{29}}$

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Question 113 Marks

Find the distance of the point (2, 3, -5) from the plane x + 2y - 2z - 9 = 0.

Answer

We know that the distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is given by
$\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|(2)+2(3)-2(-5)-9|}{\sqrt{1^2+2^2+(-2)^2}}$
$=\frac{|2+6+10-9|}{\sqrt{1+4+4}}$
$=\frac{9}{3}$
$=3\text{ units}$

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Question 123 Marks

Find the equation of the plane through the untersection of the planes 3x - y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).

Answer

The equation of the family of planes through the intersection of planes 3x - y + 2z = 4 and x + y + z = 2 is,
$(3\text{x}-\text{y}+2\text{z}-4)+\lambda(\text{x}+\text{y}+\text{z}-2)=0\ ....(\text{i})$
If it passes through (2, 2, 1), then
$(6-2+2-4)+\lambda(2+2+1-2)=0$
$\lambda=-\frac{2}{3}$
Substituting $\lambda=-\frac{2}{3}$ in (i) we get, 7x - 5y + 4z = 0 as the equation of the required plane.

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Question 133 Marks

Find the angle between the plane:
x + y - 2z = 3 and 2x - 2y + z = 5

Answer

We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between x + y - 2z = 3 and 2x - 2y + z = 5 is given by
$\cos\theta=\frac{(1)(2)+(1)(-2)+(-2)(1)}{\sqrt{1^2+1^2+(-2)^2}\sqrt{2^2+(-2)^2+1^2}}$
$=\frac{2-2-2}{\sqrt{1+1+4}\sqrt{4+4+1}}=\frac{-2}{\sqrt{6}\sqrt{9}}$
$=\frac{-2}{\sqrt{6}\sqrt{9}}=\frac{-2}{3\sqrt{6}}$
$\theta=\cos^{-1}\Big(\frac{-2}{3\sqrt{6}}\Big)$

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Question 143 Marks

Find the angle between the line $\frac{\text{x}+1}{2}=\frac{\text{y}}{3}=\frac{\text{z}-3}{6}$ and the plane 10x + 2y - 11z = 3.

Answer

The given line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}}$
We know that the angle $\theta$ between the line and the plane is given by.
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})\cdot(10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}||10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}}|}$
$=\frac{20+6-66}{\sqrt{4+9+36}\sqrt{100+4+121}}$
$=\frac{-40}{(7)(15)}=\frac{-8}{21}$
$\theta=\sin^{-1}\Big(\frac{-8}{21}\Big)$

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Question 153 Marks

Find the coordinates of the point where the line through (3, -4, -5) and (2, -3, 1) corsses the plane 2x + y + z = 7.

Answer

The equation of the through the points (3, -4, -5) and (2, -3, 1) is
$\frac{\text{x}-3}{2-3}=\frac{\text{y}+4}{-3+4}=\frac{\text{z}+5}{1+5}$
$\Rightarrow\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}$
The coordinates of any point on this line are of the form
$\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}=\lambda$
$\Rightarrow\text{x}=-\lambda+3,\text{ y}=\lambda-4,\text{ z}=6\lambda-5$
$\Rightarrow5\lambda=10$
$\Rightarrow\lambda=2$
So, the coordinates of the point are
$(-\lambda+3,\lambda-4,6\lambda-5)$
$=(-2+3,2-4,6(2)-5)$
$=(1,-2,7)$

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Question 163 Marks

Find the distance of the point (2, 3, 5) from the xy-plane.

Answer

We know that, the distance (D) of a the point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is given by
$\text{D}=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\ ...(\text{i})$
So, distance of point (2, 3, 5) from xy-plane (we know that equation of xy-plane is z = 0) is
$=\Bigg|\frac{(2)(0)+(3)(0)+(5)(1)+0}{\sqrt{(0)^2+(0)^2+(1)^2}}\Bigg|$ [Using (i)]
$=\frac{0+0+5}{\sqrt{0+0+1}}$
$=5\text{ unit}$
Distance of the point (2, 3, 5) from xy-plane = 5 unit

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Question 173 Marks

Find the angle between the line $\vec{\text{r}}=(2\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ and the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=5$

Answer

We know that the angle $\theta$ between the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
Here,
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
So, $\sin\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{2+3+4}{\sqrt{4+9+16}\sqrt{1+1+1}}$
$=\frac{9}{\sqrt{29}\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{29}}$
$\theta=\sin^{-1}\Big(\frac{3\sqrt{3}}{\sqrt{29}}\Big)$

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Question 183 Marks

Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$3\text{x}-6\text{y}-2\text{z}=7$ and $2\text{x}+\text{y}-\lambda\text{z}=5$

Answer

We know that the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are perperndicular to each other only if
a₁a₂ + b₁b₂ + c₁c₂= 0
The given planes are 3x - 6y - 2z = 7 and $2\text{x}+\text{y}-\lambda\text{z}=5$
⇒ a₁ = 3; b₁ = -6; c₁ = -2; a₂ = 2; b₂ = 1; $\text{c}_2=-\lambda$
It is given that the given planes are perpendicular.
⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0
$\Rightarrow(3)(2)+(-6)(1)+(-2)(-\lambda)=0$
$\Rightarrow6-6+2\lambda=0$
$\Rightarrow2\lambda=0$
$\Rightarrow\lambda=0$

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Question 193 Marks

Find the distance of the point $2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}$ from the plane $\vec{\text{r}}\cdot(3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}})-9=0$

Answer

We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ is given by
$\text{p}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Here, $\vec{\text{a}}=-2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}; \vec{\text{n}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}};\text{d}=9$
So, the required distance, p
$=\frac{\big|(2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})\cdot(3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}})-9\big|}{\big|3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}-9\big|}$
$=\frac{6+4-48-9}{\sqrt{9+16+144}}$
$=\frac{|-47|}{13}$
$=\frac{47}{13}\text{ units}$

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Question 203 Marks

Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y - z = 1 and 3x - 4y + z = 5.

Answer

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

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Question 213 Marks

Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.

Answer

Let $\alpha,\beta$ and $\gamma$ be the angles made by $\vec{\text{n}}$ with x, y and z-axes, respectively.
It is given that
$\alpha=\beta=\gamma$
$\Rightarrow\cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\text{l}=\text{m}=\text{n},$ where l, m, n are direction cosines of $\vec{\text{n}}$.
But $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\text{l}^2+\text{l}^2+\text{l}^2=1$
$\Rightarrow3\text{l}^2=1$
$\Rightarrow\text{l}^2=\frac{1}{3}$
$\Rightarrow\text{l}^2=\frac{1}{\sqrt{3}}$
So, $\text{l}=\text{m}=\text{n}=\frac{1}{\sqrt{3}}$
It is given that the length of the perpendicular of the plane from the origin, $\text{p}=5\sqrt{3}$
The normal form of the plane is $\text{lx}+\text{my}+\text{nz}=\text{p}$
$\Rightarrow\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=5\sqrt{3}$
$\Rightarrow\text{x}+\text{y}+\text{z}=5\sqrt{3}\ (\sqrt{3})$
$\Rightarrow\text{x}+\text{y}+\text{z}=15.$

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Question 223 Marks

Find the equation of a passing through the point (-1, -1, 2) and perpendicular to the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5.

Answer

The equation of any plane passing through (-1, -1, 2) is,
a(x + 1) + b(y + 1) + c(z - 2) = 0 ....(i)
It is given that (i) is perpendicular to each of the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5. Then,
3a + 2b - 3c = 0 ....(i)
5a - 4b + c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}+1&\text{y}+1&\text{z}-2\\3&2&-3\\5&-4&1\end{vmatrix}=0$
⇒ -10(x + 1) - 18(y + 1) - 22(z - 2) = 0
⇒ 5(x + 1) + 9(y + 1) + 11(z - 2) = 0
⇒ 5x + 5 + 9y + 9 + 11z - 22 = 0
⇒ 5x + 9y + 11z - 8 = 0

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Question 233 Marks

Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x - 3y + 5z + 7 = 0. Also, find the distance between the two planes.

Answer

Equation of plane is parallel to 2x - 3y + 5z + 7 = 0 is of the form 2x - 3y + 5z = d
Above plane is passing through (3, 4, -1)
So, substitute above point in the equation, we get
6 - 12 - 5 = d
d = -11
So, palne equation is 2x - 3y + 5z = -11
Distance between planes is given by
$\Big|\frac{-7+11}{\sqrt{4+9+25}}\Big|=\frac{4}{\sqrt{38}}$

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Question 243 Marks

Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})=1$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}})=4$

Answer

We know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{n}}_2=-\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})\cdot(-\hat{\text{i}}+\vec{\text{j}}+0\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big|}$
$=\frac{-2-3}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}$
$=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$

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Question 253 Marks

A plabne makes intercepts -6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.

Answer

We know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
So, the equation of the plane which makes intercepts -6, 3, 4 on the x-axis, y-axis and z-axis, respecticely is,
$\frac{\text{x}}{-6}+\frac{\text{y}}{3}+\frac{\text{z}}{4}=1$
$\Rightarrow-2\text{x}+4\text{y}+3\text{z}=12$
$\Rightarrow-2\text{x}+4\text{y}+3\text{z}+12=0$
$\therefore$ Length of the perpendicular from (0, 0, 0) to the plane 2x - 4y - 3z + 12 = 0
$=\Bigg|\frac{2\times0-4\times0-3\times0+12}{\sqrt{12^2+(-4)^2+(-3)^2}}\Bigg|$
$=\bigg|\frac{12}{\sqrt{4+16+9}}\bigg|$
$=\frac{12}{\sqrt{29}}\text{ units}$
Thus, the length of the perpendicular from the origin to the plane is $\frac{12}{\sqrt{29}}\text{ units}.$

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Question 263 Marks

Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=2$

Answer

The equation of the family of plane parallel to $\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=2$ is,
$\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=\text{d}\ ...(\text{i})$
If it passes through (a, b, c) then
$(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=\text{d}$
$=\text{a}+\text{b}+\text{c}$
Substituting a + b + c = d in (i) we get
$\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
$\text{x}+\text{y}+\text{z}=\text{a}+\text{b}+\text{c}$ as the equation of the required plane.

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Question 273 Marks

Find a unit normal vector to the plane x + 2y + 3z - 6 = 0

Answer

The given equation of the plane is
x + 2y + 3z - 6 = 0
x + 2y + 3z = 6
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=6$
Or $\vec{\text{r}}\cdot\vec{\text{n}}=6$
When, $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ ...(\text{i})$
Now, $|\vec{\text{n}}|=\sqrt{1^2+2^2+3^2}$
$=\sqrt{1+4+9}$
$=\sqrt{14}$
Unit vector to the plane, $\hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{14}}$
$=\frac{1}{\sqrt{14}}\hat{\text{i}}+\frac{2}{\sqrt{14}}\hat{\text{j}}+\frac{3}{\sqrt{14}}\hat{\text{k}}$

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Question 283 Marks

Find the vector equation of the plane with intercepts 3, -4 and 2 on x, y and z-axis respectively.

Answer

The equation of the plane in the intercept form is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1,$ where a, b and c are the intercepts on the x, y and z-axis, respectively.
It is given that intercepts made by the plane on the X, Y and Z-axis are 3, -4 and 2, respectively.
$\therefore$ a = 3, b = -4, c = 2
Thus, the equation of the plane is
$\frac{\text{x}}{3}+\frac{\text{y}}{(-4)}+\frac{\text{z}}{2}=1$
$\Rightarrow4\text{x}-3\text{y}+6\text{z}=12$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(4\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}})=12$
This is the vector form of the equation of the given plane.

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Question 293 Marks

Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$2\text{x}-4\text{y}+3\text{z}=5$ and $\text{x}+2\text{y}+\lambda\text{z}=5$

Answer

We know that the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are perperndicular to each other only if
a₁a₂ + b₁b₂ + c₁c₂= 0
The given planes are 2x - 4y + 3z = 5 and $\text{x}+2\text{y}+\lambda\text{z}=5$
⇒ a₁ = 2; b₁ = -4; c₁ = 3; a₂ = 1; b₂ = 2; $\text{c}_2=\lambda$
It is given that the given planes are perpendicular.
⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0
$\Rightarrow(2)(1)+(-4)(2)+(3)+(\lambda)=0$
$\Rightarrow2-8+3\lambda=0$
$\Rightarrow3\lambda=6$
$\Rightarrow\lambda=2$

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Question 303 Marks

Find the direction cosines of the unit vector perpendicular to the plane $\vec{\text{r}}.\big(6\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+1=0$ passing through the origin.

Answer

Given equation of the plane $\vec{\text{r}}.\big(6\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+1=0$
Thus, the direction ratios normal to the plane are 6, -3 and -2
Hence, the direction cosines to the normal to the plane are
$=\frac{6}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-3}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-2}{\sqrt{6^2+(-3)^2+(-2)^2}}$
$=\frac{6}{7},\frac{-3}{7},\frac{-2}{7}$
$=\frac{-6}{7},\frac{3}{7},\frac{2}{7}$
The direction cosines of the unit vector perpendicular to the plane are same as the direction cosines of the normal to the plane.
Thus, the direction cosined of the unit vector perpendicular to the plane are: $\frac{-6}{7},\frac{3}{7},\frac{2}{7}$

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Question 313 Marks

Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.

Answer

The given equation of the plane is
2x − 3y + 4z = 12
Dividing both sides by 12, we get
$\Rightarrow\frac{2\text{x}}{\text{12}}+\frac{-3\text{y}}{\text{12}}+\frac{4\text{z}}{\text{12}}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{\text{6}}+\frac{\text{y}}{-\text{4}}+\frac{\text{z}}{\text{3}}=1\ ....(1)$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(2)$
Comparing (1) and (2), we get
a = 6, b = -4 and c = 3.

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Question 323 Marks

Show that the following planes are at right angles.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=3$

Answer

We know that the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ are perpendicular to each other only if $\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{n}}_2=-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Now,
$\vec{\text{n}}_1\cdot\vec{\text{n}}_2=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)\cdot\big(-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=-2+1+1=0$
So, the given planes are perpendicular.

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Question 333 Marks

Find the equation of the plane passing throught the point (2, 4, 6) and making equal intercepts on the coordinate axes.

Answer

Intercepts on the coordinate axes are equal.
We know that, if a, b, c are Intercepts on coordinate axes by a plane, then equationb of the plane is given by,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
Here, it is given that a = b = c = p (say)
$\frac{\text{x}}{\text{p}}+\frac{\text{y}}{\text{p}}+\frac{\text{z}}{\text{p}}=1$
$\frac{\text{x}+\text{y}+\text{z}}{\text{p}}=1$
$\text{x}+\text{y}+\text{z}=\text{p}\ ...(\text{i})$
It is given that plane is passing through the point (2, 4, 6), so using equation (i)
x + y + z = p
2 + 4 + 6 = p
12 = p
Put, value of p in equation (i)
x + y + z = 12
So, the required equation of the plane is given by,
x + y + z = 12

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Question 343 Marks

Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$

Answer

$\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$
Plane is passing through $(\hat{\text{i}}-\hat{\text{j}})$ and parallel to
$\text{b}(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$ and $\text{c}(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&1\\4&-2&3\end{vmatrix}$
$\text{n}=5\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{r}}\cdot\text{n}=(\hat{\text{i}}-\hat{\text{j}})\cdot(5\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}})$
$=5-1=4$
$\text{r}\cdot(5\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}})=4$

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Question 353 Marks

Write the normal form of the equation of the plane 2x - 3y + 6z + 14 = 0

Answer

The given equation of the plane is,
2x - 3y + 6z + 14 = 0
2x - 3y + 6z + 14 = 0 ...(i)
Now, $\sqrt{2^2+(-3)^2+(6)^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Dividing (i) by 7, we get
$\frac{2}{7}\text{x}-\frac{3}{7}\text{y}+\frac{6}{7}\text{z}=2$
This is the normal form of the given equation of the plane.

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Question 363 Marks

Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=7$ and $\vec{\text{r}}\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=26$

Answer

We know that the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ are perpendicular to each other only if $\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
Here, $\vec{\text{n}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}, \vec{\text{n}}_2=\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}}$
The given planes are perpendicular.
$\Rightarrow\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
$\Rightarrow(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=0$
$\Rightarrow\lambda+4-21=0$
$\Rightarrow\lambda-17=0$
$\Rightarrow\lambda=17$

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Question 373 Marks

Find the equation of rthe planes parallel to the plane x + 2y - 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).

Answer

The equation of the plane parallel to the given plane is
x + 2y - 2z + k = 0 ....(i)
It is given that plane (i) is at a distance of 2 unit from (2, 1, 1).
$\Rightarrow\frac{|2+2-2+\text{k}|}{\sqrt{1^2+2^2+(-2)^2}}=2$
$\Rightarrow\frac{|2+\text{k}|}{3}=2$
$\Rightarrow|2+\text{k}|=6$
$\Rightarrow2+\text{k}=6,{ 2}+\text{k}=-6$
$\Rightarrow\text{k}=4,\text{k}=-8$
Substi9tuting these two values one by one in (i) we get
x + 2y - 2z + 4 = 0 and x + 2y - 2z - 8 = 0, which are the equatioms of the required planes.

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Question 383 Marks

The line $\vec{\text{r}}=\hat{\text{i}}+\lambda(2\hat{\text{i}}-\text{m}\hat{\text{j}}-3\hat{\text{k}})$ is parallel to the plane $\vec{\text{r}}\cdot(\text{m}\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})=4.$ Find m.

Answer

The given line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-\text{m}\hat{\text{j}}-3\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\text{m}\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
If the line is parallel to the plane, the normal to the plane is perpendicular to the line.
$\Rightarrow\vec{\text{b}}\perp\vec{\text{n}}$
$\Rightarrow\vec{\text{b}}\cdot\vec{\text{n}}=0$
$\Rightarrow(2\hat{\text{i}}-\text{m}\hat{\text{j}}-3\hat{\text{k}})\cdot(\text{m}\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})=0$
$\Rightarrow2\text{m}-3\text{m}-3=0$
$\Rightarrow-\text{m}-3=0$
$\Rightarrow\text{m}=-3$

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Question 393 Marks

Write the equation of the plane whose intercepts on the coordinate axes are 2, -3 and 4

Answer

Given, intercepts on the coordinate axes are 2, -3 and 4
We know that,
The equation of a plane whose intercept onb the coordinate axes are a, b and c
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{i})$
Here, a = 2, b = -3, c = 4
So,
Equation of required plane is,
$\frac{\text{x}}{2}+\frac{\text{y}}{-3}+\frac{\text{z}}{4}=1$
$\frac{6\text{x}-4\text{y}+3\text{z}}{12}=1$
${6\text{x}-4\text{y}+3\text{z}=}{12}$

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Question 403 Marks

Find the angle between the plane:
2x - y + z = 4 and x + y + 2z = 3

Answer

We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - y + z = 4 and x + y + 2z = 3 is given by
$\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+1^2}}$
$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}=\frac{3}{\sqrt{6}\sqrt{6}}$
$=\frac{3}{6}=\frac{1}{2}$
$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$

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Question 413 Marks

Show that the following point are coplanar:
(0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1)

Answer

The equation of the plane passing through points (0, 4, 3), (-1, -5, -3), (-2, -2, 1) is,

$\begin{vmatrix}\text{x}-0&\text{y}-4&\text{z}-3\\-1-0&-5-4&-3-3\\-2-0&-2-4&1-3\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}&\text{y}-4&\text{z}-3\\-1&-9&-6\\-2&-6&-2\end{vmatrix}=0$

$\Rightarrow-18\text{x}+10(\text{y}-4)-12(\text{z}-3)=0$

$\Rightarrow9\text{x}-5(\text{y}-4)+6(\text{z}-3)=0$

$\Rightarrow9\text{x}-5\text{y}+\text{z}+2=0$

Substituting the last points (1, 1, -1) (it means x = 1; y = 1; z = -1) in this plane equation, we get

9(1) - 5(1) + 6(-1) + 2 = 0

⇒ 4 - 4 = 0

⇒ 0 = 0

So, the plane equation is satisfied by the points (1, 1, -1)

So, the given pointsa are coplanar.

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Question 423 Marks

Obtain the equation of the plane passing through the point (1, - 3, -2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

Answer

The equation of any plane passing through (1, -3, -2) is,
a(x - 1) + b(y + 3) + c(z + 2) = 0 ....(i)
It is given that (i) is perpendicular to each of the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8. Then,
a + 2b + 2c = 0 ....(ii)
3a + 3b + 2c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-1&\text{y}+3&\text{z}+2\\1&2&2\\3&3&2\end{vmatrix}=0$
⇒ -2(x - 1) + 4(y + 3) - 3(z + 2) = 0
⇒ -2x + 2 + 4y + 12 - 3z - 6 = 0
⇒ 2x - 4y + 3z - 8 = 0

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Question 433 Marks

Find the angle between the plane:
x - y + z = 5 and x + 2y + z = 9

Answer

We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between x - y + z = 5 and x + 2y + z = 9 is given by
$\cos\theta=\frac{(1)(1)+(-1)(2)+(1)(1)}{\sqrt{1^2+(-1)^2+1^2}\sqrt{1^2+2^2+1^2}}$
$=\frac{1-2+1}{\sqrt{1+1+1}\sqrt{1+4+1}}$
$=\frac{0}{\sqrt{3}\sqrt{6}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$

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Question 443 Marks

Find the angle between the line $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{-1}=\frac{\text{z}-3}{2}$ and the plane 3x + 4y + z + 5 = 0

Answer

The given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and the given plane id normal to the vector $\vec{\text{n}}=3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}|\vec{|\text{n}}|}$
$=\frac{(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})\cdot(3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}})}{|3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}||3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{9-4+2}{\sqrt{9+1+4}\sqrt{9+16+1}}=\frac{7}{\sqrt{14}\sqrt{26}}$
$=\frac{7}{\sqrt{2}\sqrt{7}\sqrt{2}\sqrt{13}}=\frac{\sqrt{7}}{\sqrt{52}}=\sqrt{\frac{7}{52}}$
$\theta=\sin^{-1}\Big(\sqrt{\frac{7}{52}}\Big)$

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Question 453 Marks

Find the equation of the plane passing through the following point:
(-5, 0, -6), (-3, 10, -9) and (-2, 6, -6)

Answer

The equation of the plane passing through points (-5, 0, -6), (-3, 10, -9) and (-2, 6, -6) is given by,
$\begin{vmatrix}\text{x}+5&\text{y}-0&\text{z}+6\\-3+5&10-0&-9+6\\-2+5&6-0&-6+6\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}+5&\text{y}&\text{z}+6\\2&10&-3\\3&6&0\end{vmatrix}=0$
$\Rightarrow18(\text{x}+5)-\text{y}-18(\text{z}+6)=0$
$\Rightarrow2(\text{x}+5)-\text{y}-2(\text{z}+6)=0$
$\Rightarrow2\text{x}+10-\text{y}-2\text{z}-12=0$
$\Rightarrow2\text{x}-\text{y}-2\text{z}-2=0$

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Question 463 Marks

Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})=9$

Answer

We know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{n}}_2=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6{\hat{\text{k}}\big|}\big|\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{2-16-12}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{-16}{(7)(3)}$
$=\frac{-16}{21}$
$\theta=\cos^{-1}\Big(\frac{-16}{21}\Big)$

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Question 473 Marks

Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}.$ Hence, of otherwise, deduce the length of the perpendicular.

Answer

Let M be the foot of the perpendicular of the point P(5, 4, 2) on the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}$
Therefore, its equation is
$\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}=\text{r}$
Then, M is in the form (2r - 1, 3r + 3, -r + 1)
Direction ratios of MP are 2r - 1 - 5, 3r + 3 - 4, -r + 1 - 2 or 2r - 6, 3r - 1, -r - 1.
Since MP is perpendicular to the given line (2, 3, -1), 2 (2r - 6) + 3(3r - 1) -1(-r - 1) = 0 (Because a₁, a₂, + b₁, b₂, + c₁, c₂, = 0)
⇒ 4r - 12 + 9r - 3 + 1 + 1 = 0
⇒ 14r - 14 = 0
⇒ r = 1
So, M = (2r - 1, 3r + 3, -r + 1) = (2 (1) - 1, 3(1) + 3, -1 + 1) = (1, 6, 0)
Length of the perperndicular,
$\text{MP}=\sqrt{(1-5)^2+(6-4)^2+(0-2)^2}\\=\sqrt{16+4+4}=\sqrt{24}=2\sqrt{6}\text{ units}$

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Question 483 Marks

Find the angle between the plane:
2x - 3y + 4z = 1 and -x + y = 4

Answer

We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - 3y + 4z = 1 and -x + y + 0z = 4 is given by
$\cos\theta=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^2+(-3)^2+4^2}\sqrt{(-1)^2+1^2+0^2}}$
$=\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$

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Question 493 Marks

Show that the following point are coplanar:
(0, -1, 0), (2, 1, -1), (1, 1, 1) and (3, 3, 0)

Answer

The equation of the plane passing through points (0, -1, 0), (2, 1, -1) and (1, 1, 1) is given by,

$\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\2-0&1+1&-1-0\\1-0&1+1&1-0\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}&\text{y}+1&\text{z}\\2&2&-1\\1&2&1\end{vmatrix}=0$

$\Rightarrow4\text{x}-3(\text{y}+1)+2\text{z}=0$

$\Rightarrow4\text{x}-3\text{y}+2\text{z}-3=0$

Substituting the last points (3, 3, 0) (it means x = 3; y = 3; z = 0) in this plane equation, we get

4(3) - 3(3) + 2(0) - 3 = 0

⇒ 12 - 12 = 0

⇒ 0 = 0

So, the plane equation is satisfied by the points (3, 3, 0)

So, the given pointsa are coplanar.

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Question 503 Marks

Show that the following planes are at right angles.
x - 2y + 4z = 10 and 18x + 17y + 4z = 49

Answer

We know that the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are perperndicular to each other only if
a₁a₂ + b₁b₂ + c₁c₂= 0
The given planes are x - 2y + 4z = 10 and 18x + 17y + 4z = 49
⇒ a₁ = 1; b₁ = -2; c₁ = 4; a₂ = 18; b₂ = 17; c₂ = 4
Now,
a₁a₂ + b₁b₂ + c₁c₂
= (1)(18) + (-2)(17) + (4)(4)
= 18 - 34 + 16
= 0
So, the given planes are perpendicular.

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3 Marks - Maths STD 12 Science Questions - Vidyadip