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

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3 Marks

Question 1013 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 1023 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 1033 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 1043 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 - y + 3z - 1 = 0 and 2x - y + 3z + 3 = 0

Answer

The direction ratios of normal to the plane, L₁: a₁x + b₁y + c₁z = 0,
are a₁, b₁, c₁ and L₂: a₂x + b₂y + c₂z = 0 are a₂, b₂, c₂
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between L₁ and L₂ is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the planes are 2x - y + 3z - 1 = 0 and 2x - y + 3z + 3 = 0
Here, a₁ = 2, b₁ = -1, c₁ = 3 and a₂ = 2, b₂ = -1, c₂ = 3
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{2}=1,\ \frac{\text{b}_1}{\text{b}_2}=\frac{-1}{-1}=1\text{ and } \frac{\text{c}_1}{\text{c}_2}=\frac{3}{3}=1$
$\therefore\ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Thus, the given lines are parallel to each other.

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Question 1053 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 1063 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 1073 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 1083 Marks

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

Answer

The equation of any plane through the intersection of the planes,
3x - y + 2z - 4 = 0 and x + y + z - 2 = 0, is
$(3\text{x}-\text{y}+2\text{z}-4)+\alpha(\text{x}+\text{y}+\text{z}-2)=0,\text{ where }\alpha\in\text{R}\ \ ....(1)$
The plane passes through the point (2, 2, 1).
Therefore, this point will satisfy equation (1)
$\therefore\ (3\times2-2+2\times1-4)+\alpha(2+2+1-2)=0$
$\Rightarrow\ 2+3\alpha=0$
$\Rightarrow\ \alpha=-\frac{2}{3}$
Substituting $\alpha=-\frac{2}{3}$ in equation (1), we obtain
$(3\text{x}-\text{y}+2\text{z}-4)-\frac{2}{3}(\text{x}+\text{y}+\text{z}-2)=0,$
⇒ 3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0
⇒ (9x - 3y + 6z - 12) - 2(x + y + z - 2) = 0
⇒ 7x - 5y + 4z - 8 = 0
This is the required equation of the plane.

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Question 1093 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 1103 Marks

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

Answer

Direction ratios of the line joining the points A(3, -4, -5) and B(2, -3, 1) are 2 - 3, -3 - (-4), 1 - (-5) ⇒ -1, 1, 6
$\therefore$ Equation of the line AB are
$\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}\ \ \ .....(\text{i})$
Equation of the plane is 2x + y + z = 7 ...(ii)
Now to find the point where line (i) crosses plane (ii),
From eq. (i) $\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}=\lambda\ (\text{say})$
$\Rightarrow\ \ \text{x}-3=-\lambda,\ \text{y}+4=\lambda,\ \text{z}+5=6\lambda$
$\Rightarrow\ \ \text{x}=3-\lambda,\ \text{y}=-4+\lambda,\ \text{z}=-5+6\lambda\ \ \ ....(\text{iii})$
Putting the values of x, y, z in eq. (ii), we get
$2(3-\lambda)+(-4+\lambda)+(-5+6\lambda)=7$
$\Rightarrow\ \ 6-2\lambda-4+\lambda-5+6\lambda=7\ \ \Rightarrow\ 5\lambda=10\ \ \Rightarrow\ \lambda=2$
Putting $\lambda=2$ in eq. (iii), point of intersection of line (i) and plane (ii) is
x = 3 - 2 = 1, y = -4 + 2 = -2, z = -5 + 12 = 7
Thus, required point of intersection is (1, -2, 7).

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Question 1113 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 1123 Marks

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ - m₂n₁, l₁n₂ - l₂n₁, l₁m₂ - l₂m₁.

Answer

l₁, m₁, n₁ and l₂, m₂, n₂ are direction cosines of two mutually perpendicular of two given lines L₁ and L₂.(say)
Let $\hat{\text{n}}_1\ \text{and}\ \hat{\text{n}}_2$ be the unit vectors along these lines L₁ and L₂.
$\therefore\ \ \vec{\text{n}}_1=\text{l}_1\hat{\text{i}}+\text{m}_1\hat{\text{j}}+\text{n}_1\hat{\text{k}}\ \text{and}\ \vec{\text{n}}_2=\text{l}_2\hat{\text{i}}+\text{m}_2\hat{\text{j}}+\text{n}_2\hat{\text{k}}$
Let L be the line perpendicular to both the lines L₁ and L₂ and let $\hat{\text{n}}$ be a unit vector along line L perpendicular both lines L₁ and L₂.
$\therefore$ Cross product of two vectors
$=\hat{\text{n}}_1\times\hat{\text{n}}_2=\Big|\hat{\text{n}}_1\Big|.\Big|\hat{\text{n}}_2\Big|\sin90^{\circ}\hat{\text{n}}$ $\Big[\because\ \text{L}_1\perp\text{L}_2\ (\text{given},\ \therefore\ \text{angle between them is }90^{\circ}\Big]$
$\Rightarrow\ \ \hat{\text{n}}_1\times\hat{\text{n}}_2=\hat{\text{n}}$
$ \Rightarrow\ \hat{\text{n}}=\hat{\text{n}}_1\times\hat{\text{n}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\end{vmatrix}$
$\Rightarrow\ \ \hat{\text{n}}=(\text{m}_1\text{n}_2-\text{m}_2\text{n}_1)\hat{\text{i}}-(\text{l}_1\text{n}_2-\text{l}_2\text{n}_1)\hat{\text{j}}+(\text{l}_1\text{m}_2-\text{l}_2\text{m}_1)\hat{\text{k}}$
Since, $\hat{\text{n}}$ is a unit vector, therefore its components are its direction cosines.
Thus, direction cosines of $\hat{\text{n}}$ are m₁n₂ - m₂n₁, l₁n₂ - l₂n₁, l₁m₂ - l₂m₁
⇒ direction cosines of line L are m₁n₂ - m₂n₁, l₁n₂ - l₂n₁, l₁m₂ - l₂m₁.

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Question 1133 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 1143 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 1153 Marks

Find the vector equation of the line which is parallel to the vector $3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and which passes throught the point (1, -2, 3).

Answer

Let $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$
So, vector equation of the line, which is parallel to the vector $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and passes through the vector $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$ is $\vec{\text{r}}=\vec{\text{b}}+\lambda\vec{\text{a}}.$
$\therefore\vec{\text{r}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}+\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}{\hat{\text{j}}}+\text{z}\hat{\text{k}})-(\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}})=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}-1)\hat{\text{i}}+(\text{y}+2)\hat{\text{j}}+(\text{z}-3)\hat{\text{k}}=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$

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Question 1163 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 1173 Marks

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1) and (4, 3, -1).

Answer

Here,
A(0, 0, 0) and B(2, 1, 1)
C(3, 5, -1) and D(4, 3, -1)
Direction ratios of line AB
a₁ = 2, b₁ = 1, c₁= 1
Direction ratios of line CD
a₂ = 2, b₂ = -2, c₂= 0
Now,
a₁a₂+ b₁b₂+ c₁c₂
= (2)(1) + (1)(-2) + (1)(0)
= 2 - 2 + 0
= 0
Since, a₁a₂+ b₁b₂+ c₁c₂= 0, lines are perpendicular.

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Question 1183 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 1193 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 1203 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 1213 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 1223 Marks

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.

Answer

The given equation are
2x − y + 3z = 4 ....(1)
The second equation of the plane is
2x − y + 3z = 18 .....(2)
We know that distance between two planes ax + by + cz = d₁ and ax + by + cz = d₂ is $\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$

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Question 1233 Marks

The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle $\alpha.$ Prove that the equation of the plane in its new position is $\text{ax}+\text{by}\pm\bigg(\sqrt{\text{a}^2+\text{b}^2}\tan\alpha\bigg)\text{z}=0.$

Answer

Equation of the plane is ax + by = 0 .....(i)
$\therefore$ Equation of the plane after new position is
$\frac{\text{ax}\cos\alpha}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{\text{by}\cos\alpha}{\sqrt{\text{b}^2+\text{a}^2}}\pm\text{z}\sin\alpha=0$
Dividing both sides of above equation by cos $\alpha,$ we get
$\Rightarrow\frac{\text{ax}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{\text{by}}{\sqrt{\text{b}^2+\text{c}^2}}+\text{z}\tan\alpha=0$
$\Rightarrow\text{ax}+\text{by}\pm\text{z}\tan\alpha\sqrt{\text{a}^2+\text{b}^2}=0$ $\big($On multiplying with $\sqrt{\text{a}^2+\text{b}^2}\big)$
Hence proved.

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Question 1243 Marks

Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point $(\alpha,\beta,\gamma)$

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, c) and C(0, 0, c)
Given that the centroid of the triangle $=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\frac{\text{a}}{3}=\alpha,\frac{\text{b}}{3}=\beta,\frac{\text{c}}{3}=\gamma$
$\Rightarrow\text{a}=3\alpha,\text{b}=3\beta,\text{c}=3\gamma\ ...(\text{i})$
The equation of the plane whose intercepts on the coordinate axes are a, b and c are
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1$ [From (i)]
$\Rightarrow\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=3$

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Question 1253 Marks

Find the vector and the cartesian equations of the line that passes through the points (3, -2, -5), (3, -2, 6).

Answer

Let $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be the position vectors of the points A(3, -2, -5) and B(3, -2, 6) respectively.
$\therefore\ \ \vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\ \text{and}\ \vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$
$\therefore$ A vector along the line $=\overrightarrow{\text{AB}}=$ Position vector of point B - Position vector of point A
$\Rightarrow\ \ \overline{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}-3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$\because$ Vector equation of the line is $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$
$\therefore\ \ \vec{\text{r}}=3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}+\lambda\Big(11\hat{\text{k}}\Big)$
And another vector equation for the same line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{AB}}= \vec{\text{r}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}+\lambda\Big(11\hat{\text{k}}\Big)$
Cartesian equation
Direction ratios of line AB are 3 - 3, -2 + 2, 6 + 5 = 0, 0, 11
$\therefore$ Equation of the line is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}=\frac{\text{x}-3}{0}=\frac{\text{}\text{y}+2}{0}=\frac{\text{z}+5}{11}$

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Question 1263 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 1273 Marks

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

Answer

Equation of any plane parallel to the plane $\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=2$ is

$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=\lambda\ \ ....(\text{i})$

Plane (i) passes through (a, b, c)

$\therefore$ Putting $\vec{\text{r}}=(\text{a},\ \text{b},\ \text{c})=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ in eq. (i), we get

$\Big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\Big).\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=\lambda$

$\Rightarrow\ \ \text{a}(1)+\text{b}(1)+\text{c}(1)=\lambda$

$\Rightarrow\ \ \lambda=\text{a}+\text{b}+\text{c}$

Putting the value of $\lambda$ in eq. (i), to get the required plane is

$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=\text{a}+\text{b}+\text{c}.$

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Question 1283 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 1293 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 1303 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 1313 Marks

If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.

Answer

Given: Origin O(0, 0, 0) and point P(1, 2, -3)
To find: Equation of the plane passing through P(1, 2, - 3) = (x₁, y₁, z₁)
$\therefore$ Direction ratios of normal OP to the plane are 1 - 0, 2 - 0, -3 - 0 ⇒ 1, 2, -3 = (a, b, c)
$\therefore$ Equation of the required plane is a(x - x₁) + b(y - y₁) + c(z - z₁) = 0
⇒ 1(x - 1) + 2(y - 2) - 3(z + 3) = 0
⇒ x - 1 + 2y - 4 - 3z - 9 = 0
⇒ x + 2y - 3z - 14 = 0.

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Question 1323 Marks

Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5).

Answer

We know that direction ratios of the line joining the points A(4, 7, 8) and B(2, 3, 4) are
x₂ - x₁, y₂ - y₁, z₂ - z₁
⇒ 2 - 4, 3 - 7, 4 - 8
⇒ -2, -4, -4 = a₁, b₁, c_{1 (say)}
Again, direction ratios of the line joining the points C(-1, -2, 1) and D(1, 2, 5) are
x₂ - x₁, y₂ - y₁, z₂ - z₁
⇒ 1 - (-1), 2 - (-2), 5 - 1
⇒ 2, 4, 4 = a₂, b₂, c₂ (say)
For lines AB and CD,
$\frac{\text{a}_1}{\text{a}_2}=\frac{-2}{2},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-4}{4},\ \frac{\text{c}_1}{\text{c}_2}=\frac{-4}{4}=-1$
$\therefore\ \ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Therefore, line AB is parallel to line CD.

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Question 1333 Marks

Find the equation of a line parallel to x-axis and passing through the origin.

Answer

We know that a unit vector along x-axis is $\hat{\text{i}}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\therefore$ Direction cosines of x-axis are coefficient of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ in the unit vector i.e., 1, 0, 0 = l, m, n
$\therefore$ Equation of the required line passing through the origin (0, 0, 0) and parallel to x-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}\ \Rightarrow\ \frac{\text{x}}{1}=\frac{\text{y}}{0}=\frac{\text{z}}{0}$
Vector equation of the required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\ \ \vec{\text{r}}=\vec{0}+\lambda\hat{\text{i}}\ \ \ \ \ [\vec{\text{a}}=\vec{0}\ \text{and}\ \vec{\text{b}}=\hat{\text{i}}]$
$\Rightarrow\ \ \ \vec{\text{r}}=\lambda\hat{\text{i}}$

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Question 1343 Marks

Find the coordinates of the foot of the perpendicular drawn from the origin.
3y + 4z - 6 = 0

Answer

Let the coordinates of the foot of perpendicular P from the origin to the plane b(x₁, y₁, z₁).
3y + 4z - 6 = 0
⇒ 0x + 3y + 4z = 6 ....(1)
The direction ratios of the normal are 0, 3, and 4.
$\therefore\ \ \sqrt{0+3^2+4^2}=5$
Dividing both sides of equation (1) by 5, we obtain
$0\text{x}+\frac{3}{5}\text{y}+\frac{4}{5}\text{z}=\frac{6}{5}$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the perpendicular are
$\Big(0,\ \frac{3}{5}.\frac{6}{5},\ \frac{4}{5}.\frac{6}{5}\Big)\ \text{i.e.},\ \Big(0,\ \frac{18}{25},\ \frac{24}{25}\Big).$

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Question 1353 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 1363 Marks

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\vec{\text{r}}.\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=6.$

Answer

The required line passes through the point $\text{A}(1, 2, 3) =\vec{\text{a}}$
$\therefore\ \vec{\text{a}}=\text{Position vector of point A}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Let $\vec{\text{b}}$ be any vector along the required line.
$\therefore$ Vector equation of required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\ \vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\vec{\text{b}}\ \ \ ...(\text{i})$
Since required line is parallel to the plane $\vec{\text{r}}.\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)=5$
$\therefore\ \vec{\text{b}}.\vec{\text{n}_1}=0\ \text{and}\ \vec{\text{b}}.\vec{\text{n}_2}=0$
Comparing with $\vec{\text{r}}.\vec{\text{n}_1}=\text{d}_1$ we have, $\vec{\text{n}_1}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
And Comparing with $\vec{\text{r}}.\vec{\text{n}_2}=\text{d}_2$ we have, $\vec{\text{n}_2}=3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Since $\vec{\text{b}}$ isperpendicular to both $\vec{\text{n}_1}\ \text{and}\ \vec{\text{n}_2}$
$\therefore\ \vec{\text{b}}=\vec{\text{n}_1}\times\vec{\text{n}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&2\\3&1&1\end{vmatrix}$
Expanding along first row,
$\vec{\text{b}}=\hat{\text{i}}(-1-2)-\hat{\text{j}}(1-6)+\hat{\text{k}}(1+3)=-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
Putting this value of $\vec{\text{b}}$ in eq. (i), vector equation of required line,
$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\Big(-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\Big).$

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Question 1373 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 1383 Marks

Show that the points $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ and $3(\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are equidistant from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$ and lies on opposite side of it.

Answer

To show that these given points $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ and $3(\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are equidistant from the plane
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0,$ we first find out the mid-point of the points which is $(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}).$
On substituting $\vec{\text{r}}$ by the mid-point in plane, we get
$\text{L.H.S.}=(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+\hat{\text{j}}-7\hat{\text{k}})+9$
$=10+2-21+9=0$
$=\text{R.H.S.}$
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.

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Question 1393 Marks

If a line has the direction ratios -18, 12, -4, then what are its direction cosines?

Answer

We know that if a, b, c are direction ratios of a line, then direction cosines of the line are:
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\ \ \ ....(\text{i})$
Here direction ratios of the line are -18, 12, -4
Putting the values in eq. (i),
$\frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}},\ \frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}},\ \frac{\text{-4}}{\sqrt{(-18)^2+(12)^2+(-4)^2}}$
$\Rightarrow\ \frac{-18}{\sqrt{324+144+16}},\ \frac{12}{\sqrt{324+144+16}},\ \frac{\text{-4}}{\sqrt{324+144+16}}$
$\Rightarrow\ \frac{-18}{\sqrt{484}},\ \frac{12}{\sqrt{484}},\ \frac{\text{-4}}{\sqrt{484}}\ \Rightarrow\ \frac{-18}{22},\ \frac{12}{22},\ \frac{-4}{22}$
$\Rightarrow\ \frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11}$
Hence, direction cosines of required lines are $\frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11}.$

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Question 1403 Marks

A line makes an angle of 60° with each of X-axis and Y-axis. Find the acute angle made by the line with Z-axis.

Answer

It is given that a line makes an angle of 60° with both x-axis and y-axis.
Suppose the line makes an angle of $\alpha$ with the z-axis.
$\Rightarrow\text{l}=\cos60^\circ=\frac{1}{2}\text{m}$
$=\cos60^\circ=\frac{1}{2}\text{n}=\cos\alpha$
We know $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\Big(\frac{1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2+(\cos\alpha)^2=1$
$\Rightarrow\frac{1}{4}+\frac{1}{4}+\cos ^2\alpha=1$
$\Rightarrow\cos\alpha=\frac{1}{\sqrt{2}}$
$\Rightarrow\alpha=45^\circ$
Thus, the line makes an angle of 45° with the z-axis.

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Question 1413 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 1423 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 1433 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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