Solve the Following Question.(2 Marks)

Question 12 Marks

Find the distance of the point (13, 13, -13) from the plane 3x + 4y – 12z = 0.

Answer

The distance of the point $\left(x_1, y_1, z_1\right)$ from the plane $a x+b y+c z+d=0$ is $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

$\therefore$ the distance of the point $(1,1,-1)$ from the plane $3 x+4 y-12 z=0$ is $\frac{3(1)+4(1-12(-1))}{\sqrt{3^2+4^2+(-12)^2}}$

$=\left|\frac{3+4+12}{\sqrt{9+16+144}}\right|$

$=\frac{19}{\sqrt{169}}$

$=\frac{19}{13}$

$=19$ units.

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Question 22 Marks

Find the distance of the point $3 \hat{i}+3 \hat{j}+\hat{k}$ from the plane $\bar{r} \cdot(2 \hat{i}+3 \hat{j}+6 \hat{k})=21$

Answer

The distance of the point A( $\bar{a})$ from the plane $\bar{r} \cdot \bar{n}=p$ is given by $d=\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|} \ldots \ldots(1)$

Here, $\bar{a}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{k}, \bar{n}=2 \hat{i}+3 \hat{j}+6 \hat{k}, p=21$

$\therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$

= (3)(2) + (3)(3) + (1)(-6) = 6 + 9 – 6 = 9

Also, $|\bar{n}|=\sqrt{3^2+3^2+(-6)^2}=\sqrt{-12}=0$

$\therefore$ from (1), the required distance

$=\frac{|-12-21|}{12}$

$=0$ units.

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Question 32 Marks

The foot of the perpendicular drawn from the origin to a plane is M(1, 2,0). Find the vector equation of the plane.

Answer

The vector equation of the plane passing through $\mathrm{A}(\bar{a})$ and perpendicular to $\bar{n}$ is

$\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$

M(1, 2, 0) is the foot of the perpendicular drawn from origin to the plane. Then the plane is passing through M and is perpendicular to OM.

If $\bar{m}$ is the position vector of $M_t$ then $\bar{m}=\hat{i}$.

Normal to the plane is

$\begin{aligned} & \overline{\mathrm{n}}=\overline{\mathrm{OM}}=\hat{\mathrm{i}} \\ & \overline{\mathrm{m}} \cdot \overline{\mathrm{n}}=\hat{\mathrm{i}}, \hat{i}=5\end{aligned}$

∴ the vector equation of the required plane is

$\bar{r} \cdot(\hat{i}+2 \hat{j})=5$

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Question 42 Marks

Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the

plane $\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0$

Answer

The cartesian equation of the plane $\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0$ is $6 x+8 y+7 z=0$ The

required plane is parallel to it ∴ its cartesian equation is 6x + 8y + 7z = p …(1) A (7, 8, 6) lies on it and hence satisfies its equation ∴ (6)(7) + (8)(8) + (7)(6) = p i.e., p = 42 + 64 + 42 = 148. ∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

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Question 52 Marks

Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0.

Answer

The Cartesian equation of the plane passing through (x1, y1, z1), the direction ratios of whose normal are a, b, c, is a(x – x1) + b(y – y1) + c(z – z1) = 0 ∴ the cartesian equation of the required plane is o(x + 1) + 2(y + 2) + 5(z – 3) = 0 i.e. 0 + 2y – 4 + 10z – 15 = 0 i.e. y + 2 = 0.

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Question 62 Marks

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.

Answer

The equation of the plane is 2x + 3y + 6z = 49 Dividing each term by

$\begin{aligned} & \sqrt{2^2+3^2+(-6)^2} \\ & =\sqrt{49} \\ & =7\end{aligned}$

we get

$\frac{2}{7} x+\frac{3}{7} y-\frac{6}{7} z=\frac{49}{7}=7$

This is the normal form of the equation of plane. ∴ the direction cosines of the perpendicular drawn from the origin to the plane are

$I=\frac{2}{7}, m=\frac{3}{7}, n=\frac{6}{7}$

and length of perpendicular from origin to the plane is p = 7. the coordinates of the foot of the perpendicular from the origin to the plane are (lp, ∓, np)i.e.(2, 3, 6)

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Question 72 Marks

Find the perpendicular distance of the origin from the plane 6x + 2y + 3z – 7 = 0

Answer

The distance of the point $\left(x_1, y_1, z_1\right)$ from the plane $a x+b y+c z+d$ is $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

∴ the distance of the point (1, 1, -1) from the plane 6x + 2y + 3z – 7 = 0 is

$\left|\frac{6(1)+2(1-3(-1)+7)}{\sqrt{3^2+4^2+(-12)^2}}\right|$

$=\left|\frac{6+4+6+7}{\sqrt{9+16+144}}\right|$

$=\frac{23}{\sqrt{169}}$

$=\frac{23}{13}$

= 1units.

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Question 82 Marks

Find the direction cosines of the line $\bar{r}=\left(-2 \hat{i}+\frac{5}{2} \hat{j}-\hat{k}\right)+\lambda(2 \hat{i}+3 \hat{j})$.

Answer

The line $\bar{r}=\left(-2 \hat{i}+\frac{5}{2} \hat{j}-\hat{k}\right)+\lambda(2 \hat{i}+3 \hat{j})$ is

parallel to $\bar{b}=2 \hat{i}+3 \hat{j}$.

$\therefore$ direction ratios of the line are $2,3,0$.

$\therefore$ direction cosines of the line are

$\frac{2}{\sqrt{2^2+3^2+0}}, \frac{3}{\sqrt{2^2+3^2+0}}, 0$

i.e. $\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0$.

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Question 92 Marks

Find the vector equation of the line which passes through the origin and the point (5, -2, 3).

Answer

Let $\bar{b}$ be the position vector of the point $B(5,-2,3)$.

Then $\bar{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}$

Origin has position vector $\overline{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}$.

The vector equation the line passing through $A(\bar{a})$ and $B(\bar{b})$ is $\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})$ where $\lambda$

is a scalar.

$\therefore$ the vector equation of the required line is $\bar{r}=\overline{0}+\lambda(\bar{b}-\overline{0})=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})$

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Question 102 Marks

Find the angle between the line $\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=8$

Answer

The angle between the line $\bar{r}=\bar{a}+\lambda \bar{b}$ and the plane $\bar{r} \cdot \bar{n}=d$ is given by $\sin \theta=\left|\frac{\bar{b} \cdot \bar{n}}{|\bar{b}| \cdot|\bar{n}|}\right|$ Here $\bar{b}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{n}=2 \hat{i}-\hat{j}+\hat{k}$
$
\begin{aligned}
& \therefore \bar{b} \cdot \bar{n}=(\hat{i}+\hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})=2-1+1=2 \\
& |\bar{b}|=\sqrt{1+1+1}=\sqrt{3} \text { and }|\bar{n}|=\sqrt{4+1+1}=\sqrt{6} \\
& \therefore \sin \theta=\left|\frac{\bar{b} \cdot \bar{n}}{|\bar{b}| \cdot|\bar{n}|}\right|=\frac{2}{\sqrt{3} \cdot \sqrt{6}}=\frac{\sqrt{2}}{3} \\
& \therefore \theta=\sin ^{-1}\left(\frac{\sqrt{2}}{3}\right)
\end{aligned}
$

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Question 112 Marks

Find the angle between planes $\bar{r} \cdot(\hat{i}+\hat{j}-2 \hat{k})=8$ and $\bar{r} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=3$

Answer

Normal to the given planes are $\bar{n}_1=\hat{i}+\hat{j}-2 \hat{k}$ and $\bar{n}_2=-2 \hat{i}+\hat{j}+\hat{k}$
The acute angle $\theta$ between normal is given by
$
\begin{aligned}
& \cos \theta=\left|\frac{\bar{n}_1 \cdot \bar{n}_2}{\left|\bar{n}_1\right| \cdot\left|\bar{n}_2\right|}\right| \\
& \therefore \cos \theta=\left|\frac{\mid(\hat{i}+\hat{j}-2 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{6} \cdot \sqrt{6}}\right|=\left|\frac{-3}{6}\right|=\frac{1}{2} \\
& \therefore \cos \theta=\frac{1}{2} \quad \therefore \theta=60^{\circ}=\frac{\pi}{3}
\end{aligned}
$
The acute angle between normals $\bar{n}_1$ and $\bar{n}_2$ is $60^{\circ}$.
$\therefore \quad$ The angle between given planes is $60^{\circ}$.

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Question 122 Marks

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane
$
2 x+y-2 z=18
$

Answer

The Direction ratios of the normal to the plane $2 x+y-2 z=18$ are $2,1,-2$.
$\therefore \quad$ Direction cosines are $\frac{2}{3}, \frac{1}{3},-\frac{2}{3}$
The normal form of the given Cartesian equation is $\frac{2}{3} x+\frac{1}{3} y-\frac{2}{3} z=6$
$
\therefore \quad p=6
$
The coordinates of the foot of the perpendicular are $(l p, m p, n p)=\left(6\left(\frac{2}{3}\right), 6\left(\frac{1}{3}\right), 6\left(-\frac{2}{3}\right)\right) \equiv(4,2,-4)$

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Question 132 Marks

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

Answer

First we write the given Cartesian equation in normal form.
i.e. in the form $l x+m y+n z=p$
Direction ratios of the normal are $1,-3,4$.
$\therefore \quad$ Direction cosines are $\frac{1}{\sqrt{26}}, \frac{-3}{\sqrt{26}}, \frac{4}{\sqrt{26}}$
Given equation can be written as $\frac{1}{\sqrt{26}} x-\frac{-3}{\sqrt{26}} y+\frac{4}{\sqrt{26}} z=\frac{6}{\sqrt{26}}$
$\therefore \quad$ The distance of the origin from the plane is $\frac{6}{\sqrt{26}}$

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Question 142 Marks

Find the vector equation of the plane which is at a distance of 6 unit from the origin and to which the vector $2 \hat{i}-\hat{j}+2 \hat{k}$ is normal.

Answer

Here $p=6$ and $\bar{n}=2 \hat{i}-\hat{j}+2 \hat{k}$
$
\therefore|\bar{n}|=3
$
$
\therefore \hat{n}=\frac{\bar{n}}{|\bar{n}|}=\frac{2 \hat{i}-\hat{j}+2 \hat{k}}{3}
$
The required equation is $\therefore \bar{r} \cdot \hat{n}=p$
$
\begin{aligned}
& \therefore \bar{r} \cdot\left(\frac{2 \hat{i}-\hat{j}+2 \hat{k}}{3}\right)=6 \\
& \therefore \bar{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})=18
\end{aligned}
$

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Question 152 Marks

Find the vector equation of the plane passing through the point $\mathrm{A}(-1,2,-5)$ an parallel to vectors $4 \hat{i}-\hat{j}+3 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$.

Answer

The vector equation of the plane passing through point $\mathrm{A}(\bar{a})$ and parallel to $\bar{b}$ and $\bar{c}$ is $\bar{r} \cdot(\bar{b} \times \bar{c})=\bar{a} \cdot(\bar{b} \times \bar{c})$
Here $\bar{a}=-\hat{i}+\widehat{2 j}-5 \hat{k}, \bar{b}=4 \hat{i}-\hat{j}+3 \hat{k}, \bar{c}=\hat{i}+\hat{j}-\hat{k}$
$
\begin{aligned}
& \bar{b} \times \bar{c}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
4 & -1 & 3 \\
1 & 1 & -1
\end{array}\right|=-2 \hat{i}+7 \hat{j}+5 \hat{k} \\
& \bar{a} \cdot(\bar{b} \times \bar{c})=(-\hat{i}+2 \hat{j}-5 \hat{k}) \cdot(-2 \hat{i}+7 \hat{j}+5 \hat{k})=-9
\end{aligned}
$
The required equation is $\bar{r} \cdot(-2 \hat{i}+7 \hat{j}+5 \hat{k})=-9$

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Question 162 Marks

The foot of the perpendicular drawn from the origin to a plane is $\mathrm{M}(2,1,-2)$. Find vector equation of the plane.

Answer

OM is normal to the plane.
$\therefore$ The direction ratios of the normal are $2,1,-2$.
The plane passes through the point having position vector $2 \hat{i}+\hat{j}-2 \hat{k}$ and vector $\overline{\mathrm{OM}}=2 \hat{i}+\hat{j}-2 \hat{k}$ is normal to it.
Its vector equation is $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$.
$\begin{aligned} & \bar{r} \cdot(2 \hat{i}+\hat{j}-2 \hat{k})=(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(2 \hat{i}+\hat{j}-2 \hat{k}) \\ & \bar{r} \cdot(2 \hat{i}+\hat{j}-2 \hat{k})=9\end{aligned}$

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Question 172 Marks

$\text{Show that lines :}\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k}) \text { and } \bar{r}=(4 \hat{i}-3 \hat{j}+2 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})$ intersect each other.

Answer

Lines $\bar{r}=\bar{a}_1+\lambda_1 \bar{b}_1$ and $\bar{r}=\bar{a}_2+\lambda_2 \bar{b}_2$ intersect each other if and only if
$
\left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=0
$
Here $\bar{a}_1=\hat{i}+\hat{j}-\hat{k}, \bar{a}_2=4 \hat{i}-3 \hat{j}+2 \hat{k}, \quad \bar{b}_1=2 \hat{i}-2 \hat{j}+\hat{k}, \bar{b}_2=\hat{i}-2 \hat{j}+2 \hat{k}$,
$
\begin{aligned}
& \bar{a}_2-\bar{a}_1=3 \hat{i}-4 \hat{j}+3 \hat{k} \\
& \left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\left|\begin{array}{lll}
3 & -4 & 3 \\
2 & -2 & 1 \\
1 & -2 & 2
\end{array}\right|=3(-2)+4(3)+3(-2)=-6+12-6=0
\end{aligned}
$
Given lines intersect each other.

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Question 182 Marks

Find the distance of the point $\mathrm{P}(0,2,3)$ from the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.

Answer

Let $\mathrm{M}$ be the foot of the perpendicular drawn from the point $\mathrm{P}(0,2,3)$ to the given line.
$\mathrm{M}$ lies on the line. Let co-ordinates of $\mathrm{M}$ be $(5 \lambda-3,2 \lambda+1,3 \lambda-4)$.
The direction ratios of PM are $(5 \lambda-3)-0,(2 \lambda+1)-2,(3 \lambda-4)-3$ i.e. $5 \lambda-3,2 \lambda-1,3 \lambda-7$
The direction ratios of given line are 5, 2, 3 and PM is perpendicular to the given line.
$
\begin{aligned}
& \therefore \quad 5(5 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-7)=0 \\
& \therefore \quad \lambda=1 \\
& \therefore \quad \text { The co-ordinates of } \mathrm{M} \text { are }(2,3,-1) .
\end{aligned}
$
$\therefore$ The co-ordinates of $\mathrm{M}$ are $(2,3,-1)$.
The distance of $\mathrm{P}$ from the line is $\mathrm{PM}=\sqrt{(2-0)^2+(3-2)^2+(-1-3)^2}=\sqrt{21}$ unit

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Question 192 Marks

Find the co-ordinates of points on the line $\frac{x+1}{2}=\frac{y-2}{3}=\frac{z+3}{6}$, which are at 3 unit distance from the base point $\mathrm{A}(-1,2,-3)$.

Answer

Let $\mathrm{Q}(2 \lambda-1,3 \lambda+2,6 \lambda-3)$ be a point on the line which is at 3 unit distance from the point $\mathrm{A}(-1,2,-3) \therefore \mathrm{AQ}=3$
$
\begin{aligned}
& \left.\therefore \quad \sqrt{(2 \lambda)^2+(3 \lambda)^2+(6 \lambda)^2}=3 \quad \therefore \quad(2 \lambda)^2+(3 \lambda)^2+(6 \lambda)^2\right)=9 \quad \therefore 49 \lambda^2=9 \\
& \therefore \quad \lambda=-\frac{3}{7} \text { or } \frac{3}{7}
\end{aligned}
$
$\therefore$ There are two points on the line which are at a distance of 3 units from $P$. Their co-ordinates are $\mathrm{Q}(2 \lambda-1,3 \lambda+2,6 \lambda-3)$
Hence, the required points are
$
\left(-\frac{1}{7}, 3 \frac{2}{7},-\frac{3}{7}\right) \text { and }\left(-1 \frac{6}{7}, \frac{5}{7},-5 \frac{4}{7}\right)
$

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Question 202 Marks

Find the angle between lines $\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$
$\text { and } \bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})$

Answer

Let $\bar{b}$ and $\bar{c}$ be vectors along given lines.
$
\therefore \quad \bar{b}=2 \hat{i}-2 \hat{j}+\hat{k} \text { and } \bar{c}=\hat{i}+2 \hat{j}+2 \hat{k}
$
Angle between lines is same as the angle between $\bar{b}$ and $\bar{c}$.
The angle between $\bar{b}$ and $\bar{c}$ is given by,
$
\begin{aligned}
& \cos \theta=\frac{\bar{b} \cdot \bar{c}}{|\bar{b}| \cdot|\bar{c}|}=\frac{(2 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{3 \times 3}=\frac{0}{9}=0 \\
& \therefore \cos \theta=0 \quad \therefore \theta=90^{\circ}
\end{aligned}
$
Lines are perpendicular to each other.

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Question 212 Marks

Find the Cartesian equations of the line passing through $\mathrm{A}(1,2,3)$ and $\mathrm{B}(2,3.4)$.

Answer

The Cartesian equations of the line passing through $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)$ are $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Here $\left(x_1, y_1, z_1\right)=(1,2,3)$ and $\left(x_2, y_2, z_2\right)=(2,3,4)$.
$\therefore$ Required Cartesian equations are $\frac{x-1}{2-1}=\frac{y-2}{3-2}=\frac{z-3}{4-3}$.
$
\begin{aligned}
& \therefore \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1} . \\
& \therefore x-1=y-2=z-3 .
\end{aligned}
$

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Question 222 Marks

Find the vector equation of the line passing through $2 \hat{i}+\hat{j}-\hat{k}$ and parallel to the line joining points $-\hat{i}+\hat{j}+4 \hat{k}$ and $\hat{i}+2 \hat{j}+2 \hat{k}$.

Answer

Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be points with position vectors $\bar{a}=2 \hat{i}+\hat{j}-\hat{k}, \bar{b}=-\hat{i}+\hat{j}+4 \hat{k}$ and $\bar{c}=\hat{i}+2 \hat{j}+2 \hat{k}$ respectively.
$
\overline{B C}=\bar{c}-\bar{b}=(\hat{i}+2 \hat{j}+2 \hat{k})-(-\hat{i}+\hat{j}+4 \hat{k})=2 \hat{i}+\hat{j}-2 \hat{k}
$
The required line passes through $2 \hat{i}+\hat{j}-\hat{k}$ and is parallel to $2 \hat{i}+\hat{j}-2 \hat{k}$
Its equation is $\bar{r}=(2 \hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-2 \hat{k})$

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Question 232 Marks

Find the vector equation of the line passing through the point having position vector $2 \hat{i}+\hat{j}-3 \hat{k}$ and perpendicular to vectors $\hat{i}+\hat{j}+\hat{k}$. and $\hat{i}+2 \hat{j}-\hat{k}$

Answer

Let $a=2 \hat{i}+\hat{j}-3 \hat{k}, \bar{b}=\hat{i}+\hat{j}+\hat{k}$ and $\mathrm{c}=\hat{i}+2 \hat{j}-\hat{k}$
We know that $\bar{b} \times \bar{c}$ is perpendicular to both $\bar{b}$ and $\bar{c}$.
$\therefore \bar{b} \times \bar{c}$ is parallel to the required line.
$
\bar{b} \times \bar{c}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
1 & 2 & -1
\end{array}\right|=-3 \hat{i}+2 \hat{j}+\hat{k}
$
Thus required line passes through $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and parallel to $-3 \hat{i}+2 \hat{j}+\hat{k}$.
$\therefore$ Its equation is $\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(-3 \hat{i}+2 \hat{j}+\hat{k})$

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Question 242 Marks

The equation of the line passing through A $(\bar{a})$ and $\mathrm{B}(\bar{b})$ is $\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})$.

Answer

Let $L$ be the line which passes through $\mathrm{A}(\bar{a})$ and $\mathrm{B}$ $(\bar{b})$.
Let $\mathrm{P}(\bar{r})$ be a variable point on the line L other than $\mathrm{A}$.
$\therefore \overline{A P}$ and $\lambda \overline{A B}$ are collinear.
$\therefore \overline{A P}=\lambda \overline{A B}$, where $\lambda$ is a scalar.
$\therefore \bar{r}-\bar{a}=\lambda(\bar{b}-\bar{a}) \quad \bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})$
$\therefore \bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})$ is the required equation of the line.

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Question 252 Marks

Find the vector equation of the line passing through the point having position vector

$-\hat{i}-\hat{j}+2 \hat{k}$ and parallel to the line $\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$.

Answer

Let $\mathrm{A}$ be point having position vector $\bar{a}=-\hat{i}-\hat{j}+2 \hat{k}$

The required line is parallel to the line

$\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k}+2(3 \hat{i}+2 \hat{j}+\hat{k})$

$\therefore$ it is parallel to the vector

$\bar{b}=3 \hat{i}+2 \hat{j}+\hat{k}$

The vector equation of the line passing through $A(\bar{a})$ and parallel to $\bar{b}$ is $\bar{r}=\bar{a}+\lambda \bar{b}$ where

λ is a scalar. ∴ the required vector equation of the line is

$\overline{\mathrm{r}}=(-\hat{i}-\hat{j}+2 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$

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Question 262 Marks

Find the vector equation of line passing through the point having position vector

$5 \hat{i}+4 \hat{j}+3 \hat{k}$ and having direction ratios $-3,4,2$.

Answer

Let $\mathrm{A}$ be the point whose position vector is $\bar{a}=5 \hat{i}+4 \hat{j}+3 \hat{k}$.

Let $\bar{b}$ be the vector parallel to the line having direction ratios $-3,4,2$

Then, $\bar{b}=-3 \hat{i}+4 \hat{j}+2 \hat{k}$

The vector equation of the line passing through $\mathrm{A}(\bar{a})$ and parallel to $\bar{b}$ is $\bar{r}=\bar{a}+\lambda \bar{b}$,

where λ is a scalar. ∴ the required vector equation of the line is

$\bar{r}=5 \hat{i}+4 \hat{j}+3 \hat{k}+\lambda(-3 \hat{i}+4 \hat{j}+2 \hat{k})$.

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Question 272 Marks

Find the vector equation of the line passing through points having position vectors

Answer

$3 \hat{i}+4 \hat{j}-7 \hat{k}$ and $6 \hat{i}-\hat{j}+\hat{k}$.

The vector equation of the line passing through the $A(\bar{a})$ and $B(\bar{b})$ is $\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a}), \lambda$

is a scalar

∴ the vector equation of the line passing through the points having position vectors

$3 \hat{i}+4 \hat{j}-7 \hat{k}$ and $6 \hat{i}-\hat{j}+\hat{k}$ is

is $\bar{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda[(6 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}+4 \hat{j}-7 \hat{k})]$

i.e. $\bar{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda(3 \hat{i}-5 \hat{j}+8 \hat{k})$.

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Question 282 Marks

Find the vector equation of the line passing through the point having position vector

$-2 \hat{i}+\hat{j}+\hat{k}$ and parallel to vector $4 \hat{i}-\hat{j}+2 \hat{k}$

Answer

The vector equation of the line passing through $\mathrm{A}(\bar{a})$ and parallel to the vector $\bar{b}$ is $\bar{r}=\bar{a}$

∴ the vector equation of the line passing through the point having position vector

$-2 \hat{i}+\hat{j}+\hat{k}$ and parallel to the vector $4 \hat{i}-\hat{j}+2 \hat{k}$ is

$\bar{r}=(-2 \hat{i}+\hat{j}+\hat{k})+\lambda(4 \hat{i}-\hat{j}+2 \hat{k})$

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