Maths Solve the Following Question.(3 Marks)

$\therefore$ the lines intersect at $\mathrm{O}$ whose position vector is $\overline{0}$

Since z = 0 for both the lines, both the lines lie in XY- plane. Hence, we have to find equation of XY-plane. Z-axis is perpendicular to XY-plane.

$\therefore$ normal to $X Y$ plane is $\hat{k}$,

$0(\overline{0})$ lies on the plane.

By using $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$, the vector equation of the required plane is $\bar{r} \cdot \hat{k}=\overline{0} \cdot \bar{k}$

i.e. $\bar{r} \cdot \hat{k}=0$,

Hence, the given lines intersect each other and the vector equation of the plane determine

by them is $\bar{r} \cdot \hat{k}=0$

is $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n} \ldots(1)$

The position vectors $\bar{a}$ and $\bar{b}$ of the given points $A$ and $B$ are $\bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$ and

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

If $M$ is the midpoint of segment $A B$, the position vector $\bar{m}$ of $M$ is given by

$\begin{aligned} \bar{m}=\frac{\bar{a}+\bar{b}}{2} & =\frac{(2 \hat{i}+3 \hat{j}+6 \hat{k})+(4 \hat{i}+3 \hat{j}-2 \hat{k})}{2} \\ & =\frac{6 \hat{i}+6 \hat{j}+4 \hat{k}}{2}=3 \hat{i}+3 \hat{j}+2 \hat{k}\end{aligned}$

The plane passes through $\mathrm{M}(\bar{m})$.

$\mathrm{AB}$ is perpendicular to the plane

If $\bar{n}$ is normal to the plane, then $\bar{n}=\overline{\mathrm{AB}}$

$\begin{aligned} & \therefore \bar{n}=\bar{b}-\bar{a}=(4 \hat{i}+3 \hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ & =2 \hat{i}-8 \hat{k} \\ & \therefore \bar{m} \cdot \bar{n}=(3 \hat{i}+3 \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-8 \hat{k}) \\ & =(3)(2)+(3)(0)+(2)(-8) \\ & =6+0-16=-10 \text {. } \\ & \end{aligned}$

$\therefore$ from (1), the vector equation of the required plane is

$\begin{aligned} & \bar{r} \cdot \bar{n}=\bar{m} \cdot \bar{n} \\ & \text { i.e. } \bar{r} \cdot(2 \hat{i}-8 \hat{k})=-10 \\ & \text { i.e. } \bar{r} \cdot(\hat{i}-4 \hat{k})=-5\end{aligned}$

is $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n} \ldots(1)$

We can take $\bar{a}=\overline{0}$ since the plane passes through the origin.

The point $M$ with position vector $\bar{m}=\hat{i}+4 \hat{j}+k$ lies on the line and hence it lies on the

plane.

$\therefore \overline{\mathrm{OM}}=\bar{m}=\hat{i}+4 \hat{j}+\hat{k}$ lies on the plane.

The plane contains the given line which is parallel to $\bar{b}=\hat{i}+2 \hat{j}+\hat{k}$

Let $\bar{n}$ be normal to the plane. Then $\bar{n}$ is perpendicular to $\overline{\mathrm{OM}}$ as well as $\bar{b}$

$\therefore \bar{n}=\overline{\mathrm{OM}} \times \bar{b}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 1 \\ 1 & 2 & 1\end{array}\right|$

$=(4-2) \hat{i}-(1-1) \hat{j}+(2-4) \hat{k}$

$=2 \vec{i}-2 \hat{k}$

$\therefore$ from (1), the vector equation of the required plane is

$\begin{aligned} & \bar{r} \cdot(2 \hat{i}-2 \hat{k})=\overline{0}-\bar{n}=0 \\ & \text { i.e. } \bar{r} \cdot(\hat{i}-\hat{k})=0 .\end{aligned}$

$\sin \theta=\left|\frac{\bar{b} \cdot \bar{n}}{|\bar{b}||\bar{n}|}\right|$

$\ldots(1)$

Here, $\overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{n}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\therefore \bar{b} \cdot \bar{n}=(\hat{i}-\hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})$

= (2)(2) + (3)(-1) + (-6)(1) = 4 – 3 – 6 = -5

Also, $|\bar{b}|=\sqrt{1^2+1^2+(-1)^2}=\sqrt{2}=1$

$|\bar{n}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{4}$

$\therefore$ from (1), we have

$\sin \theta=\left|\frac{2 \sqrt{2}}{-3}\right|=\frac{2 \sqrt{2}}{3}$

$\therefore \theta=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$

vectors $\bar{b}$ and $\bar{c}$ is

$\overline{\mathrm{r}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$

Here, $\overline{\mathrm{a}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

$\begin{aligned} & \overline{\mathrm{b}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}} \\ & \overline{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ & \therefore \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{k}} \\ 4 & 0 & 3 \\ 1 & 1 & 0\end{array}\right|\end{aligned}$

$\begin{aligned} & =(-1-3) \hat{i}-(4-3) \hat{j}+(3+1) \hat{k} \\ & =-4 \hat{i}-\hat{j}+4 \hat{k} \\ & \therefore \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \bar{c})=(-2 \hat{i}+7 \hat{j}+5 \hat{k}) \cdot(-4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{k})\end{aligned}$

= (-2)(-4) + (7)(-1) + (5)(4) = 8 – 7 + 8 = 35

$\therefore$ From (1), the vector equation of the required plane is $\overline{\mathrm{r}} \cdot(-3 \hat{\mathrm{i}}-3 a t \mathrm{j}+4 \hat{\mathrm{k}})=35$.

collinear is

$\overline{\mathrm{r}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}) \ldots(1)$

The required plane makes intercepts 1, 1, 1 on the coordinate axes. ∴ it passes through the three non collinear points A = (1, 0, 0), B = (0, 1, 0), C = (0, , 1)

$\begin{aligned} & \therefore \overline{\mathrm{a}}=\hat{\mathrm{i}}, \overline{\mathrm{b}}=\hat{\mathrm{j}}, \overline{\mathrm{c}}=\hat{\mathrm{k}} \\ & \overline{\mathrm{AB}}=\overline{\mathrm{b}}-\overline{\mathrm{a}}=\hat{\mathrm{j}}-\hat{\mathrm{i}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ & \therefore \overline{\mathrm{AC}}=\overline{\mathrm{c}}-\overline{\mathrm{a}}=\hat{\mathrm{k}}-\hat{\mathrm{i}}=-\hat{\mathrm{i}}+\hat{\mathrm{k}}\end{aligned}$

$\begin{aligned} & \therefore \overline{\mathrm{AB}} \times \overline{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right| \\ & =(1-0) \hat{\mathrm{i}}-(-1-0) \hat{j}+(0+1) \hat{\mathrm{k}} \\ & =\hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\end{aligned}$

Also

$\begin{aligned} & \text { a. }(\overline{A B} \times \overline{A C}) \\ & =\hat{i}(\hat{i}+\hat{j}+\hat{k}) \\ & =1 \times 1+0 \times 1+0 \times 1 \\ & =1\end{aligned}$

$\therefore$ from(1)the vector equation of the required plane is $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$.

$\mathrm{C}(\bar{c})$ is $\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}) \ldots(1)$

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

$\begin{aligned} \therefore \overline{\mathrm{AB}}=\bar{b}-\bar{a} & =(2 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}-2 \hat{j}+\hat{k}) \\ & =\hat{i}+\hat{j}-4 \hat{k}\end{aligned}$

$\begin{aligned} \overline{\mathrm{AC}}=\bar{c}-\bar{a} & =(\hat{j}+5 \hat{k})-(\hat{i}-2 \hat{j}+\hat{k}) \\ & =-\hat{i}+3 \hat{j}+4 \hat{k}\end{aligned}$

$\begin{aligned} \therefore \overline{\mathrm{AB}} \times \overline{\mathrm{AC}} & =\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -4 \\ -1 & 3 & 4\end{array}\right| \\ & =(4+12) \hat{i}-(4-4) \hat{j}+(3+1) \hat{k}\end{aligned}$

$=16 \hat{i}+4 \hat{k}$

Now, $\begin{aligned} \bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}) & =(\hat{i}-2 \hat{j}+\hat{k}) \cdot(16 \hat{i}+4 \hat{k}) \\ & =(1)(16)+(-2)(0)+(1)(4)=20\end{aligned}$

$\therefore$ from (1), the vector equation of the required plane is

$\bar{r} \cdot(16 \hat{i}+4 \hat{k})=20$.

normal and p is the length of perpendicular drawn from origin to the plane.

Given pane is r. $(6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+24 \hat{\mathrm{k}})=13 \ldots(1)$

$\bar{n}=6 \hat{i}+8 \hat{j}+24 \hat{k}$ is normal to the plane

$\therefore|\bar{n}|=\sqrt{6^2+8^2+24^2}=\sqrt{76}=13$

Dividing both sides of (1) by 13 , get

$\overline{\mathrm{r} .}\left(\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{13}\right)=\frac{76}{13}$

i.e. $\bar{r} \cdot\left(\frac{3}{13} \hat{\mathrm{i}}+\frac{4}{13} \hat{\mathrm{j}}+\frac{12}{13} \hat{k}\right)=\frac{1}{2}$

This is the normal form of the equation of plane.

Comparing with $\bar{r} \cdot \hat{n}=p_{\text {, }}$

(i) the length of the perpendicular from the origin to plane is $\frac{1}{2}$.

(ii) direction cosines of the normal are $\frac{3}{13}, \frac{4}{13}, \frac{12}{13}$

the plane, then the vector equation of the plane $\bar{r} \cdot \hat{n}=\mathrm{p}$

Here, $\overline{\mathbf{n}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $p=5$

$\therefore|\bar{n}|=\sqrt{2^2+1^2+(2)^2}$

$\begin{aligned} & =\sqrt{9} \\ & =3 \\ & \hat{\mathrm{n}}=\frac{\overline{\mathrm{n}}}{|\bar{n}|} \\ & =\frac{1}{3}(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\end{aligned}$

$\therefore$ the vector equation of the required plane is

$\overline{\mathrm{r}} \cdot\left[\frac{1}{3}(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\right]=5$

i.e. $\bar{r} \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=15$.

The coordinates of any point on this line are given by x = λ + 1, y = -2λ + 2, z = 2λ + 3 Let M(λ + 1, -2λ + 2, 2λ + 3) … (1) be the point on the line whose distance from A(1, 2, 3) is 3 units.

$\begin{aligned} & \therefore \sqrt{(\lambda+1-1)^2+(-2 \lambda+2-2)^2+(2 \lambda+3-3)^2}=3 \\ & \therefore \sqrt{\lambda^2+4 \lambda^2+4 \lambda^2}=3 \\ & \therefore \sqrt{9 \lambda^2}=3 \\ & \therefore 9 \lambda^2=9 \\ & \therefore \lambda^2=1 \quad \therefore \lambda= \pm 1\end{aligned}$

When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)] i. e. M = (2, 0, 5) When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)] i. e. M = (0, 4, 1) Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).

$\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ intersect, if

$\therefore\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=0$

$\ldots(1)$

$\begin{aligned} & \text { Here, }\left(x_1, y_1, z_1\right) \equiv(1,-1,1) \\ & \left(x_2, y_2, z_2\right) \equiv(2,-m, 2)_r \\ & a_1=2, b_1=3, c_1=4 \\ & a_2=1, b_2=2, c_2=1\end{aligned}$

Substituting these values in (1), we get

$\begin{aligned} & \left|\begin{array}{rrr}2-1 & -m+1 & 2-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0 \\ & \therefore\left|\begin{array}{rrr}1 & 1-m & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\end{aligned}$

∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0 ∴ -5 + 2 – 2m + 1 = 0 ∴ -2m = 2 ∴ m = -1.

∴ the direction ratios of the line are 1, 1, 0. The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0. ∴ its directiton ratios are 0, 1, 0.

Let $\bar{a}$ and $\bar{b}$ be the vectors in the direction of the lines $x=y, z=0$ and $x=0, z=0$.

Then $\bar{a}=\vec{i}+\vec{j}, \quad \bar{b}=\vec{j}$

$\therefore \bar{a} \cdot \bar{b}=(\hat{i}+\hat{j}) \cdot \hat{j}=(1)(0)+(1)(1)+(0)(0)=1$

$|\bar{a}|=\sqrt{1^2+1^2}=\sqrt{2}$

$|\bar{b}|=|\hat{j}|=1$

If θ is the acute angle between the lines, then

$\cos \theta=\left|\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}\right|=\left|\frac{1}{\sqrt{2} \times 1}\right|=\frac{1}{\sqrt{2}}=\cos 45^{\circ}$.

$\therefore \theta=45^{\circ}$.

the lines

$\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{2}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{1}$ respectively.

$\begin{aligned} \text { Then } \bar{a} & =\hat{i}-\hat{j}+2 \hat{k}, \bar{b}=2 \hat{i}+\hat{j}+\hat{k} \\ \therefore \bar{a} \cdot \bar{b} & =(\hat{i}-\hat{j}+2 \hat{k}) \cdot(2 \hat{i}+\hat{j}+\hat{k}) \\ & =(1)(2)+(-1)(1)+(2)(1) \\ & =2-1+2=3\end{aligned}$

$\begin{aligned} & |\bar{a}|=\sqrt{1^2+(-1)^2+2^2}=\sqrt{6} \\ & |\bar{b}|=\sqrt{2^2+1^2+1^2}=\sqrt{6}\end{aligned}$

If $\theta$ is the angle between the lines, then

$\begin{aligned} & \cos \theta=\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{3}{\sqrt{6} \sqrt{6}}=\frac{1}{2}=\cos 60^{\circ} \\ & \therefore \theta=60^{\circ} .\end{aligned}$

$\begin{aligned} & x=3+2 \lambda, y=2-\lambda, z=1+0(\lambda) \\ & x-3=2 \lambda, y-2=-\lambda, z=1 \\ & \therefore \frac{x-3}{2}=\frac{y-2}{-1}=\lambda_z z=1\end{aligned}$

∴ the cartesian equations of the line are

$\frac{x-3}{2}=\frac{y-2}{-1}, z=1$

This line is passing through the point A(-5, -4, -5) and having direction ratios 3, 5, 6.

Let $\bar{a}$ be the position vector of the point A w.r.t, the origin and $\bar{b}$ be the vector parallel to

the line.

Then $\bar{a}=-5 \hat{i}-4 \hat{j}-5 \hat{k}$ and $\bar{b}=3 \hat{i}+5 \hat{j}+6 \hat{k}$.

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.

$\therefore$ the vector equation of the required line is $\bar{r}=(-5 \hat{i}-4 \hat{j}-6 \hat{k})+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})$

ratios 3, 5, 6 as it is parallel to the given line. It passes through the point (-2, 4, -5). The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{2-z_1}{c}$

$\therefore$ the required cartesian equations of the line are

$\frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6}$

i.e. $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$

$\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$, where $\lambda$ is a scalar.

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

$3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and parallel to the vector

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

$\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$, where $\lambda$ is a scalar.

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

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

$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+\lambda(6 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

∴ the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0 is

$\frac{3(1)+4(1)-12(-1)+20}{\sqrt{3^2+4^2+(-12)^2}}$

$=\left|\frac{3+4+12+20}{\sqrt{9+16+144}}\right|=\frac{39}{\sqrt{169}}$

$=\frac{39}{13}=3$ units

$\begin{aligned} & \text { Here, } \bar{a}=4 \hat{i}-3 \hat{j}+\hat{k}, \bar{n}=2 \hat{i}+3 \hat{j}-6 \hat{k}, \mathrm{p}=21 \\ & \therefore \bar{a} \cdot \bar{n}=(4 \hat{i}-3 \hat{j}+\hat{k}) \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\end{aligned}$

= (4)(2) + (-3)(3) + (1)(-6) = 8 – 9 – 6 = -7

Also, $\sqrt{2^2+3^2+(-6)^2}=\sqrt{49}=7$

∴ from (1), the required distance

$=\frac{|-7-21|}{7}=$ 4units

$\sin \theta=\left|\frac{\bar{b} \cdot \bar{n}}{|b \vec{b}||n|}\right|$

$\ldots(1)$

$\begin{aligned} \text { Here, } \bar{b} & =2 \hat{i}+3 \hat{j}-6 \hat{k}, \bar{n}=2 \hat{i}-\hat{j}+\hat{k} \\ \therefore \bar{b} \cdot \bar{n} & =(2 \hat{i}+3 \hat{j}-6 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k}) \\ & =(2)(2)+(3)(-1)+(-6)(1) \\ & =4-3-6=-5\end{aligned}$

Also, $|\bar{b}|=\sqrt{2^2+3^2+(-6)^2}=\sqrt{49}=7$

$|\bar{n}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$

$\therefore$ from $(1)$, we have

$\begin{aligned} & \sin \theta=\left|\frac{-5}{7 \sqrt{6}}\right|=\frac{5}{7 \sqrt{6}} \\ & \therefore \theta=\sin ^{-1}\left(\frac{5}{7 \sqrt{6}}\right) .\end{aligned}$

$\cos \theta=\left|\frac{n_1 \cdot n_2}{\left|n_1 \|\right| n_2 \mid}\right|$

$\ldots(1)$

Here, $\bar{n}_1=\hat{i}+\hat{j}+2 \hat{k}, \bar{n}_2=2 \vec{i}-\hat{j}+\vec{k}$

$\begin{aligned} \therefore \bar{n}_1 \cdot \bar{n}_2 & =(\hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k}) \\ & =(1)(2)+(1)(-1)+(2)(1)\end{aligned}$

$=2-1+2=3$

Also, $\left|\bar{n}_1\right|=\sqrt{1^2+1^2+2^2}=\sqrt{6}$

$\left|\bar{n}_2\right|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$

$\therefore$ from (1), we have

$\cos \theta=\left|\frac{3}{\sqrt{6} \cdot \sqrt{6}}\right|=\frac{3}{6}=\frac{1}{2} \cos 60^{\circ}$

$\therefore \theta=60^{\circ}$.

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

M(1, 0, 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 $M$ then $\bar{m}=\hat{i}$

Normal to the plane is

$\begin{aligned} & \bar{n}=\overline{\mathrm{OM}}=\hat{i} \\ & \bar{m} \cdot \bar{n}=\hat{i} \cdot \hat{i}=1\end{aligned}$

$\therefore$ the vector equation of the required plane is $\bar{r} \cdot \hat{i}=1$

$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$

The required plane is parallel to XY-plane. ∴ it is perpendicular to Z-axis i.e. Z-axis is normal to the plane. Z-axis has direction ratios 0, 0, 1. The plane passes through (7, 8, 6). ∴ the cartesian equation of the required plane is 0(x – 7) + 0(y – 8) + 1 (z – 6) = 0 i.e. z = 6.

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

Here, $\bar{a}=\hat{i}+\hat{j}+\hat{k}_1 \bar{n}=4 \hat{i}+5 \hat{j}+6 \hat{k}$

$\therefore \bar{a} \cdot \bar{n}=(\hat{i}+\hat{j}+\hat{k}) \cdot(4 \hat{i}+5 \hat{j}+6 \hat{k})$

= (1)(4) + (1)(5) + (1)(6) = 4 + 5 + 6 = 15

$\therefore$ the vector equation of the required plane is $\bar{r} \cdot(4 \hat{i}+5 \hat{j}+6 \hat{k})=15$.

to the plane, then the vector equation of the plane is $\bar{r} \cdot \hat{n}=\mathrm{p}$

Here, $\bar{n}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $p=42$

$\therefore|\bar{n}|=\sqrt{2^2+1^2+(-2)^2}=\sqrt{9}=3$

$\hat{n}=\frac{n}{|\bar{n}|}=\frac{1}{3}(2 \hat{i}+\hat{j}-2 \hat{k})$

$\therefore$ the vector equation of the required plane is

$=\left[\frac{1}{3}(2 \hat{i}+\hat{j}-2 \hat{k})\right]=42$

i.e. $\bar{r} \cdot(2 \hat{i}+\hat{j}-2 \hat{k})=126$

$\frac{x-2}{1}=\frac{y-4}{2}=\frac{z+4}{-2}$

The coordinates of the origin O are (0, 0, 0)

For $x=0, \frac{x-2}{1}=\frac{0-2}{1}=-2$

For $y=0, \frac{y-4}{2}=\frac{0-4}{2}=-2$

For $z=0, \frac{z+4}{-2}=\frac{0+4}{-2}=-2$

∴ coordinates of the origin O satisfy the equation of the line. Hence, the line passes through the origin.

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Here, $\left(x_1, y_1, z_1\right)=(2,2,1)$ and $\left(x_2, y_2, z_2\right)=(1,3,0)$

∴ the required cartesian equations are

$\begin{aligned} & \frac{x-2}{1-2}=\frac{y-2}{3-2}=\frac{z-1}{0-1} \\ & \text { i.e. } \frac{x-2}{-1}=\frac{y-2}{1}=\frac{z-1}{-1}\end{aligned}$

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

∴ the cartesian equations of the line passing through the point (-1, 2, 1) and having direction ratios 2, 3, 1 are

$\frac{x-(-1)}{2}=\frac{y-2}{3}=\frac{z-1}{1}$

i.e. $\frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{1}$

by

$\bar{b} \times \bar{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 1\end{array}\right|$

$\begin{aligned} & =\hat{i}(1+1)-\hat{j}(1-2)+\hat{k}(-1-2) \\ & =2 \hat{i}+\hat{j}-3 \hat{k}\end{aligned}$

Since the line is perpendicular to the vector $\bar{b}$ and $\bar{c}$, it is parallel to $\bar{b} \times \bar{c}$. The vector

equation of the line passing through $A(\bar{a})$ and parallel to $\bar{b} \times \bar{c}$ is

$\bar{r}=\bar{a}+\lambda(\bar{b} \times \bar{c})$, where $\lambda$ is a scalar.

Here, $\bar{a}=\hat{i}+2 \hat{j}+3 \hat{k}$

Hence, the vector equation of the required line is

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