Let $\bar{A}$ be a vector parallel to line of intersection of planes $P_1$ and $P_2$ through origin. $P_1$…
Let $\bar{A}$ be a vector parallel to line of intersection of planes $P_1$ and $P_2$ through origin. $P_1$ is parallel to the vectors $2 \hat{j}+3 \hat{k}$ and $4 \hat{j}-3 \hat{k}$ and $P_2$ is parallel to $\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}$, then the angle between $\bar{A}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ is
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{\pi}{6}$
$\frac{3 \pi}{4}$
Solution
Vector equation of the plane passing through the point $A(\bar{a})$ and parallel to non-zero vectors $\bar{b}$ and $\overline{\mathrm{c}}$ is $\overline{\mathrm{r}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$
Plane $P_1$ is passing through the origin and parallel to vectors $\overline{b_1}=2 \hat{j}+3 \hat{k}$ and $\overline{c_1}=4 \hat{j}-3 \hat{k}$
$\therefore \quad \overline{\mathrm{b}_1} \times \overline{\mathrm{c}_1}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
0 & 2 & 3 \\
0 & 4 & -3
\end{array}\right|=-18 \hat{\mathrm{i}}$
$\therefore \quad$ Equation of $\mathrm{P}_1$ is: $\mathrm{r} \cdot(-18 \mathrm{i})=0$
Plane $P_2$ is passing through the origin and parallel to vectors $\overline{b_2}=\hat{j}-\hat{k}$ and $\overline{c_2}=3 \hat{i}+3 \hat{j}$
$\therefore \quad \overline{b_2} \times \overline{c_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & -1 \\
3 & 3 & 0
\end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}$
$\therefore \quad$ Equation of $\mathrm{P}_2$ is $: \mathrm{r}_2 \cdot(3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=0$
Note that $\overline{\mathrm{A}}$ is parallel to the cross product of $-18 \hat{i}$ and $3 \hat{i}-3 \hat{j}-3 \hat{k}$
$\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-18 & 0 & 0 \\
3 & -3 & -3
\end{array}\right|=-54 \hat{\mathrm{j}}+54 \hat{\mathrm{k}}$
Let $\theta$ be the required angle.
$\therefore \quad \theta=$ Angle between $54(-\hat{\mathrm{j}}+\hat{\mathrm{k}})$ and $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
$\begin{aligned}
\therefore \quad \cos \theta & =\frac{54 \times(-1-2)}{54 \sqrt{0+1+1} \sqrt{4+1+4}} \\
& = \pm \frac{3}{3 \sqrt{2}} \\
& = \pm \frac{1}{\sqrt{2}} \\
\therefore \quad \theta=\frac{\pi}{4} & , \frac{3 \pi}{4}
\end{aligned}$