Question

Let $\overrightarrow{A}$ be vector parallel to line of intersection of planes $P_1$ and $P_2$, Plane $P_1$ is parallel to the vectors $2\hat{j}+3\hat{k}$ and $4\hat{j}-3\hat{k}$ and that $P_2$ is parallel to $\hat{j}-\hat{k}$ and $3\hat{i}+3\hat{j}$ then the angle between vector $\overrightarrow{A}$ and a given vector $2\hat{i}+\hat{j}-2\hat{k}$

Answer 1

We have,

The plane $P_1$ is parallel to the vector $2\hat{i}+3\hat{k}$ and $4\hat{j}-3\hat{k}$

Then the normal of $P_1=(2\hat{i}+3\hat{k})\times (4\hat{j}-3\hat{k})$

$\left|\begin{array}{ccc}i& j& k\\ 2& 0& 3\\ 0& 4& -3\end{array}\right|$

$=\hat{i}(0-12)-\hat{j}(-6-0)+\hat{k}(8-0)$

$=-12\hat{i}+6\hat{j}+8\hat{k}$

The normal of $P_2=(\hat{j}-\hat{k})\times (3\hat{i}+3\hat{j})$

$\left|\begin{array}{ccc}i& j& k\\ 0& 1& -1\\ 3& 3& 0\end{array}\right|$

$\hat{i}(0+3)-\hat{j}(0+3)+\hat{k}(0-3)$

$=3\hat{i}-3\hat{j}-3\hat{k}$

The line of intersection of the planes will be $=(-12\hat{i}+6\hat{j}+8\hat{k})\times (3\hat{i}-3\hat{j}-3\hat{k})$

$\left|\begin{array}{ccc}i& j& k\\ -12& 6& 8\\ 3& -3& -3\end{array}\right|$

$=(-18+24)\hat{i}-(36-24)\hat{j}+(36-18)\hat{k}$

$=6\hat{i}-12\hat{j}+18\hat{k}$

Now, the angle between $\overrightarrow{A}=6\hat{i}-12\hat{j}+18\hat{k}and2\hat{i}+\hat{j}-2\hat{k}$ is given by

$\cos \theta =\frac{\left(6\right)\left(2\right)+(-12)\left(1\right)+\left(18\right)(-2)}{\sqrt{6^2+{(-12)}^2+{\left(18\right)}^2}\sqrt{2^2+1^2+{(-2)}^2}}$

$\cos \theta =\frac{12-12-36}{\sqrt{504}\sqrt{9}}$

$\cos \theta =\frac{-36}{\sqrt{4\times 126}\left(3\right)}$

$\cos \theta =\frac{-36}{2\sqrt{126}\times 3}$

$\cos \theta =\frac{-6}{3\sqrt{14}}$

$\cos \theta =\frac{-2}{\sqrt{14}}$

$\theta ={\cos }^{-1}\left(\frac{-2}{\sqrt{14}}\right)$

Hence, this is the answer.