Admiddevadxvi

2023-02-21

Let $\overrightarrow{A}$ be 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 $\hat{j}-\hat{k}$ and $3\hat{i}+3\hat{j}$ then the angle between vector $\overrightarrow{A}$ and $2\overrightarrow{i}+\overrightarrow{j}-2\hat{k}$

Jakob Howell

Beginner2023-02-22Added 4 answers

Let vector $\overrightarrow{AO}$ be parallel to line of planes ${P}_{1}$ and ${P}_{2}$ through origin.

Normal to plane ${p}_{1}$ is

${\overrightarrow{n}}_{1}=[(2\overrightarrow{j}+3\overrightarrow{k})\times 4\hat{j}-3\hat{k})]=-18\hat{i}$

Normal to plane ${p}_{2}$ is

$\overrightarrow{{n}_{2}}=(\hat{j}-\hat{k})\times (3\hat{i}+3\hat{j})=3\hat{i}-3\hat{j}-3\hat{k}$

So, $\overrightarrow{OA}$ is parallel to $\pm (\overrightarrow{{n}_{1}}\times \overrightarrow{{n}_{2}})=54\hat{j}-54\hat{k}$

Angle between $54(\hat{j}-\hat{k})$ and $(2\hat{i}+\hat{j}-2\hat{k})$ is

$\mathrm{cos}\theta =\pm (\frac{54+108}{3.54\sqrt{2}})=\pm \frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\theta =\frac{\pi}{4},\frac{3\pi}{4}$

Normal to plane ${p}_{1}$ is

${\overrightarrow{n}}_{1}=[(2\overrightarrow{j}+3\overrightarrow{k})\times 4\hat{j}-3\hat{k})]=-18\hat{i}$

Normal to plane ${p}_{2}$ is

$\overrightarrow{{n}_{2}}=(\hat{j}-\hat{k})\times (3\hat{i}+3\hat{j})=3\hat{i}-3\hat{j}-3\hat{k}$

So, $\overrightarrow{OA}$ is parallel to $\pm (\overrightarrow{{n}_{1}}\times \overrightarrow{{n}_{2}})=54\hat{j}-54\hat{k}$

Angle between $54(\hat{j}-\hat{k})$ and $(2\hat{i}+\hat{j}-2\hat{k})$ is

$\mathrm{cos}\theta =\pm (\frac{54+108}{3.54\sqrt{2}})=\pm \frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\theta =\frac{\pi}{4},\frac{3\pi}{4}$