Find the angle between the line \overrightarrow r  = (\widehat i - \widehat k + 3\widehat k) + \lambda (3\widehat i  - \widehat j + 2\widehat k) and the plane \overrightarrow r .(\widehat i + \widehat j + \widehat k) = 3
Find the angle between the line \overrightarrow r  = (\widehat i - \widehat k + 3\widehat k) + \lambda (3\widehat i  - \widehat j + 2\widehat k) and the plane \overrightarrow r .(\widehat i + \widehat j + \widehat k) = 3

Answer:

The equation of plane is:

r=(2i^j^+3k^)+λ(3i^j^+2k^)

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

On comparing with \vec{r}=\vec{a}+\lambda \vec{b}

Here, \vec{b}=(3 i-j+2 k)

The equation of plane

r·(i+j+k^)=3

\vec{r} \cdot(i+j+\hat{k})=3

On comparing with \vec{r} \cdot \vec{n}=0

n=(i+j+k^)

\vec{n}=(i+j+\hat{k})

Let ‘ \theta ‘ be the anyle between line and plane

    \[\begin{aligned} \sin \theta &=\left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right| \\ \therefore \sin \theta &=\left|\frac{(3 i-j+2 k) \cdot(\hat{i}+\vec{j}+\vec{k})}{\sqrt{9+1+4} \sqrt{1+1+1}}\right| \\ \therefore \sin \theta &=\frac{3-1+2}{\sqrt{14} \cdot \sqrt{3}}=\frac{4}{\sqrt{14} \sqrt{3}}=\frac{4}{\sqrt{42}} \\ \therefore \theta &=\sin ^{-1}\left(\frac{4}{\sqrt{42}}\right) \end{aligned}\]