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Find the angle between the line \overrightarrow r = (3\widehat i + \widehat k) + \lambda (\widehat j + \widehat k) and the plane \overrightarrow r .(2\widehat i - \widehat j + 2\widehat k) = 1
CBSE | Class 12 | Exercise 28G | Maths | RS Aggarwal | The Planes
Find the angle between the line \overrightarrow r = (3\widehat i + \widehat k) + \lambda (\widehat j + \widehat k) and the plane \overrightarrow r .(2\widehat i - \widehat j + 2\widehat k) = 1

Answer:

The equation of line is:

r→=(3i^+k^)+λ(j^+k^)

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

Comparing with \vec{r}=\vec{a}+\lambda \vec{b}

\vec{b}=(j+\dot{k}).

The equation of plane is:

\vec{r}.(2 \hat{i}-\hat{j}+2 \hat{k})=1Comparing with\vec{r} \cdot \vec{n}=0\vec{n}=(2 \hat{i}-\hat{j}+2 \hat{k})Let '\theta$ ‘ be the angle between line and plane

    \[\begin{aligned} \sin \theta &=\left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{h}|}\right| \\ \therefore \sin \theta &=\left|\frac{(\bar{j}+\hat{k}) \cdot(2 \hat{i}-j+2 \hat{k})}{\sqrt{1+1} \cdot \sqrt{4+1+4}}\right| \\ \therefore \sin \theta &=\left|\frac{(0 \times 2)+(1 \times-1)+(1 \times 2)}{\sqrt{2}}\right|=\left|\frac{-1+2}{3 \sqrt{2}}\right|=\frac{1}{3 \sqrt{2}} \\ \sin \theta &=\frac{1}{3 \sqrt{2}} \\ \theta &=\sin ^{-1}\left(\frac{1}{3\sqrt{2}}\right) \end{aligned}\]

 

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