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The angle between the line \(r=(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3…
The angle between the line \(r=(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})\) and the plane \(\mathrm{r} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=8\) is
$\sin ^{-1}\left(\frac{2 \sqrt{7}}{\sqrt{5}}\right)$ $\sin ^{-1}\left(\frac{3 \sqrt{7}}{\sqrt{5}}\right)$ $\sin ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)$ $\sin ^{-1}\left(\frac{\sqrt{7}}{3 \sqrt{5}}\right)$ Solution
The angle between the line $\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ and the plane $\overline{\mathrm{r}}-\overline{\mathrm{n}}=\mathrm{dr}$ is $\sin \theta=\left|\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}| \cdot|\overline{\mathrm{n}}|}\right|$
Here $\overline{\mathrm{b}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\overline{\mathrm{n}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\begin{aligned}
\overline{\mathrm{b}} \cdot \overline{\mathrm{n}} &=(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=3+2+0=5 \\
|\mathrm{~b}| &=\sqrt{3^{2}+1^{2}}=\sqrt{10} \text { and }|\overline{\mathrm{n}}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{14} \\
\sin \theta &=\left|\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}}{|\overline{\mathrm{b}}| \cdot|\overline{\mathrm{n}}|}\right|=\left|\frac{5}{\sqrt{10} \cdot \sqrt{14}}\right| \Rightarrow \sin \theta=\frac{\sqrt{5}}{2 \sqrt{7}} \Rightarrow \theta=\sin ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)
\end{aligned}$
Asked in: MHT CET 2020 (16 Oct Shift 1)
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