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The plane — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsThe plane3 Marks
Question
Find the angle between the line $\vec{\text{r}}=(2\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ and the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=5$
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Answer

We know that the angle $\theta$ between the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
Here,
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
So, $\sin\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{2+3+4}{\sqrt{4+9+16}\sqrt{1+1+1}}$
$=\frac{9}{\sqrt{29}\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{29}}$
$\theta=\sin^{-1}\Big(\frac{3\sqrt{3}}{\sqrt{29}}\Big)$

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