If the position vectors of the vertices $A, B$ and $C$ of $\triangle A B C$ are $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ respectively, then $\angle B=$

Question

If the position vectors of the vertices $A, B$ and $C$ of $\triangle A B C$ are $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ respectively, then $\angle B=$, $\cos ^{-1}\left(\frac{7}{3 \sqrt{10}}\right)$, $\cos ^{-1}\left(\frac{8}{\sqrt{105}}\right)$, $\cos ^{-1}\left(\frac{1}{\sqrt{42}}\right)$, $\cos ^{-1}\left(-\frac{7}{3 \sqrt{10}}\right)$

MARKS App · Accepted Answer

$\begin{aligned} & \because \mathbf{B A}=+3 \hat{\mathbf{i}}-6 \hat{\mathbf{k}} \text { and } \mathbf{B C}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \ & \begin{array}{l} \therefore \cos (\angle B)=\frac{|\mathbf{B A} \cdot \mathbf{B}|}{|\mathbf{B A}| \mid \mathbf{B C}} \ \quad=\frac{12+12}{\sqrt{9+36} \sqrt{16+1+4}}=\frac{24}{\sqrt{45 \times 21}} \ \quad=\frac{24}{3 \sqrt{105}}=\frac{8}{\sqrt{105}} \ \Rightarrow \angle B=\cos ^{-1}\left(\frac{8}{\sqrt{105}}\right) \end{array} \end{aligned}$ Hence, option (2) is correct.

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