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The position vectors of vertices of a $\Delta A B C$ are $\begin{array}{lll}4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}, & \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \quad \text { and } & -\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}}\end{array}$
respectively, then $\angle A B C$ is equal to
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respectively, then $\angle A B C$ is equal to
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2813 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Here $, \overrightarrow{\mathbf{A B}}=-3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{B C}}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{A B}} \cdot \overrightarrow{\mathbf{B C}}=6+6-12=0$
$$
\angle A B C=\frac{\pi}{2}
$$
and $\overrightarrow{\mathbf{A B}} \cdot \overrightarrow{\mathbf{B C}}=6+6-12=0$
$$
\angle A B C=\frac{\pi}{2}
$$
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