The position vectors of the vertices of a Del | vedclass

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(D) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4\hat{i}-2\hat{j}$,$\vec{b} = \hat{i}+4\hat{j}-3\hat{k}$,and $\vec{c} = -\hat{i

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4\hat{i}-2\hat{j}$,$\vec{b} = \hat{i}+4\hat{j}-3\hat{k}$,and $\vec{c} = -\hat{i}+5\hat{j}+\hat{k}$.We need to find $\angle ABC$,which is the angle between vectors $\vec{BA}$ and $\vec{BC}$.First,calculate $\vec{BA} = \vec{a} - \vec{b} = (4-1)\hat{i} + (-2-4)\hat{j} + (0-(-3))\hat{k} = 3\hat{i} - 6\hat{j} + 3\hat{k}$.Next,calculate $\vec{BC} = \vec{c} - \vec{b} = (-1-1)\hat{i} + (5-4)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} + \hat{j} + 4\hat{k}$.Now,find the dot product $\vec{BA} \cdot \vec{BC} = (3)(-2) + (-6)(1) + (3)(4) = -6 - 6 + 12 = 0$.Since the dot product of $\vec{BA}$ and $\vec{BC}$ is $0$,the vectors are perpendicular to each other.Therefore,$\angle ABC = \frac{\pi}{2}$.

The position vectors of the vertices of a $\Delta ABC$ are $4\hat{i}-2\hat{j}$,$\hat{i}+4\hat{j}-3\hat{k}$,and $-\hat{i}+5\hat{j}+\hat{k}$ respectively. Then,$\angle ABC$ is equal to:

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