If triangle ABC is a right-angled triangle in | vedclass

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(C) Since $	riangle ABC$ is right-angled at $A$, we have $\vec{AB} \cdot \vec{AC} = 0$.Given expression: $E = \vec{AB} \cdot \vec{AC} + \vec{BA} \cdo

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(C) Since $	riangle ABC$ is right-angled at $A$, we have $\vec{AB} \cdot \vec{AC} = 0$.Given expression: $E = \vec{AB} \cdot \vec{AC} + \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$.Since $\vec{AB} \cdot \vec{AC} = 0$, the expression becomes $E = \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$.Note that $\vec{BA} \cdot \vec{BC} = |\vec{BA}| |\vec{BC}| \cos B = |\vec{BA}|^2$ (projection of $\vec{BA}$ on $\vec{BC}$ is $\vec{BA}$ itself in right triangle).Actually, in right triangle $ABC$ with hypotenuse $BC$, $\vec{BA} \cdot \vec{BC} = |\vec{BA}|^2$ and $\vec{CA} \cdot \vec{CB} = |\vec{CA}|^2$.Thus, $E = |\vec{BA}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$.Calculate $|\vec{BC}|^2 = |\vec{c} - \vec{b}|^2 = |(5-3)\hat{i} + (1-(-2))\hat{j} + (-3-1)\hat{k}|^2 = |2\hat{i} + 3\hat{j} - 4\hat{k}|^2$.$|\vec{BC}|^2 = 2^2 + 3^2 + (-4)^2 = 4 + 9 + 16 = 29$.

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