The position vectors of points A, B, C are gi | vedclass

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(A) The position vectors of the vertices are $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(a, -3, 1)$.Since $\angle C = \pi/2$,the vectors $\vec{CA}$ and $\vec{

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$2$ and $1$

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(A) The position vectors of the vertices are $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(a, -3, 1)$.Since $\angle C = \pi/2$,the vectors $\vec{CA}$ and $\vec{CB}$ must be perpendicular,meaning their dot product is zero.$\vec{CA} = (2-a)\hat{i} + (-1 - (-3))\hat{j} + (1-1)\hat{k} = (2-a)\hat{i} + 2\hat{j} + 0\hat{k}$.$\vec{CB} = (1-a)\hat{i} + (-3 - (-3))\hat{j} + (-5-1)\hat{k} = (1-a)\hat{i} + 0\hat{j} - 6\hat{k}$.For $\angle C = 90^\circ$,$\vec{CA} \cdot \vec{CB} = 0$.$(2-a)(1-a) + (2)(0) + (0)(-6) = 0$.$(2-a)(1-a) = 0$.This gives $a = 2$ or $a = 1$.Thus,the values of $a$ are $2$ and $1$.

The position vectors of points $A, B, C$ are given by $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $a\hat{i} - 3\hat{j} + \hat{k}$ respectively. If these points form a right-angled triangle with $\angle C = \pi/2$,find the value of $a$.

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