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(D) The position vectors are $\vec{A} = \hat{i}+2\hat{j}+\hat{k}$,$\vec{B} = \hat{i}+3\hat{j}-2\hat{k}$,and $\vec{C} = 2\hat{i}+\hat{j}-\hat{k}$.$\vec

Vedclass · Accepted Answer

$\frac{1}{6}(7\hat{i}+12\hat{j}+\hat{k})$

Vedclass · Answer

(D) The position vectors are $\vec{A} = \hat{i}+2\hat{j}+\hat{k}$,$\vec{B} = \hat{i}+3\hat{j}-2\hat{k}$,and $\vec{C} = 2\hat{i}+\hat{j}-\hat{k}$.$\vec{AB} = \vec{B} - \vec{A} = 0\hat{i}+\hat{j}-3\hat{k}$ and $\vec{AC} = \vec{C} - \vec{A} = \hat{i}-\hat{j}-2\hat{k}$.$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -3 \ 1 & -1 & -2 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(0+3) + \hat{k}(0-1) = -5\hat{i}-3\hat{j}-\hat{k}$.Area of $	riangle ABC = \frac{1}{2}|\vec{AB} 	imes \vec{AC}| = \frac{1}{2}\sqrt{(-5)^2+(-3)^2+(-1)^2} = \frac{\sqrt{35}}{2}$.Volume of tetrahedron = $\frac{1}{3} 	imes 	ext{Area}(ABC) 	imes h = \frac{\sqrt{805}}{6\sqrt{2}}$.$\frac{1}{3} 	imes \frac{\sqrt{35}}{2} 	imes h = \frac{\sqrt{805}}{6\sqrt{2}} \implies h = \frac{\sqrt{805}}{\sqrt{35}\sqrt{2}} = \sqrt{\frac{23}{2}}$.Let $D$ be the vertex. The altitude $DE = h = \sqrt{\frac{23}{2}}$.In $	riangle ADE$,$AD^2 = AE^2 + DE^2$. Given $AD = \frac{\sqrt{110}}{3}$,$AD^2 = \frac{110}{9}$.$AE^2 = \frac{110}{9} - \frac{23}{2} = \frac{220-207}{18} = \frac{13}{18}$.$E$ lies on the median from $A$ to $BC$. Let $F$ be the midpoint of $BC$,$F = \frac{B+C}{2} = \frac{3\hat{i}+4\hat{j}-3\hat{k}}{2} = 1.5\hat{i}+2\hat{j}-1.5\hat{k}$.Vector $\vec{AF} = F-A = 0.5\hat{i}+0\hat{j}-2.5\hat{k} = \frac{1}{2}(\hat{i}-5\hat{k})$.Unit vector along $AF$ is $\frac{\hat{i}-5\hat{k}}{\sqrt{26}}$.$\vec{AE} = AE \cdot \frac{\vec{AF}}{|AF|} = \sqrt{\frac{13}{18}} \cdot \frac{\hat{i}-5\hat{k}}{\sqrt{26}} = \frac{\sqrt{13}}{3\sqrt{2}} \cdot \frac{\hat{i}-5\hat{k}}{\sqrt{2}\sqrt{13}} = \frac{\hat{i}-5\hat{k}}{6}$.Position vector of $E = \vec{A} + \vec{AE} = (\hat{i}+2\hat{j}+\hat{k}) + \frac{1}{6}(\hat{i}-5\hat{k}) = \frac{6\hat{i}+12\hat{j}+6\hat{k}+\hat{i}-5\hat{k}}{6} = \frac{7\hat{i}+12\hat{j}+\hat{k}}{6}$.

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