If overline{OA} = hat{i} + 2hat{j} + 3hat{k} | vedclass

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(A) Let $C$ be a point on the line such that $\overrightarrow{BC}$ is parallel to $\overrightarrow{OA} 	imes \overrightarrow{OB}$ and $|\overrightarr

Vedclass · Accepted Answer

$6\hat{i} + 11\hat{j} - 7\hat{k}$

Vedclass · Answer

(A) Let $C$ be a point on the line such that $\overrightarrow{BC}$ is parallel to $\overrightarrow{OA} 	imes \overrightarrow{OB}$ and $|\overrightarrow{BC}| = \sqrt{189}$.First, calculate the cross product $\vec{v} = \overrightarrow{OA} 	imes \overrightarrow{OB}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 4 & 0 & 1 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(1-12) + \hat{k}(0-8) = 2\hat{i} + 11\hat{j} - 8\hat{k}$.Since $\overrightarrow{BC}$ is parallel to $\vec{v}$, we have $\overrightarrow{BC} = \lambda(2\hat{i} + 11\hat{j} - 8\hat{k})$.The magnitude is $|\overrightarrow{BC}| = |\lambda| \sqrt{2^2 + 11^2 + (-8)^2} = |\lambda| \sqrt{4 + 121 + 64} = |\lambda| \sqrt{189}$.Given $|\overrightarrow{BC}| = \sqrt{189}$, we get $|\lambda| = 1$, so $\lambda = \pm 1$.The position vector of $C$ is $\vec{OC} = \vec{OB} + \overrightarrow{BC} = (4\hat{i} + \hat{k}) + \lambda(2\hat{i} + 11\hat{j} - 8\hat{k})$.For $\lambda = 1$, $\vec{OC} = (4+2)\hat{i} + (0+11)\hat{j} + (1-8)\hat{k} = 6\hat{i} + 11\hat{j} - 7\hat{k}$.

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