Aspire Faculty ID #16320 · Topic: NIMCET 2010 · Just now
NIMCET 2010

The position vectors of $A,B,C,D$ are $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}+5\hat{j}$, $3\hat{i}+2\hat{j}-3\hat{k}$, $\hat{i}-6\hat{j}-\hat{k}$ Angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is:

Solution

Compute vectors: $\overrightarrow{AB} = (2-1)\hat{i}+(5-1)\hat{j}+(0-1)\hat{k}=\hat{i}+4\hat{j}-\hat{k}$ $\overrightarrow{CD} = (1-3)\hat{i}+(-6-2)\hat{j}+(-1+3)\hat{k}=-2\hat{i}-8\hat{j}+2\hat{k}$ They are scalar multiples: $\overrightarrow{CD} = -2(\overrightarrow{AB})$ ⇒ Angle = $\pi$

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