Let $ABCD\$be the parallelogram whose sides $AB$ and $AD$ are represented by the vectors $2\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} - 5\widehat{\mathbf{k}}$ and $\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$ respectively. Then if $\overrightarrow{\mathbf{a}}\$is a unit vector parallel to $\overrightarrow{\mathbf{AC}}$, then $\overrightarrow{\mathbf{a}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Let ${\overrightarrow{\mathbf{R}}}_{1} = 2\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} - 5\widehat{\mathbf{k}}$
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and ${\overrightarrow{\mathbf{R}}}_{2} = \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$
$	herefore\ \ \overrightarrow{\mathbf{R}}$ (along $\overrightarrow{\mathbf{AC}}) = {\overrightarrow{\mathbf{R}}}_{1} + {\overrightarrow{\mathbf{R}}}_{2}$
$= 3\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}}$
$	herefore\ \overrightarrow{\mathbf{a}}\ (unit\ vector\ along\ AC) = \frac{\overrightarrow{\mathbf{R}}}{|\overrightarrow{\mathbf{R}}|}$
$= \frac{3\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}}}{\sqrt{9 + 36 + 4}}$
$= \frac{1}{7}(3\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}})$

Explanation

Reviewed by Dr. Abubacker

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