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If the position vectors of the points $A$ and $B$ are $2 \hat{i}+3 \hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2 \hat{k}$ respectively, then the unit vector along $\overrightarrow{\mathrm{BA}}$ and in the direction of $\overrightarrow{\mathrm{AB}}$ is
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The correct answer is:
$\frac{1}{\sqrt{26}}(-\hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
$\overrightarrow{\mathrm{AB}}=-\hat{i}-4 \hat{j}+3 \hat{k}$ So, $\overrightarrow{\mathrm{BA}}=\hat{i}-4 \hat{j}+3 \hat{k}$
Unit vector along $\overrightarrow{\mathrm{BA}}$ and in direction $\overrightarrow{\mathrm{AB}}$ is $=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{BA}}|}$
$=\frac{-\hat{i}-4 \hat{j}+3 \hat{k}}{\sqrt{1+16+9}}=\frac{-\hat{i}-4 \hat{j}+3 \hat{k}}{\sqrt{26}}$
Unit vector along $\overrightarrow{\mathrm{BA}}$ and in direction $\overrightarrow{\mathrm{AB}}$ is $=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{BA}}|}$
$=\frac{-\hat{i}-4 \hat{j}+3 \hat{k}}{\sqrt{1+16+9}}=\frac{-\hat{i}-4 \hat{j}+3 \hat{k}}{\sqrt{26}}$
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