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  4. The perpendicular distance of the point hati+ hatj+ hatk from the line joining the points 3 hati+4 hatj- hatk and 2 hati+ hatj- hatk is

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Q. The perpendicular distance of the point $\hat{i}+\hat{j}+\hat{k}$ from the line joining the points $3 \hat{i}+4 \hat{j}-\hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$ is

Vector Algebra

Solution:

$\Theta$ Area of $\triangle PAB =\frac{1}{2}|\overrightarrow{ PA } \times \overrightarrow{ PB }|=\frac{1}{2}( AB )( PM )$
$\Theta \overrightarrow{ PA }=2 \hat{ i }+3 \hat{ j }-2 \hat{ k }$ and $\overrightarrow{ PB }=\hat{ i }-2 \hat{ k }$
image
$ \overrightarrow{ PA } \times \overrightarrow{ PB }=\begin{vmatrix}
\hat{ i } & \hat{ j } & \hat{ k } \\2 & 3 & -2 \\1 & 0 & -2\end{vmatrix}$
$=-6 \hat{ i }+2 \hat{ j }-3 \hat{ k } \text { and } \overrightarrow{ AB }=-\hat{ i }-3 \hat{ j } $
$\therefore 7=\sqrt{10} \cdot PM \Rightarrow PM =\frac{7}{\sqrt{10}} $