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Q.
The distance from the point $(1, 6 , 3 )$ to the line $\vec{r} = \left(\hat{j}+2 \hat{k}\right)+\lambda \left(\hat{i}+2 \hat{j}+3 \hat{k}\right)$ is
Three Dimensional Geometry
Solution:
Distance $PM$ = ?
$P(\alpha, \beta, \gamma) = (1, 6, 3)$
$M (x,y,z) = (\lambda, 2\lambda+1, 3\lambda +2)$
dr’s of PM are $(\lambda-1, 2 \lambda-5, 3\lambda-1)$
Dr’s of line are $(1, 2, 3) $
Now, $PM \bot AB$
$\therefore \lambda=1$
$\therefore M(1, 3, 5)$
$\therefore P(1, 6, 3), M (1, 3, 5)$
$\therefore PM= \sqrt{\left(6-3\right)^{2}+\left(3-5\right)^{2}}=\sqrt{13}$ Short Cut Method :
Using fact : The $\bot$ distance from $(\alpha, \beta, \gamma)$ on the line
$\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ is given by
$\sqrt{\left(\alpha-x_{1}\right)^{2}+\left(\beta-y_{1}\right)^{2}+\left(\gamma-z_{1}\right)^{2}-\left\{l\left(\alpha-x_{1}\right)+m\left(\beta-y_{1}\right)+n \left(\gamma-z_{1}\right)\right\}^{2}}$
Now, $(\alpha, \beta, \gamma) = (1, 6, 3)$,
$(x_{1}, y_{1}, z_{1}) = (0, 1, 2)$,
$(l, m, n) = \left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)$
$\therefore $ Required distance
$=\sqrt{\left(1-0\right)^{2}+\left(6-1\right)^{2}+\left(3-2\right)^{2}-\left[\frac{1}{\sqrt{14}}\left(1+2\left(5\right)+3\left(1\right)\right)\right]^{2}}$
$=\sqrt{27-14}=\sqrt{13}$
