Mathematics

The distance of point $\left(1,0,2\right)$ from the point of intersection of the line $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}$ and the plane $x-y+z=16$ is

$13$

SOLUTION

$\displaystyle \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$

$x=3\lambda +2$  $y=4\lambda -1$  $z=12\lambda +2$

$(x,y,z)\rightarrow x-y+z=16$ should satisfy

$3\lambda +2-(4\lambda -1)+12\lambda +2=16$

$3\lambda +2-4\lambda +1+12\lambda +2=16$

$11\lambda =11$

$\lambda =1$

$(x,y,z)=(5,3,14)$ - point of intersection of line

Distance b/w (1,0,2) & (5,3,14)

$d=\sqrt{(5-1)^{2}+(3-0)^{2}+(14-2)^{2}}$

$=\sqrt{16+9+144}$

$=\sqrt{169}$

$d=13$

$\therefore$ option C = 13

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Single Correct Medium Published on 09th 09, 2020
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