Three Dimensional Geometry Question 5
Question 5 - 2024 (27 Jan Shift 1)
The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3} \quad$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is
(1) 12
(2) 14
(3) 18
(4) 21
Show Answer
Answer (2)
Solution
$B=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$
$(7,-2,11)$
Point $\mathrm{B}$ lies on $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$
$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$
$-3 \lambda-6=0$
$\lambda=-2$
$\mathrm{AB}=\sqrt{(7-3)^{2}+(4+2)^{2}+(11+1)^{2}}$
$=\sqrt{16+36+144}$
$=\sqrt{196}=14$
