Q. The point on the line $\frac{x - 2}{1} = \frac{y + 3}{- 2} = \frac{\mathcal{z}\mathbf{+}5}{- 2}$ at a distance of 6 from the point $(2, - 3,\ - 5)$ is
A) $(3, - 5,\ - 3)$
B) $(4, - 7,\ - 9)$
C) $(0,2,\ - 1)$
D) $( - 3,\ 5,\ 3)$
Explanation
(b)
Direction cosines of the given line are $\frac{1}{3}, - \frac{2}{3}, - \frac{2}{3}$
Hence, the equation of line can be point in the form $\frac{x - 2}{1/3} = \frac{y + 3}{- 2/3} = \frac{\mathcal{z} + 5}{- 2/3} = r$
Therefore, any point on the line is $\left( 2 + \frac{r}{3}, - 3 - \frac{2r}{3}, - 5 - \frac{2r}{3} \right)$, where $r = \pm 6$
Points are $(4,\ - 7,\ - 9)$ and $(0,\ 1,\ - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.