Q.37. The distance of the point P(-2,4,-5) from the line \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} is
A. \frac{\sqrt{37}}{10}
B. \sqrt{\frac{37}{10}}
C. \frac{37}{\sqrt{10}}
D. \frac{37}{10}
Answer :- B. \sqrt{\frac{37}{10}}
Explanation :- Since the point is (-2, 4, -5),
\therefore a = -2, b = 4, c = -5Given equation of line is
\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} x_1=-3, y_1=4, z_1=-8Direction ratios (d.r.s) of the line are 3, 5, 6
Direction cosines (d.c.s) are \frac{3}{\sqrt{70}}, \frac{5}{\sqrt{70}}, \frac{6}{\sqrt{70}}
Perpendicular distance of the point from the line is
\sqrt{\left[(a-x_1)^2 + (b-y_1)^2 + (c-z_1)^2\right]} = \sqrt{1^2 + 0 + 3^2 - \left[\frac{1(3)}{\sqrt{70}}+\frac{0(5)}{\sqrt{70}}+\frac{3(6)}{\sqrt{70}}\right]^2} = \sqrt{1 + 9-\left(\frac{3}{\sqrt{70}}+\frac{18}{\sqrt{70}}\right)^2}= \sqrt{\frac{37}{10}} units