Q.36. 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),
a = -2, b = 4, c = -5Given equation of line is \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}
x1 = -3, y1=4, z1=-8
d.r.s of the lines are 3,5,6
d.c.s are \frac{3}{\sqrt{70}}, \frac{5}{\sqrt{70}}, \frac{6}{\sqrt{70}}Perpendicular distance from point is
\sqrt{(a-x_1)^2 + (b-y_1)^2 + (c-z_1)^2} - \left[\left( a-x_1 \right)l + \left( b-y_1 \right)m + \left( c-z_1 \right)n\right]^2 = \sqrt{1^2 + 0 + 3^2 - \left[\left(\frac{1(3)}{\sqrt{70}}\right) + \left(\frac{0(5)}{\sqrt{70}}\right) + \left(\frac{3(6)}{\sqrt{70}}\right)\right]^2} = \sqrt{1 + 9 - \left(\left(\frac{3}{\sqrt{70}}\right) + \left(\frac{18}{\sqrt{70}}\right)\right)^2} = \sqrt{\frac{37}{10}}. \text{units}