Question

If a line has direction ratios 2, -3, 7 then find the direction cosines.

(a) l=\(\frac{2}{\sqrt{62}},m=-\frac{7}{\sqrt{62}},n=\frac{7}{\sqrt{62}}\)

(b) l=\(\frac{2}{\sqrt{6}},m=-\frac{3}{\sqrt{6}},n=\frac{7}{\sqrt{6}}\)

(c) l=-\(\frac{2}{\sqrt{62}},m=-\frac{3}{\sqrt{62}},n=-\frac{7}{\sqrt{62}}\)

(d) l=\(\frac{2}{\sqrt{62}},m=-\frac{3}{\sqrt{62}},n=\frac{7}{\sqrt{62}}\)

The question was posed to me in an interview for internship.

This question is from Direction Cosines and Direction Ratios of a Line in chapter Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right choice is (d) l=\(\frac{2}{\sqrt{62}},m=-\frac{3}{\sqrt{62}},n=\frac{7}{\sqrt{62}}\)

Easy explanation: For a given line, if a, b, c are the direction ratios and l, m, n are the direction cosines of the line then

l=±\(\frac{a}{\sqrt{a^2+b^2+c^2}}\)

m=±\(\frac{b}{\sqrt{a^2+b^2+c^2}}\)

n=±\(\frac{c}{\sqrt{a^2+b^2+c^2}}\)

∴l=\(\frac{2}{\sqrt{2^2+(-3)^2+7^2}}, \,m=-\frac{3}{\sqrt{2^2+(-3)^2+7^2}}, \,n=\frac{7}{\sqrt{2^2+(-3)^2+7^2}}\)

Hence, l=\(\frac{2}{\sqrt{62}}, \,m=-\frac{3}{\sqrt{62}}, \,n=\frac{7}{\sqrt{62}}\).