Find the direction cosines of the line passing through two points (4, -5, -6) and (-1, 2, 8).

(a) \frac{5}{270},\frac{7}{\sqrt{270}},\frac{14}{\sqrt{270}}

(b) –\frac{7}{\sqrt{270}}, \frac{7}{\sqrt{270}},\frac{7}{\sqrt{270}}

(c) –\frac{5}{\sqrt{270}}, \frac{7}{\sqrt{270}},\frac{14}{\sqrt{270}}

(d) –\frac{5}{\sqrt{20}}, \frac{7}{\sqrt{720}},\frac{14}{\sqrt{270}}

I have been asked this question in a national level competition.

My doubt stems from Direction Cosines and Direction Ratios of a Line in division Three Dimensional Geometry of Mathematics – Class 12

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Right option is (c) –\frac{5}{\sqrt{270}}, \frac{7}{\sqrt{270}},\frac{14}{\sqrt{270}}

For explanation: The direction cosines of two lines passing through two points is given by:

\frac{x_2-x_1}{PQ}, \frac{y_2-y_1}{PQ}, \frac{z_2-z_1}{PQ}

and PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

In the given problem we have, P(4,-5,-6) and Q(-1,2,8)

∴PQ = \sqrt{(-1-4)^2+(2+5)^2+(8+6)^2}

=\sqrt{25+49+196}=\sqrt{270}

Hence, the direction ratios are l=\frac{(-1-4)}{\sqrt{270}}=-\frac{5}{\sqrt{270}}

m=\frac{(2+5)}{\sqrt{270}}=\frac{7}{\sqrt{270}}

n=\frac{(8+6)}{\sqrt{270}}=\frac{14}{\sqrt{270}}.