If a line has direction ratios , then what is the sum of the squares of its direction cosines?

Correct answer: D. 1 Direction ratios are Then, direction cosine, aligned & l= (a+b) (a+b)^2+(b+c)^2+(c+a)^2 & m= (b+c) (a+b)^2+(b+c)^2+(c+a)^2 aligned n= (c+a) (a+b)^2+(b+c)^2+(c+a)^2 Sum of squares of direction cosines l^2+m^2+n^2=((a+b)^2+(b+c)^2+(c+a)^2)/((a+b)^2+(b+c)^2+(c+a)^2)=1

NDA 2020

Exam: NDA
Year: 2020
Subject: Mathematics
Chapter: Three Dimensional Geometry

NDA 2020 · Mathematics · Three Dimensional Geometry

If a line has direction ratios , then what is the sum of the...

If a line has direction ratios

< a+b, b+c, c+a >

, then what is the sum of the squares of its direction cosines?

Options

A $(a+b+c)^2$
B $2(a+b+c)$
C3
D1

Correct answer

D. 1

Step-by-step solution

Direction ratios are

< a+b, b+c, c+a >

Then, direction cosine,

\begin{aligned} & l=\frac{(a+b)}{\sqrt{(a+b)^2+(b+c)^2+(c+a)^2}} \\ & m=\frac{(b+c)}{\sqrt{(a+b)^2+(b+c)^2+(c+a)^2}}\end{aligned}

n=\frac{(c+a)}{\sqrt{(a+b)^2+(b+c)^2+(c+a)^2}}

Sum of squares of direction cosines

l^2+m^2+n^2=\frac{(a+b)^2+(b+c)^2+(c+a)^2}{(a+b)^2+(b+c)^2+(c+a)^2}=1$

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