Question

Let $A=(1,2,2)$, $B=(2,3,6)$and $C=(3,4,12)$. The direction cosines of a line equally inclined with $OA,OB$ and $OC$ , where $O$ is the origin, are

• A. $\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0$
• B. $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$
• C. $\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}$

Let the direction cosines of the line be $(a,b,c)$
Since it's equally inclined to $OA,OB,OC$
$\frac{a+2b+2c}{3}=\frac{2a+3b+6c}{7}=\frac{3a+4b+12c}{13}=k$
$\Rightarrow a+2b+2c=3k...\left(1\right)2a+3b+6c=7k....\left(2\right)3a+4b+12c=13k....\left(3\right)$
Eliminating $a$ from $\left(1\right)$ and $\left(2\right)$, we get
$b-2c=-k....\left(3\right)$
Similarly, eliminating $a$ from $\left(1\right)$ and $\left(3\right)$, we get
$2b-6c=-4k...\left(4\right)$
From $\left(3\right)$ and $\left(4\right)$, we get $b=c=k...\left(5\right)$
Using the above value in $\left(1\right)$ we get $a=-k....\left(6\right)$
From $\left(5\right)$ and $\left(6\right)$, $a:b:c=-1:1:1$
Also, $a^2+b^2+c^2=1\Rightarrow k=\pm \frac{1}{\sqrt{3}}$
Hence, direction cosines are $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ or $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$