Question

If 1,2,1 are the direction ratios of a line then the direction cosines of this line is equal to

• A. $\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}$
• B. $\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}}$
• C. $\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$
• D. $\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}$

Answer 1

The correct option is B $\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}}$

We know that the direction ratios are proportional to direction cosines i.e. l= k.a, m= k.b , n= k.c where k is any constant, l ,m & n are direction cosines and a,b & c are direction ratios.
We also know that $l^2+m^2+n^2=1$
So,$(k.a{)}^2+(k.b{)}^2+(k.c{)}^2=1$
$k^2(a^2+b^2+c^2)=1$
$k^2=\frac{1}{(a^2+b^2+c^2)}$
Or, $k=\pm \sqrt{\frac{1}{(a^2+b^2+c^2)}}$

In the question given direction ratios or a,b,c are 1, 2, 1.
So, $k=\pm \sqrt{\frac{1}{\left(\right(1{)}^2+(2{)}^2+(1{)}^2)}}$
Or, $k=\pm \sqrt{\frac{1}{6}}$
So the direction cosines will be -
l=1.k
or,$l=\sqrt{\frac{1}{6}}\left(ontakingkas\sqrt{\frac{1}{6}}\right)$
Similarly ,$m=2.\sqrt{\frac{1}{6}}$ $\&$ n=$\sqrt{\frac{1}{6}}$
Also,-$\sqrt{\frac{1}{6}},-2.\sqrt{\frac{1}{6}},-\sqrt{\frac{1}{6}}$ on taking k as $-\sqrt{\frac{1}{6}}$