Write the direction ratios of the vector $\to a=\hat{i}+\hat{j}-2\hat{k}$and hence calculate its direction cosines.

Solution: Here, $\vec{a}=\hat{i}+\hat{j}-2\hat{k}$
So, direction ratios will be $\left(1,1,-2\right)$.
Let its direction cosines are (k,k,-2k).
Then,$k^2+k^2+{\left(-2k\right)}^2=1$
$6k^2=1$
$k=\pm \frac{1}{\sqrt{6}}$
We will take $k=\frac{1}{\sqrt{6}}$ for direction cosines in the same direction of vector.
So, direction cosines will be $\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)$