Question

Find the direction cosines of the following vectors:
(i) $2\hat{i}+2\hat{j}-\hat{k}$

(ii) $6\hat{i}-2\hat{j}-3\hat{k}$

(iii) $3\hat{i}-4\hat{k}$

Answer 1

(i) We have, $2\hat{i}+2\hat{j}-\hat{k}$
The direction cosines are $\frac{2}{\sqrt{2^2+2^2+{\left(-1\right)}^2}},\frac{2}{\sqrt{2^2+2^2+{\left(-1\right)}^2}},\frac{-1}{\sqrt{2^2+2^2+{\left(-1\right)}^2}}$ or, $\frac{2}{3},\frac{2}{3},\frac{-1}{3}$

(ii) We have, $6\hat{i}-2\hat{j}-3\hat{k}$
The direction cosines are $\frac{6}{\sqrt{6^2+{\left(-2\right)}^2+{\left(-3\right)}^2}},\frac{-2}{\sqrt{6^2+{\left(-2\right)}^2+{\left(-3\right)}^2}},\frac{-3}{\sqrt{6^2+{\left(-2\right)}^2+{\left(-3\right)}^2}}$ or, $\frac{6}{7},\frac{-2}{7},\frac{-3}{7}$

(iii) We have, $3\hat{i}-4\hat{k}$
The direction cosines are $\frac{3}{\sqrt{3^2+0+{\left(-4\right)}^2}},\frac{0}{\sqrt{3^2+0+{\left(-4\right)}^2}},\frac{-4}{\sqrt{3^2+0+{\left(-4\right)}^2}}$ or, $\frac{3}{5},0,\frac{-4}{5}$