Question

The projections of a vector on the three coordinate axis are $6,-3,2$ respectively. The direction cosines of the vector are

• A. $\frac{6}{2},\frac{-3}{5},\frac{2}{5}$
• B. $\frac{6}{7},\frac{-3}{7},\frac{2}{7}$
• C. $\frac{-6}{7},\frac{-3}{7},\frac{2}{7}$
• D. $6,-3,2$

Answer 1

The correct option is B $\frac{6}{7},\frac{-3}{7},\frac{2}{7}$

Let the vector $\overline{PQ}$ be $(x_1-x_2,y_1-y_2,z_1-z_2)$
We have $x_1-x_2=6,y_1-y_2=-3$ and $z_1-z_2=2$

Length of $PQ$ $=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2}$

$=\sqrt{6^2+3^2+2^2}=\sqrt{36+9+4}=7$

The direction cosines of PQ are$(\frac{x_1-x_2}{PQ},\frac{y_1-y_2}{PQ},\frac{z_1-z_2}{PQ},)$

i.e. $(\frac{6}{7},-\frac{3}{7},\frac{2}{7})$