Question

Find a vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 unit.

Answer 1

Let $a=5\hat{i}-\hat{j}+2\hat{k}$
Comparing with $X=x\hat{i}+y\hat{j}+z\hat{k},wegetx=5,y=-1,z=2$
$\therefore Magnitude|a|=\sqrt{x^2+y^2+z^2}=\sqrt{5^2+(-1{)}^2+2^2}=\sqrt{25+1+4}=\sqrt{30}$
$\therefore$ Unit vector in the direction of given vector
$\hat{a}=\frac{a}{|a|}=\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}$
Hence, the vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 units is given by (multiplying the unit vector by 8, we have the vector with magnitude 8 as the magnitude of unit vectors is 1).
$8\hat{a}=8\left(\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}\right)=\frac{40}{\sqrt{30}}\hat{i}-\frac{8}{\sqrt{30}}\hat{j}+\frac{16}{\sqrt{30}}\hat{k}$