Question

What is the direction cosine of angle which the vector $\sqrt{2\hat{i}}+\hat{j}+\hat{k}makeswithy-axis?$

Answer 1

For a vector

$ai+bj+ck$

direction cosines are given by

$t=\frac{a}{\sqrt{a^2+b^2+c^2}},m=\frac{b}{\sqrt{a^2+b^2+c^2}},n=\frac{c}{\sqrt{a^2+b^2+c^2}}$

$L=\cos \alpha ,m=\cos \beta ,n=\cos \gamma$

Where

$\alpha ,\beta ,\gamma$

are angles made by the vectors with x-axis, y-axis and z-axis respectively for the given vector $\sqrt{2i}+j+k$

$m=\cos \beta =\frac{1}{\sqrt{2+1+1}}=\frac{1}{\sqrt{4}}=\frac{1}{2}$

$\therefore \beta =\frac{\pi }{3}$.