Question

What is the cosine of the angle which the vector $\sqrt{2}i+j+k$ makes with y axis?

Answer 1

For a vector
ai +bj +ck
direction cosinesare given by:
$l=\frac{a}{\sqrt{a62+b^2+c^2}}.m=\frac{b}{\sqrt{a^2+b^2+c^2}},n=\frac{c}{\sqrt{a^2+b^2+c^2}}$
$l=cos\alpha ,m=cos\beta ,n=cos\gamma$,
where
$\alpha ,\beta ,\gamma$
are angles made by the vectors with x-axis, y -axis and z-axis respectivley.
for the given vector $\sqrt{2}i+j+k,$
$m=cos\beta =\frac{1}{\sqrt{2+1+1}}=\frac{1}{\sqrt{4}}=\frac{1}{2}$
therefore
$\beta =\frac{\pi }{3}$
Thus angle made with y -axis is 60 deg.