Question

What is the cosine of the angle which the vector √2i+j+k makes with y-axis?

Answer 1

Check this attachment :-)

Answer 2

Answer:

The angle is 60 degree.

Step-by-step explanation:

Given : The cosine of the angle which the vector $\sqrt{2}i+j+k$ makes with y-axis.

To find : What is the cosine of the angle?

Solution :

For a vector $ai+bj+ck$

Cosines for the vector is

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

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

Where, $\alpha, \beta,\gamma$ are the angles made by the vectors with x-axis,y-axis and z-axis respectively.

For the given vector,

$\sqrt{2}i+j+k$,

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

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

Therefore, $\frac{1}{2}=\cos \beta$

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

Thus, the angle made by y-axis is 60 degree.