Question

show that a=icap-jcap/root 2 is a vector​

Answer 1

We're given,

$\vec{a}=\dfrac{\hat{\! i}-\hat{\! j}}{\sqrt2}$

We just take the RHS as,

$\vec{a}=\dfrac{1}{\sqrt2}\ \hat{\! i}-\dfrac{1}{\sqrt2}\ \hat{\! j}$

Now let's check whether  $\vec{a}$  has both magnitude and direction.

$|\vec{a}|\ =\ \sqrt{\left(\dfrac{1}{\sqrt2}\right)^2+\left(-\dfrac{1}{\sqrt2}\right)^2+2\cdot\dfrac{1}{\sqrt2}\cdot-\dfrac{1}{\sqrt2}\cdot\cos90\textdegree}\\ \\ \\ |\vec{a}|\ =\ \sqrt{\dfrac{1}{2}+\dfrac{1}{2}}\\ \\ \\ |\vec{a}|\ =\ 1$

Let  $\vec{a}$  makes an angle  $\alpha$  with x - axis.

Then,

$\alpha=\tan^{-1}\left(\dfrac{-\frac{1}{\sqrt2}\sin90\textdegree}{\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\cos90\textdegree}\right)\\ \\ \\ \alpha=\tan^{-1}\left(\dfrac{-\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}\right)\\ \\ \\ \alpha=\tan^{-1}\left(-1\right)\\ \\ \\ \alpha=-45\textdegree$

So,  $\vec{a}$  exists.