Question

What is the angle that the vector A=2i+3j makes with y axis.

Answer 1

A = 2i + 3j

let angle form by Vector A = ∅
then ,
cos∅ = ( 2i + 3j )( 0 + j)/√(2² +3²)
= 3/√13

∅ = cos^-1(3/√13 )

Answer 2

Angle that vector A = 2i + 3j makes with y axis is $\bold{\beta=\cos ^{-1} \frac{3}{\sqrt{13}}}$

Solution:

$\vec{A}=2 i+3j$;can be written in form of x and y as $\vec{A}=x i+y j$

Let us find the magnitude A=2i+3j

To find the magnitude let us take the square root of $\sqrt{2^{2}+3^{2}}=\sqrt{13}$  

Now as we can see the magnitude, let us plot the angle and line

Now let us find the value of \beta, to find the value  

We use $\cos \beta=\frac{y}{A}=\frac{3}{\sqrt{13}}$

Similarly finding the value of $\sin \beta=\frac{y}{A}=\frac{2}{\sqrt{13}}$

Now both the above cases can give the value of the angle beta, but in term of cos,

So the value of $\bold{\beta=\cos ^{-1} \frac{3}{\sqrt{13}}}$