Question

find the angle that vector A= 2i cap + 3 j cap - k cap makes with y axis?

Answer 1

The y-axis is parallel to the y-unit vector, so we’ll use <0,1> to represent it, or you could write that as 0i+1j I really don’t like that notation though.

We know that geometrically, the dot product of two vectors can be seen as the projection of one across the other, multiplied by both their magnitudes. This can be expressed more concisely as.

a⋅b=|a||b|cos(θ)

Where a and b are two arbitrary vectors of the same two dimensions and theta is the angle between them. Let’s take the vector of interest <2,3> to be a and the vector representing the y axis <0,1> to be b. It doesn’t matter though, dot products are commutative.

Based on this, θ=arccos(a⋅b|a||b|)

To save you the time, the angle is .588 radians or 33.7 degrees.

There you go!

Answer 2

▶ Solution :

Given :

A vector quantity is provided as

$\bf{\red{\vec{A}=2\hat{i}+3\hat{j}-\hat{k}}}$

To Find :

Angle made by given vector with y-axis.

Concept :

▪ This question is completely based on concept of dot product of vectors.

▪ Dot product of two vector quantities A and B is given by

$\boxed{\bf{\pink{\vec{A}{\tiny{\bullet}}\vec{B}=|\vec{A}||\vec{B}|\cos\theta}}}$

Calculation :

$\implies\sf\:|\vec{A}|=\sqrt{(2)^2+(3)^2+(1)^2}=\sqrt{14}\:unit\\ \\ \implies\sf\:|\vec{B}|=1\:unit\\ \\ \mapsto\sf\:(2\hat{i}+3\hat{j}-\hat{k})(\hat{j})=(\sqrt{14})(1)\cos\theta\\ \\ \mapsto\sf\:3(\hat{j}\:\hat{j})=\sqrt{14}\cos\theta\\ \\ \mapsto\sf\:\cos\theta=\dfrac{3(1)}{\sqrt{14}}\\ \\ \mapsto\boxed{\bf{\purple{\theta=\cos^{-1}(\dfrac{3}{\sqrt{14}})}}}$