 # Projection of a Vector on a Line

In this article, we will learn about how to project a vector on a line and the angle between two vectors. The two vectors here are the vector to be projected and the vector of the line on which the projection is done.

### Suggested Videos        ## Angle Between Two Vectors

Look at the figures given below: If a vector $$\vec{AB}$$ makes an angle $$\theta$$ with a given directed line l, in the anticlockwise direction, then the projection of $$\vec{AB}$$ on l is a vector $$\vec{p}$$ with magnitude |$$\vec{AB}$$| $$\cos \theta$$.

Also, the direction of $$\vec{p}$$ is the same (or opposite) to that of the line l, depending upon whether $$\cos \theta$$ is positive or negative. The vector $$\vec{p}$$ is the projection vector and has magnitude |$$\vec{p}$$|. It is also called the projection of vector $$\vec{AB}$$ on the directed line l.

In each of the figures shown above, the projection vector of $$\vec{AB}$$ along the line l is the vector: $$\vec{AC}$$.

### Browse more Topics under Vector Algebra

You can download Vector Algebra Cheat Sheet by clicking on the download button below  ## Important Observations

• If $$\hat{p}$$ is the unit vector along a line l, then the projection of a vector $$\vec{a}$$ on the line l is,

$$\vec{a}$$.$$\hat{p}$$

• The projection of a vector $$\vec{a}$$ on other vector $$\vec{b}$$, is

$$\vec{a}$$.$$\hat{b}$$
Or, $$\vec{a}$$.$$\frac{\vec{b}}{|\vec{b}|}$$
Or, $$\frac{1}{|\vec{b}|}$$($$\vec{a}$$.$$\vec{b}$$)

• If $$\theta$$ = 0, then the projection vector of $$\vec{AB}$$ is $$\vec{a}$$ itself. Also, if $$\theta$$ = $$\pi$$, then the projection vector of $$\vec{AB}$$ is $$\vec{BA}$$.
• If $$\theta$$ = $$\frac{\pi}{2}$$ or $$\theta$$ = $$\frac{3\pi}{2}$$, then the projection vector of $$\vec{AB}$$ is zero.

### Remark

If $$\alpha$$, $$\beta$$, and $$\gamma$$ are the direction angles of the vector $$\vec{a}$$ = a1$$\hat{i}$$ + a2$$\hat{j}$$ + a3$$\hat{k}$$, then its direction cosines are,

$$\cos{\alpha}$$ = $$\frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}$$ = $$\frac{a_1}{|\vec{a}|}$$

Similarly, $$\cos{\beta}$$ = $$\frac{a_2}{|\vec{a}|}$$
And, $$\cos{\gamma}$$ = $$\frac{a_3}{|\vec{a}|}$$

It is important to note that, |$$\vec{a}$$|$$\cos{\alpha}$$, |$$\vec{a}$$|$$\cos{\beta}$$, and |$$\vec{a}$$|$$\cos{\gamma}$$ are the projections of $$\vec{a}$$ along the OX, OY, and OZ respectively. Or, the scalar components a1, a2, and a3 of the vector $$\vec{a}$$ are the projections of $$\vec{a}$$ along the x, y, and z-axis respectively.

Lastly, if $$\vec{a}$$ is a unit vector, then the vector can be expressed as

$$\vec{a}$$ = $$\cos{\alpha}$$$$\hat{i}$$ + $$\cos{\beta}$$$$\hat{j}$$ + $$\cos{\gamma}$$$$\hat{k}$$

Let’s look at an example to understand the angle between two vectors well.

### Example

Question: Find the angle between two vectors $$\vec{a}$$ and $$\vec{b}$$ with magnitudes 1 and 2 respectively. Also, given $$\vec{a}$$.$$\vec{b}$$ = 1.

Solution: We know that, $$\vec{a}$$.$$\vec{b}$$ = 1
|$$\vec{a}$$| = 2 and |$$\vec{b}$$| = 1
Therefore, the angle between two vectors $$\theta$$ is,
$$\theta$$ = cos-1 ($$\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$$) = cos-1 ($$\frac{1}{2}$$) = $$\frac{\pi}{3}$$

## More Solved Examples for You

Question 1: Find the angle between two vectors $$\hat{i}$$ – 2$$\hat{j}$$ + 3$$\hat{k}$$ and 3$$\hat{i}$$ – 2$$\hat{j}$$ + $$\hat{k}$$.

Answer : Let, $$\vec{a}$$ = $$\hat{i}$$ – 2$$\hat{j}$$ + 3$$\hat{k}$$ and $$\vec{b}$$ = 3$$\hat{i}$$ – 2$$\hat{j}$$ + $$\hat{k}$$

Now, $$\vec{a}$$.$$\vec{b}$$ = ($$\hat{i}$$ – 2$$\hat{j}$$ + 3$$\hat{k}$$).(3$$\hat{i}$$ – 2$$\hat{j}$$ + $$\hat{k}$$)

Now, using properties of multiplication of vectors, we get $$\vec{a}$$.$$\vec{b}$$ = 1 x 3 + (- 2) x (- 2) + 3 x 1 = 10.

Next, the magnitude of $$\vec{a}$$ is, |$$\vec{a}$$| = $$\sqrt{1^2 + (-2)^2 + 3^2}$$ = $$\sqrt{14}$$

And, the magnitude of $$\vec{b}$$ is, |$$\vec{b}$$| = $$\sqrt{3^2 + (-2)^2 + 1^2}$$ = $$\sqrt{14}$$

We know that, $$\vec{a}$$.$$\vec{b}$$ = |$$\vec{a}$$||$$\vec{b}$$|$$\cos{\theta}$$

Substituting the values, we get
10 = $$\sqrt{14}$$ x $$\sqrt{14}$$ $$\cos{\theta}$$
Or, $$\cos{\theta}$$ = $$\frac{10}{14}$$ = $$\frac{5}{7}$$
Therefore, $$\theta$$ = cos-1 ($$\frac{5}{7}$$)

Question 2: How to find the angle between two vectors with the use of the cross product?

Answer: Firstly, the cross product of the 2 vectors a and b represented by a x b is the vector ab sin θ n, here, a and b are the magnitudes for the vector a and vector b, θ is the angle measured from a to b in the anti-clockwise direction and n here is said to be the unit vector, perpendicular to ‘a’ as well as‘b’.

Question 3: What is the formula for the angle between two vectors?

Answer: A simpler way to find out the angle between 2 vectors is the dot product formula. Here, (A.B =|A|x|B|xcos(X)) let vector ‘A’ be ‘2i’ and vector ‘B’ be ‘3i+4j’. According to the question, ‘X’ is the angle between the vectors. Therefore, A.B = |A|x|B|xcos(X) = 2i.

Question 4: What is the angle that forms between 2 equal vectors?

Answer: That angle is 180° – 60° = 120°. So, when 2 vectors of the same magnitude form a vector of the same magnitude, then the angle between the 2 vectors is 120°.

Question 5: How to find out if 2 vectors are parallel?

Answer: The 2 vectors A and B are parallel only if they are scalar multiples of each other.

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