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Maths > Vector Algebra > Projection of a Vector on a Line
Vector Algebra

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.

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Angle Between Two Vectors

Look at the figures given below:

angle between two vectors

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: \( \vec{AC} \).

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Projection of a Vector on a Line

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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