Projection of Vectors

Projection of Vectors

Hello JEE aspirants! Vectors are a fundamental concept in physics and mathematics, and their projections play a crucial role in solving many problems, especially those involving forces, work, and geometry. Understanding vector projections will give you an edge in JEE Main. Let's dive in!

Conceptual Explanation with Intuition

Imagine shining a light directly onto a vector $\vec{a}$ from above, and it casts a shadow on another vector $\vec{b}$. That shadow is essentially the projection of $\vec{a}$ onto $\vec{b}$. It tells us how much of $\vec{a}$ acts in the direction of $\vec{b}$.

1. Scalar Projection of $\vec{a}$ on $\vec{b}$

The scalar projection (also known as the component) is the length of the shadow. It’s a scalar value (a number, not a vector) indicating the magnitude of $\vec{a}$ in the direction of $\vec{b}$.

Explanation:

• $\vec{a} \cdot \vec{b}$ is the dot product of $\vec{a}$ and $\vec{b}$. Recall that $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos{\theta}$, where $\theta$ is the angle between the two vectors.
• $|\vec{b}|$ is the magnitude (length) of vector $\vec{b}$.
• $\hat{b}$ is the unit vector in the direction of $\vec{b}$, i.e., $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$. Therefore, $\vec{a} \cdot \hat{b} = |\vec{a}| |\hat{b}| \cos{\theta} = |\vec{a}| \cos{\theta}$ since $|\hat{b}| = 1$.

So, the formula essentially gives you the component of $\vec{a}$ along $\vec{b}$, scaled by the cosine of the angle between them.

Example: Let $\vec{a} = 2\hat{i} + 3\hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$. The scalar projection of $\vec{a}$ on $\vec{b}$ is $$\frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{1^2 + 1^2}} = \frac{2 + 3}{\sqrt{2}} = \frac{5}{\sqrt{2}}$$.

2. Vector Projection of $\vec{a}$ on $\vec{b}$

The vector projection is the vector representing the shadow. It has both magnitude (length) and direction (along $\vec{b}$).

Explanation:

• $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}$ is the scalar projection divided by the magnitude of $\vec{b}$. Since the scalar projection is $|\vec{a}| \cos{\theta}$, this term becomes $\frac{|\vec{a}| \cos{\theta}}{|\vec{b}|}$.
• Multiplying this by $\vec{b}$ gives the vector projection. Essentially, we're scaling the vector $\vec{b}$ by the scalar projection divided by $|\vec{b}|$.
• The alternative formula shows the same result using unit vector $\hat{b}$.

Example: Using the same vectors as before, $\vec{a} = 2\hat{i} + 3\hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$, the vector projection of $\vec{a}$ on $\vec{b}$ is $$\frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{(\sqrt{1^2 + 1^2})^2} (\hat{i} + \hat{j}) = \frac{5}{2} (\hat{i} + \hat{j}) = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j}$$.

3. Component of $\vec{a}$ Perpendicular to $\vec{b}$

This is the part of $\vec{a}$ that's "left over" after you've taken out the component that acts along $\vec{b}$. It's a vector perpendicular to $\vec{b}$.

Explanation: We simply subtract the vector projection from the original vector $\vec{a}$. This leaves us with a vector component that is orthogonal (perpendicular) to $\vec{b}$.

Example: Again, $\vec{a} = 2\hat{i} + 3\hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$. The component of $\vec{a}$ perpendicular to $\vec{b}$ is $$(2\hat{i} + 3\hat{j}) - (\frac{5}{2}\hat{i} + \frac{5}{2}\hat{j}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$$.

4. Work Done as Application of Dot Product

Work done is a classic application of the dot product and vector projection. If a force $\vec{F}$ causes a displacement $\vec{d}$, the work done is the component of the force along the direction of the displacement, multiplied by the magnitude of the displacement.

Explanation:

• $W$ is the work done (a scalar quantity).
• $\vec{F}$ is the force vector.
• $\vec{d}$ is the displacement vector.
• $\theta$ is the angle between the force and displacement vectors.

In other words, the work done is the scalar projection of the force vector onto the displacement vector, multiplied by the magnitude of the displacement. Or, it's the scalar projection of the displacement vector onto the force vector, multiplied by the magnitude of the force.

Example: A force $\vec{F} = 5\hat{i} + 2\hat{j}$ N acts on an object, causing a displacement of $\vec{d} = 3\hat{i} + \hat{j}$ m. The work done is $$(5\hat{i} + 2\hat{j}) \cdot (3\hat{i} + \hat{j}) = (5 \times 3) + (2 \times 1) = 15 + 2 = 17 \text{ Joules}$$.

Mastering vector projections is crucial for tackling a wide range of JEE problems. Keep practicing, visualize the concepts, and you'll be well on your way to success! Good luck!