Q.Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

Concept: Vector Projection — the scalar projection of $\vec{a}$ onto $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

Let $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.

1. Compute the dot product:

   $\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0$.

2. The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

3. The projection of $\vec{a}$ onto $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{0}{\sqrt{2}} = 0$.

Concept First: What Does Projection Mean?

When we project one vector onto another, we are asking: how much of the first vector points in the direction of the second?

Think of a stick leaning against a wall. The shadow it casts on the floor is its projection onto the floor. Similarly, the projection of vector $\vec{a}$ onto vector $\vec{b}$ is the component of $\vec{a}$ that lies along $\vec{b}$.

The formula for the scalar projection (the signed length of the shadow) of $\vec{a}$ onto $\vec{b}$ is:

If you want the vector projection (the actual vector along $\vec{b}$), you multiply that scalar by the unit vector in the direction of $\vec{b}$:

Here, the problem asks for "the projection" — in standard Indian exam language, this means the scalar projection (the magnitude of the projection, with sign). Let's proceed.

Step-by-Step Solution

1. Identify the vectors

Let $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.

2. Compute the dot product

3. Interpret the dot product result

A dot product of zero means the vectors are perpendicular (orthogonal). When two vectors are at right angles, one has no component along the other — just like a vertical pole casts no shadow on a horizontal floor directly beneath it.

4. Apply the projection formula

The magnitude of $\vec{b}$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$, but since the numerator is zero, the result is simply $0$.

5. Final answer

The projection is zero. This means $\hat{i} - \hat{j}$ has no component along $\hat{i} + \hat{j}$.

Method: Scalar Projection of One Vector onto Another

Use this whenever you need "the projection of $\vec{a}$ on $\vec{b}$" — how much of $\vec{a}$ lies along $\vec{b}$.

Steps

Step 1: Identify which vector you project ONTO.

You project $\vec{a}$ onto $\vec{b}$, so $\vec{b}$'s length goes in the denominator. Getting this right decides the whole formula.

Step 2: Compute the dot product.

Step 3: Apply the scalar-projection formula.

Divide by $|\vec{b}|$ (the vector projected onto), not $|\vec{a}|$.

Step 4: Interpret the result.

A value of $0$ means $\vec{a}\perp\vec{b}$ (no component along $\vec{b}$); a negative value means $\vec{a}$ leans opposite to $\vec{b}$. The sign carries meaning — keep it.

Common Mistakes

Mistake 1: Dividing by $|\vec{a}|$ instead of $|\vec{b}|$.

Why it's wrong: projecting onto $\vec{b}$ means $|\vec{b}|$ is the denominator. Correct approach: for the projection of $\vec a$ on $\vec b$, use $\tfrac{\vec a\cdot\vec b}{|\vec b|}$.

Mistake 2: Sign slip on the $-\hat{j}$ component.

Why it's wrong: $\vec{a}\cdot\vec{b}=(1)(1)+(-1)(1)=0$; treating $-\hat{j}$ as $+\hat{j}$ gives a non-zero, wrong projection. Correct approach: carry the negative sign, which here makes the dot product exactly $0$.

Mistake 3: Not recognising that a zero dot product gives projection $0$.

Why it's wrong: perpendicular vectors have no shadow along each other, so the projection is $0$ regardless of $|\vec b|$. Correct approach: once $\vec a\cdot\vec b=0$, conclude the projection is $0$.