JEE PYQ: Vector Algebra - Question ID 0395fc890c7c (JEE Main 2022)

In vector algebra, the projection of a vector $\vec{A}$ onto another vector $\vec{B}$ is a scalar quantity given by: $\text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$ However, the problem states that the projection is zero. This implies that the numerator of the projection formula, i.e., the dot product $\vec{A} \cdot \vec{B}$, must be zero. This is because the magnitude $|\vec{B}|$ is always positive and cannot make the projection zero unless the dot product itself is zero.

The dot product of two vectors $\vec{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{B} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ is: $\vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 + a_3 b_3$ If $\vec{A} \cdot \vec{B} = 0$, the vectors are perpendicular (orthogonal) to each other.

Given: $\vec{A} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ $\vec{B} = \hat{i} + 2 \hat{j} + \alpha \hat{k}$ The projection of $\vec{A}$ on $\vec{B}$ is zero, so: $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 0 \implies \vec{A} \cdot \vec{B} = 0$

Compute the dot product $\vec{A} \cdot \vec{B}$: $\vec{A} \cdot \vec{B} = (2)(1) + (4)(2) + (-2)(\alpha) = 2 + 8 - 2\alpha$ Set the dot product to zero: $2 + 8 - 2\alpha = 0$ Simplify: $10 - 2\alpha = 0 \implies 2\alpha = 10 \implies \alpha = 5$

1. Misinterpreting the projection formula: Some students confuse the projection formula with the component of a vector along another. Remember, the projection is $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$, not $\frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$ or $|\vec{A}| \cos \theta$.
2. Ignoring the condition for zero projection: Students often forget that the projection is zero only if the dot product is zero. They might incorrectly set the magnitude $|\vec{B}|$ to zero, which is impossible since magnitudes are always non-negative.
3. Sign errors in dot product calculation: A common mistake is misplacing the sign of the $\hat{k}$ component, leading to incorrect values for $\alpha$. Always double-check the signs of each component.
4. Overcomplicating the problem: Some students introduce unnecessary steps, such as calculating the angle between vectors or the magnitude of $\vec{B}$. Stick to the condition $\vec{A} \cdot \vec{B} = 0$ for simplicity.

Exam Tip: For problems involving projections or orthogonality, always start by writing the dot product condition. This simplifies the problem and avoids unnecessary calculations.