JEE PYQ: Vector Algebra - Question ID 38ce486d407d (JEE Main 2022)

ID: 38ce486d407dJEE Main 2022Numerical Value

If A=(2i^+3j^k^)m\vec{A}=(2 \hat{i}+3 \hat{j}-\hat{k})\, \mathrm{m} and B=(i^+2j^+2k^)m\vec{B}=(\hat{i}+2 \hat{j}+2 \hat{k}) \,\mathrm{m}. The magnitude of component of vector A\vec{A} along vector B\vec{B} will be ____________ m\mathrm{m}.

Your Answer

Step-by-step Explanation

Core Formula & Concept:

To find the magnitude of the component of vector A\vec{A} along vector B\vec{B}, we use the concept of the projection of one vector onto another. The key formula is:

Component of A along B=ABB\text{Component of } \vec{A} \text{ along } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}

Here:

  • AB\vec{A} \cdot \vec{B} is the dot product of vectors A\vec{A} and B\vec{B}.
  • B|\vec{B}| is the magnitude of vector B\vec{B}.
This formula arises because the dot product AB=ABcosθ\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta, where θ\theta is the angle between the vectors. Dividing by B|\vec{B}| isolates the component of A\vec{A} in the direction of B\vec{B}.

Step-by-Step Derivation:

Given: A=(2i^+3j^k^)m,B=(i^+2j^+2k^)m\vec{A} = (2\hat{i} + 3\hat{j} - \hat{k})\, \text{m}, \quad \vec{B} = (\hat{i} + 2\hat{j} + 2\hat{k})\, \text{m} Step 1: Compute the Dot Product AB\vec{A} \cdot \vec{B}
The dot product is calculated as: AB=AxBx+AyBy+AzBz\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z Substitute the components: AB=(2)(1)+(3)(2)+(1)(2)=2+62=6m2\vec{A} \cdot \vec{B} = (2)(1) + (3)(2) + (-1)(2) = 2 + 6 - 2 = 6\, \text{m}^2 Step 2: Compute the Magnitude of B\vec{B}
The magnitude of B\vec{B} is: B=Bx2+By2+Bz2=12+22+22=1+4+4=9=3m|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\, \text{m} Step 3: Calculate the Component of A\vec{A} along B\vec{B}
Using the formula: Component of A along B=ABB=63=2m\text{Component of } \vec{A} \text{ along } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = \frac{6}{3} = 2\, \text{m} Final Answer: The magnitude of the component of A\vec{A} along B\vec{B} is 2 m.

Common Traps & Exam Tip:

  1. Confusing Dot Product with Cross Product: Students often mistakenly use the cross product formula instead of the dot product. Remember, the component along a vector requires the dot product, not the cross product.
  2. Incorrect Magnitude Calculation: Forgetting to take the square root or misapplying the formula for magnitude (e.g., adding components directly) leads to errors. Always verify the magnitude calculation.
  3. Sign Errors in Components: The negative sign in A\vec{A}'s k^\hat{k} component is easy to overlook. Double-check all signs when computing the dot product.
  4. Unit Mismatch: Ensure all components are in the same units (here, meters) to avoid dimensional inconsistencies in the final answer.
Exam Tip: For projection problems, always write down the formula first and then substitute values systematically. This reduces errors and makes your solution easier to review.