Class CBSE Class 12 Mathematics Vector Algebra Q #925
KNOWLEDGE BASED
APPLY
2 Marks 2023 VSA
24. If the projection of the vector $\hat{i}+\hat{j}+\hat{k}$ on the vector $p\hat{i}+\hat{j}-2\hat{k}$ is $\frac{1}{3}$, then find the value(s) of $p$.
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Step-by-Step Solution

  1. Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = p\hat{i} + \hat{j} - 2\hat{k}$.
  2. The projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
  3. We are given that the projection is $\frac{1}{3}$. Therefore, $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{1}{3}$.
  4. Calculate the dot product $\vec{a} \cdot \vec{b} = (1)(p) + (1)(1) + (1)(-2) = p + 1 - 2 = p - 1$.
  5. Calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{p^2 + 1^2 + (-2)^2} = \sqrt{p^2 + 1 + 4} = \sqrt{p^2 + 5}$.
  6. Substitute these values into the projection formula: $\frac{p - 1}{\sqrt{p^2 + 5}} = \frac{1}{3}$.
  7. Cross-multiply: $3(p - 1) = \sqrt{p^2 + 5}$.
  8. Square both sides: $(3(p - 1))^2 = (\sqrt{p^2 + 5})^2$, which simplifies to $9(p^2 - 2p + 1) = p^2 + 5$.
  9. Expand and rearrange: $9p^2 - 18p + 9 = p^2 + 5$, so $8p^2 - 18p + 4 = 0$.
  10. Divide by 2: $4p^2 - 9p + 2 = 0$.
  11. Factor the quadratic equation: $(4p - 1)(p - 2) = 0$.
  12. Solve for $p$: $4p - 1 = 0$ or $p - 2 = 0$. This gives $p = \frac{1}{4}$ or $p = 2$.

Correct Answer: p = 1/4, 2

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the formula for the projection of one vector onto another to solve for the unknown variable 'p'.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to execute a specific procedure (formula for projection) to arrive at the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of a standard formula from the textbook.

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