Class CBSE Class 12 Mathematics Vector Algebra Q #1425
KNOWLEDGE BASED
REMEMBER
2 Marks 2025 AISSCE(Board Exam) VSA
Find a vector of magnitude 5 which is perpendicular to both the vectors $3\hat{i}-2\hat{j}+\hat{k}$ and $4\hat{i}+3\hat{j}-2\hat{k}$.
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Detailed Solution

Step 1: Find a vector perpendicular to both given vectors

Let $\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} + 3\hat{j} - 2\hat{k}$. A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a} \times \vec{b}$.

Step 2: Calculate the cross product

We compute the cross product as follows: $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 4 & 3 & -2 \end{vmatrix} $$ $$ \vec{a} \times \vec{b} = \hat{i}((-2)(-2) - (1)(3)) - \hat{j}((3)(-2) - (1)(4)) + \hat{k}((3)(3) - (-2)(4)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(4 - 3) - \hat{j}(-6 - 4) + \hat{k}(9 + 8) $$ $$ \vec{a} \times \vec{b} = \hat{i}(1) - \hat{j}(-10) + \hat{k}(17) $$ $$ \vec{a} \times \vec{b} = \hat{i} + 10\hat{j} + 17\hat{k} $$

Step 3: Find the magnitude of the cross product

The magnitude of the vector $\vec{a} \times \vec{b}$ is: $$ |\vec{a} \times \vec{b}| = \sqrt{(1)^2 + (10)^2 + (17)^2} = \sqrt{1 + 100 + 289} = \sqrt{390} $$

Step 4: Find the unit vector in the direction of the cross product

The unit vector in the direction of $\vec{a} \times \vec{b}$ is: $$ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}} $$

Step 5: Find the vector of magnitude 5 in the direction of the cross product

A vector of magnitude 5 in the direction of $\hat{n}$ is: $$ \vec{v} = 5\hat{n} = 5\left(\frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}}\right) = \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k}) $$ Since the question asks for a vector perpendicular to both given vectors, the vector can also be in the opposite direction. Therefore, the required vector is $\pm \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k})$.

Final Answer: $\pm \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k})$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because the student needs to recall the formula for the cross product of two vectors and the formula for finding a unit vector.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (calculating the cross product and normalizing the resulting vector) 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 of vector algebra concepts, specifically the cross product and its application in finding perpendicular vectors.

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