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 $4\hat{i}+3\hat{j}-2\hat{k}$.
Prev Next

AI Tutor Explanation

Powered by Gemini

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

AI generated content. Review strictly for academic accuracy.

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.

More from this Chapter

MCQ_SINGLE
If \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), \(|\vec{a}| = \sqrt{37}\), \(|\vec{b}| = 3\) and \(|\vec{c}| = 4\), then the angle between \(\vec{b}\) and \(\vec{c}\) is
MCQ_SINGLE
The position vectors of points P and Q are \(\vec{p}\) and \(\vec{q}\) respectively. The point R divides line segment PQ in the ratio 3:1 and S is the mid-point of line segment PR. The position vector of S is:
MCQ_SINGLE
In $\Delta ABC$, $\vec{AB}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{AC}=3\hat{i}-\hat{j}+4\hat{k}$. If D is mid-point of BC, then vector $\vec{AD}$ is equal to :
MCQ_SINGLE
$\vec{a}$ and $\vec{b}$ are two non-zero vectors such that the projection of $\vec{a}$ on $\vec{b}$ is 0. The angle between $\vec{a}$ and $\vec{b}$:
SA
If $\vec{a}$ and $\vec{b}$ are unit vectors inclined with each other at an angle $\theta$, then prove that $\frac{1}{2}|\vec{a}-\vec{b}|=\sin\frac{\theta}{2}$.
View All Questions