Q1347: Find a vector of magnitude 4 units perpendicular to each of ...

Class CBSE Class 12 Mathematics Vector Algebra Q #1347

COMPETENCY BASED

UNDERSTAND

3 Marks 2024 AISSCE(Board Exam) SA

Find a vector of magnitude 4 units perpendicular to each of the vectors $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$ and hence verify your answer.

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Detailed Solution

Step 1: Find a vector perpendicular to both given vectors

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$. 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: $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(1) - (-1)(1)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(1 - 1) - \hat{j}(-2 - 1) + \hat{k}(2 + 1) = 0\hat{i} + 3\hat{j} + 3\hat{k} $$ So, $\vec{a} \times \vec{b} = 3\hat{j} + 3\hat{k}$.

Step 3: Find the unit vector in the direction of the cross product

The magnitude of $\vec{a} \times \vec{b}$ is $|\vec{a} \times \vec{b}| = \sqrt{0^2 + 3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$. The unit vector in the direction of $\vec{a} \times \vec{b}$ is $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{3\hat{j} + 3\hat{k}}{3\sqrt{2}} = \frac{\hat{j} + \hat{k}}{\sqrt{2}}$.

Step 4: Find the vector of magnitude 4 units

We want a vector of magnitude 4 in the direction of $\hat{n}$. This is given by $4\hat{n} = 4\left(\frac{\hat{j} + \hat{k}}{\sqrt{2}}\right) = \frac{4}{\sqrt{2}}(\hat{j} + \hat{k}) = 2\sqrt{2}(\hat{j} + \hat{k}) = 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}$. Also, the vector can be in the opposite direction, so we also have $-4\hat{n} = -2\sqrt{2}\hat{j} - 2\sqrt{2}\hat{k}$.

Step 5: Verify the answer

Let $\vec{v} = 2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}$. We need to verify that $\vec{v}$ is perpendicular to both $\vec{a}$ and $\vec{b}$. This means that $\vec{v} \cdot \vec{a} = 0$ and $\vec{v} \cdot \vec{b} = 0$. $$ \vec{v} \cdot \vec{a} = (0)(2) + (2\sqrt{2})(-1) + (2\sqrt{2})(1) = -2\sqrt{2} + 2\sqrt{2} = 0 $$ $$ \vec{v} \cdot \vec{b} = (0)(1) + (2\sqrt{2})(1) + (2\sqrt{2})(-1) = 2\sqrt{2} - 2\sqrt{2} = 0 $$ Thus, $\vec{v}$ is perpendicular to both $\vec{a}$ and $\vec{b}$. The same verification holds for $-2\sqrt{2}\hat{j} - 2\sqrt{2}\hat{k}$.

Final Answer: $2\sqrt{2}\hat{j} + 2\sqrt{2}\hat{k}$ and $-2\sqrt{2}\hat{j} - 2\sqrt{2}\hat{k}$

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concept of cross product and its geometric interpretation to find a vector perpendicular to two given vectors. They also need to understand how to find a unit vector and scale it to achieve a desired magnitude.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific procedure (calculating the cross product, finding the magnitude, and scaling the vector) to arrive at the solution. It involves knowing the steps to find a vector perpendicular to two given vectors.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of vectors and cross products to solve a problem, rather than just recalling definitions or formulas.