Question

If $\vec{a} = \hat{i} + \sqrt{11}\hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} + \sqrt{11}\hat{j} - 10\hat{k}$ are two vectors then the component of $\vec{b}$ perpendicular to $\vec{a}$ is

• A. $3\hat{i} - \sqrt{11}\hat{j} - 4\hat{k}$
• B. $\hat{i} - \sqrt{11}\hat{j} - 5\hat{k}$
• C. $-(\hat{i} + \sqrt{11}\hat{j} + 6\hat{k})$
• D. $-5\hat{i} + \sqrt{11}\hat{j} + 3\hat{k}$

Hint:

Resolving a vector into parallel and perpendicular components is a fundamental vector operation. A good way to check your answer is to compute the dot product of your perpendicular component and the vector $\vec{a}$. The result should be zero. Here, $\vec{b}_{\perp} \cdot \vec{a} = (-\hat{i} - \sqrt{11}\hat{j} - 6\hat{k}) \cdot (\hat{i} + \sqrt{11}\hat{j} - 2\hat{k}) = -1 - 11 + 12 = 0$. This confirms the result is correct.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Any vector $\vec{b}$ can be resolved into two components: one that is parallel to another vector $\vec{a}$, and one that is perpendicular to $\vec{a}$. The parallel component is the projection of $\vec{b}$ onto $\vec{a}$. The perpendicular component is found by subtracting the parallel component from the original vector $\vec{b}$.
Step 2: Key Formula or Approach
Let $\vec{b}_{\parallel}$ be the component of $\vec{b}$ parallel to $\vec{a}$, and $\vec{b}_{\perp}$ be the component perpendicular to $\vec{a}$. 1. The parallel component (vector projection) is given by: $\vec{b}_{\parallel} = \text{proj}_{\vec{a}}\vec{b} = \left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right)\vec{a}$. 2. The perpendicular component is then given by: $\vec{b}_{\perp} = \vec{b} - \vec{b}_{\parallel}$.
Step 3: Detailed Explanation
Given vectors: $\vec{a} = \hat{i} + \sqrt{11}\hat{j} - 2\hat{k}$ $\vec{b} = \hat{i} + \sqrt{11}\hat{j} - 10\hat{k}$ 1. Calculate the dot product $\vec{b \cdot \vec{a}$:} \[ \vec{b} \cdot \vec{a} = (1)(1) + (\sqrt{11})(\sqrt{11}) + (-10)(-2) = 1 + 11 + 20 = 32 \] 2. Calculate the squared magnitude of $\vec{a$, $|\vec{a}|^2$:} \[ |\vec{a}|^2 = (1)^2 + (\sqrt{11})^2 + (-2)^2 = 1 + 11 + 4 = 16 \] 3. Calculate the parallel component $\vec{b_{\parallel}$:} \[ \vec{b}_{\parallel} = \left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right)\vec{a} = \left(\frac{32}{16}\right)\vec{a} = 2\vec{a} \] \[ \vec{b}_{\parallel} = 2(\hat{i} + \sqrt{11}\hat{j} - 2\hat{k}) = 2\hat{i} + 2\sqrt{11}\hat{j} - 4\hat{k} \] 4. Calculate the perpendicular component $\vec{b_{\perp}$:} \[ \vec{b}_{\perp} = \vec{b} - \vec{b}_{\parallel} = (\hat{i} + \sqrt{11}\hat{j} - 10\hat{k}) - (2\hat{i} + 2\sqrt{11}\hat{j} - 4\hat{k}) \] \[ \vec{b}_{\perp} = (1-2)\hat{i} + (\sqrt{11}-2\sqrt{11})\hat{j} + (-10 - (-4))\hat{k} \] \[ \vec{b}_{\perp} = -\hat{i} - \sqrt{11}\hat{j} - 6\hat{k} \] Factoring out the negative sign gives: \[ \vec{b}_{\perp} = -(\hat{i} + \sqrt{11}\hat{j} + 6\hat{k}) \] Step 4: Final Answer
The component of vector $\vec{b}$ perpendicular to vector $\vec{a}$ is $-(\hat{i} + \sqrt{11}\hat{j} + 6\hat{k})$.