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: Formula for Perpendicular Component: The component of vector \( \bar{b} \) perpendicular to \( \bar{a} \) is given by: \[ \bar{b}_{\perp} = \bar{b} - \text{proj}_{\bar{a}}\bar{b} = \bar{b} - \left( \frac{\bar{b} \cdot \bar{a}}{|\bar{a}|^2} \right) \bar{a} \]
Step 2: Calculate Dot Product and Magnitude Squared: \[ \bar{b} \cdot \bar{a} = (1)(1) + (\sqrt{11})(\sqrt{11}) + (-10)(-2) = 1 + 11 + 20 = 32 \] \[ |\bar{a}|^2 = (1)^2 + (\sqrt{11})^2 + (-2)^2 = 1 + 11 + 4 = 16 \]
Step 3: Calculate Projection Vector: \[ \text{proj}_{\bar{a}}\bar{b} = \left( \frac{32}{16} \right) \bar{a} = 2\bar{a} = 2(\bar{i} + \sqrt{11}\bar{j} - 2\bar{k}) = 2\bar{i} + 2\sqrt{11}\bar{j} - 4\bar{k} \]
Step 4: Calculate Perpendicular Component: \[ \bar{b}_{\perp} = \bar{b} - 2\bar{a} \] \[ \bar{b}_{\perp} = (\bar{i} + \sqrt{11}\bar{j} - 10\bar{k}) - (2\bar{i} + 2\sqrt{11}\bar{j} - 4\bar{k}) \] \[ \bar{b}_{\perp} = (1-2)\bar{i} + (\sqrt{11}-2\sqrt{11})\bar{j} + (-10+4)\bar{k} \] \[ \bar{b}_{\perp} = -\bar{i} - \sqrt{11}\bar{j} - 6\bar{k} \] \[ \bar{b}_{\perp} = -(\bar{i} + \sqrt{11}\bar{j} + 6\bar{k}) \]