Question

If $\mathbf{a} = \hat{i} + \hat{j} - \hat{k}$, $\mathbf{b} = \hat{i} - \hat{j} + \hat{k}$, and $\mathbf{c}$ is a unit vector perpendicular to $\mathbf{a}$ and coplanar with $\mathbf{a}$ and $\mathbf{b}$, then the unit vector $\mathbf{d}$ perpendicular to both $\mathbf{a}$ and $\mathbf{c}$ is

• A. ±1/√6(2j^-j^+k^)
• B. ±1/√2(j^+k^)
• C. ±1/√6(i^-2j^+k^)
• D. ±1/√2(j^-k^)

Answer 1

The Correct Option is B

Step 1: Since \( \mathbf{c} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \), it can be written as: \[ \mathbf{c} = \alpha \mathbf{a} + \beta \mathbf{b} \] Also, \( \mathbf{c} \perp \mathbf{a} \), so: \[ \mathbf{c} \cdot \mathbf{a} = 0 \Rightarrow (\alpha \mathbf{a} + \beta \mathbf{b}) \cdot \mathbf{a} = 0 \Rightarrow \alpha (\mathbf{a} \cdot \mathbf{a}) + \beta (\mathbf{b} \cdot \mathbf{a}) = 0 \]

Step 2: Compute the dot products. \[ \mathbf{a} = (1, 1, -1), \quad \mathbf{b} = (1, -1, 1) \] \[ \mathbf{a} \cdot \mathbf{a} = 1^2 + 1^2 + (-1)^2 = 3, \quad \mathbf{a} \cdot \mathbf{b} = 1 \cdot 1 + 1 \cdot (-1) + (-1) \cdot 1 = 1 - 1 -1 = -1 \]

Substitute into the condition: \[ \alpha \cdot 3 + \beta \cdot (-1) = 0 \Rightarrow 3\alpha - \beta = 0 \Rightarrow \beta = 3\alpha \]

So: \[ \mathbf{c} = \alpha \mathbf{a} + 3\alpha \mathbf{b} = \alpha (\mathbf{a} + 3\mathbf{b}) \]

Compute \( \mathbf{a} + 3\mathbf{b} \): \[ \mathbf{a} + 3\mathbf{b} = (1,1,-1) + 3(1,-1,1) = (1+3, 1-3, -1+3) = (4, -2, 2) \] So: \[ \mathbf{c} = \alpha (4\hat{i} - 2\hat{j} + 2\hat{k}) \]

Step 3: Make \( \mathbf{c} \) a unit vector. \[ |\mathbf{c}| = |\alpha| \cdot \sqrt{4^2 + (-2)^2 + 2^2} = |\alpha| \cdot \sqrt{16 + 4 + 4} = |\alpha| \cdot \sqrt{24} = |\alpha| \cdot 2\sqrt{6} \Rightarrow |\alpha| = \frac{1}{2\sqrt{6}} \]

So: \[ \mathbf{c} = \frac{1}{2\sqrt{6}} (4\hat{i} - 2\hat{j} + 2\hat{k}) = \frac{1}{\sqrt{6}} (2\hat{i} - \hat{j} + \hat{k}) \]

Step 4: Now find vector \( \mathbf{d} \) perpendicular to both \( \mathbf{a} \) and \( \mathbf{c} \):
Use cross product: \[ \mathbf{d} = \mathbf{a} \times \mathbf{c} \] \[ \mathbf{a} = (1,1,-1), \quad \mathbf{c} = \frac{1}{\sqrt{6}}(2, -1, 1) \]

Compute cross product: \[ \mathbf{a} \times \mathbf{c} = \frac{1}{\sqrt{6}} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & -1 & 1 \end{vmatrix} = \frac{1}{\sqrt{6}} \left[ \hat{i}(1 \cdot 1 - (-1)(-1)) - \hat{j}(1 \cdot 1 - (-1)(2)) + \hat{k}(1 \cdot (-1) - 1 \cdot 2) \right] \] \[ = \frac{1}{\sqrt{6}} \left[ \hat{i}(1 - 1) - \hat{j}(1 + 2) + \hat{k}(-1 - 2) \right] = \frac{1}{\sqrt{6}} \left[ 0\hat{i} - 3\hat{j} - 3\hat{k} \right] = \frac{-3}{\sqrt{6}} (\hat{j} + \hat{k}) \]

Now normalize: \[ \mathbf{d} = \pm \frac{1}{\sqrt{2}} (\hat{j} + \hat{k}) \]

Answer:

\[ \boxed{ \pm \frac{1}{\sqrt{2}} (\hat{j} + \hat{k}) } \]