Question:
A unit vector coplanar with \(\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) is
A unit vector coplanar with \(\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) is
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Use linear combination and orthogonality conditions.
Updated On: Apr 23, 2026
- \(\frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}}\)
- \(\frac{\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{3}}\)
- \(\frac{\mathbf{i} + \mathbf{j} + 2\mathbf{k}}{\sqrt{6}}\)
- \(\frac{-\mathbf{j} + 2\mathbf{k}}{\sqrt{5}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \mathbf{v} = \alpha(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + \beta(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \]
Step 2: Calculation / Simplification}
\(\mathbf{v} = (\alpha+\beta)\mathbf{i} + (\alpha+2\beta)\mathbf{j} + (2\alpha+\beta)\mathbf{k}\)
Perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\):
\((\alpha+\beta) + (\alpha+2\beta) + (2\alpha+\beta) = 0\)
\(4\alpha + 4\beta = 0 \Rightarrow \alpha = -\beta\)
\(\mathbf{v} = 0\mathbf{i} + (-\beta+2\beta)\mathbf{j} + (-2\beta+\beta)\mathbf{k} = \beta(\mathbf{j} - \mathbf{k})\)
Unit vector: \(|\mathbf{v}| = |\beta|\sqrt{2} = 1 \Rightarrow \beta = \pm\frac{1}{\sqrt{2}}\)
\(\mathbf{v} = \pm \frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}}\)
Step 3: Final Answer
\[ \frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}} \]
\[ \mathbf{v} = \alpha(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + \beta(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \]
Step 2: Calculation / Simplification}
\(\mathbf{v} = (\alpha+\beta)\mathbf{i} + (\alpha+2\beta)\mathbf{j} + (2\alpha+\beta)\mathbf{k}\)
Perpendicular to \(\mathbf{i} + \mathbf{j} + \mathbf{k}\):
\((\alpha+\beta) + (\alpha+2\beta) + (2\alpha+\beta) = 0\)
\(4\alpha + 4\beta = 0 \Rightarrow \alpha = -\beta\)
\(\mathbf{v} = 0\mathbf{i} + (-\beta+2\beta)\mathbf{j} + (-2\beta+\beta)\mathbf{k} = \beta(\mathbf{j} - \mathbf{k})\)
Unit vector: \(|\mathbf{v}| = |\beta|\sqrt{2} = 1 \Rightarrow \beta = \pm\frac{1}{\sqrt{2}}\)
\(\mathbf{v} = \pm \frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}}\)
Step 3: Final Answer
\[ \frac{\mathbf{j} - \mathbf{k}}{\sqrt{2}} \]
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