MHT CET · Maths · Vector Algebra
Let \(\overline{\mathrm{u}}, \overline{\mathrm{v}}, \overline{\mathrm{w}}\) be the vectors such that \(|\overline{\mathrm{u}}|=1,|\overline{\mathrm{v}}|=2,|\overline{\mathrm{w}}|=3\). If the projection \(\overline{\mathrm{v}}\) along \(\overline{\mathrm{u}}\) is equal to that of \(\overline{\mathrm{w}}\) along \(\overline{\mathrm{u}}\) and the vectors \(\overline{\mathrm{v}}\), \(\overline{\mathrm{w}}\) are perpendicular to each other then \(|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|\) equals
- A \(\sqrt{14}\)
- B \(14\)
- C \(\sqrt{7}\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(\sqrt{14}\)
Step-by-step Solution
Detailed explanation
\((\overline{\mathrm{v}} \cdot \overline{\mathrm{u}})/|\overline{\mathrm{u}}| = (\overline{\mathrm{w}} \cdot \overline{\mathrm{u}})/|\overline{\mathrm{u}}| \implies \overline{\mathrm{v}} \cdot \overline{\mathrm{u}} = \overline{\mathrm{w}} \cdot \overline{\mathrm{u}}\) \(\overline{\mathrm{v}} \cdot \overline{\mathrm{w}} = 0\)
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