Question

Let \( \overline{\text{u}}, \overline{\text{v}}, \overline{\text{w}} \) be the vectors such that \( |\overline{\text{u}}| = 1, |\overline{\text{v}}| = 2, |\overline{\text{w}}| = 3 \). If the projection of \( \overline{\text{v}} \) along \( \overline{\text{u}} \) is equal to that of \( \overline{\text{w}} \) along \( \overline{\text{u}} \) and the vectors \( \overline{\text{v}}, \overline{\text{w}} \) are perpendicular to each other then \( |\overline{\text{u}} - \overline{\text{v}} + \overline{\text{w}}| \) equals

• A. \( \sqrt{14} \)
• B. \( 14 \)
• C. \( \sqrt{7} \)
• D. \( 2 \)

Hint:

Expanding the squared magnitude of a sum of vectors helps identify terms that cancel or become zero.

Answer 1

The Correct Option is A

Step 1: Concept Projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\). Perpendicular vectors have dot product zero.

Step 2: Meaning Given \(\vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u}\) (since \(|\vec{u}|=1\)) and \(\vec{v} \cdot \vec{w} = 0\).

Step 3: Analysis \(|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) - 2(\vec{v} \cdot \vec{w})\)
Substitute given values: \(= 1^2 + 2^2 + 3^2 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{v}) - 0\) \(= 1 + 4 + 9 = 14\).

Step 4: Conclusion Taking square root, the value is \(\sqrt{14}\). Final Answer: (A)