Question

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

• A. \(4\)
• B. \(\sqrt{7}\)
• C. \(2\)
• D. \(\sqrt{14}\)

Hint:

Square the magnitude and use the dot products; the unknown ones cancel.

Answer 1

The Correct Option is D

Step 1: Group the terms:
Write $\bar u - \bar v + \bar w = \bar u + (\bar w - \bar v)$.

Step 2: Compute:
$|\bar w - \bar v|^2 = 9 + 4 - 0 = 13$ (perpendicular vectors).
$\bar u\cdot(\bar w - \bar v) = \bar u\cdot\bar w - \bar u\cdot\bar v = 0$.
So $|\bar u + (\bar w - \bar v)|^2 = 1 + 13 + 0 = 14$.

Final Answer:
The magnitude is $\sqrt{14}$, option (D). \[ \boxed{\sqrt{14}} \]