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: Understanding the Concept:
The projection of \(\bar v\) along \(\bar u\) is \(\frac{\bar u\cdot\bar v}{|\bar u|}\). Since \(|\bar u| = 1\), equal projections mean \(\bar u\cdot\bar v = \bar u\cdot\bar w\).

Step 2: Key Formula or Approach:
\(|\bar u - \bar v + \bar w|^2 = |\bar u|^2 + |\bar v|^2 + |\bar w|^2 - 2\bar u\cdot\bar v + 2\bar u\cdot\bar w - 2\bar v\cdot\bar w\).

Step 3: Detailed Explanation:
Since \(\bar u\cdot\bar v = \bar u\cdot\bar w\), the terms \(-2\bar u\cdot\bar v + 2\bar u\cdot\bar w\) cancel.
Since \(\bar v\perp\bar w\), \(\bar v\cdot\bar w = 0\).
\[ |\bar u - \bar v + \bar w|^2 = 1 + 4 + 9 = 14 \]
So the magnitude is \(\sqrt{14}\).

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