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: Understanding the Concept:
We are given the magnitudes of three vectors and some relationships involving their dot products.
We need to find the magnitude of a linear combination of these vectors, which can be done by squaring the expression and expanding it using dot products.
Step 2: Key Formula or Approach:
1. Projection of vector \( \vec{a} \) on vector \( \vec{b} \) is \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
2. If two vectors are perpendicular, their dot product is zero.
3. The square of the magnitude of a vector is the dot product of the vector with itself: \( |\vec{a}|^2 = \vec{a} \cdot \vec{a} \).
Step 3: Detailed Explanation:
Given:
\( |\bar{u}| = 1 \), \( |\bar{v}| = 2 \), \( |\bar{w}| = 3 \).
Projection of \( \bar{v} \) along \( \bar{u} \) is \( \frac{\bar{v} \cdot \bar{u}}{|\bar{u}|} \). Since \( |\bar{u}| = 1 \), it simplifies to \( \bar{v} \cdot \bar{u} \).
Projection of \( \bar{w} \) along \( \bar{u} \) is \( \frac{\bar{w} \cdot \bar{u}}{|\bar{u}|} = \bar{w} \cdot \bar{u} \).
We are given that these projections are equal:
\[ \bar{v} \cdot \bar{u} = \bar{w} \cdot \bar{u} \quad \implies \quad \bar{u} \cdot \bar{v} = \bar{u} \cdot \bar{w} \] Also given, \( \bar{v} \) and \( \bar{w} \) are perpendicular, so:
\[ \bar{v} \cdot \bar{w} = 0 \] We need to find \( |\bar{u} - \bar{v} + \bar{w}| \). Let's square it:
\[ |\bar{u} - \bar{v} + \bar{w}|^2 = (\bar{u} - \bar{v} + \bar{w}) \cdot (\bar{u} - \bar{v} + \bar{w}) \] Expanding the dot product:
\[ = \bar{u} \cdot \bar{u} - \bar{u} \cdot \bar{v} + \bar{u} \cdot \bar{w} - \bar{v} \cdot \bar{u} + \bar{v} \cdot \bar{v} - \bar{v} \cdot \bar{w} + \bar{w} \cdot \bar{u} - \bar{w} \cdot \bar{v} + \bar{w} \cdot \bar{w} \] Using properties \( \bar{a} \cdot \bar{b} = \bar{b} \cdot \bar{a} \) and \( \bar{a} \cdot \bar{a} = |\bar{a}|^2 \):
\[ = |\bar{u}|^2 + |\bar{v}|^2 + |\bar{w}|^2 - 2(\bar{u} \cdot \bar{v}) + 2(\bar{u} \cdot \bar{w}) - 2(\bar{v} \cdot \bar{w}) \] Substitute the known values and relations:
\( |\bar{u}|^2 = 1^2 = 1 \)
\( |\bar{v}|^2 = 2^2 = 4 \)
\( |\bar{w}|^2 = 3^2 = 9 \)
\( \bar{u} \cdot \bar{w} = \bar{u} \cdot \bar{v} \), so \( - 2(\bar{u} \cdot \bar{v}) + 2(\bar{u} \cdot \bar{w}) = 0 \).
\( \bar{v} \cdot \bar{w} = 0 \).
Putting it all together:
\[ |\bar{u} - \bar{v} + \bar{w}|^2 = 1 + 4 + 9 + 0 - 2(0) \] \[ |\bar{u} - \bar{v} + \bar{w}|^2 = 14 \] Taking the square root (magnitude is always non-negative):
\[ |\bar{u} - \bar{v} + \bar{w}| = \sqrt{14} \] Step 4: Final Answer:
The value is \( \sqrt{14} \).