Let `vecu,vecv,vecw` be such that `|vecu|=1,|vecv|=2,|vecw|3`. If the projection of `vecv along vecu` is equal to that of `vecw along vecv,vecw` are perpendicular to each other then `|vecu-vecv+vecw|` equals (A) 2 (B) `sqrt(7)` (C) `sqrt(14)` (D) `14`

Options

A. 2
B. `sqrt7`
C. `sqrt14`
D. 14

Explanation

To solve the problem step by step, we will use the properties of vectors and their projections.

### Step 1: Understanding the Given Information
We have three vectors:
- \( \vec{u} \) with \( |\vec{u}| = 1 \)
- \( \vec{v} \) with \( |\vec{v}| = 2 \)
- \( \vec{w} \) with \( |\vec{w}| = 3 \)

We are given that the projection of \( \vec{v} \) along \( \vec{u} \) is equal to the projection of \( \vec{w} \) along \( \vec{u} \), and that \( \vec{v} \) and \( \vec{w} \) are perpendicular to each other.

### Step 2: Setting Up the Projection Condition
The projection of \( \vec{v} \) along \( \vec{u} \) is given by:
\[ \text{proj}_{\vec{u}} \vec{v} = \frac{\vec{v} \cdot \vec{u}}{|\vec{u}|^2} \vec{u} = \vec{v} \cdot \vec{u} \cdot \vec{u} \quad (\text{since } |\vec{u}|^2 = 1) \]

Similarly, the projection of \( \vec{w} \) along \( \vec{u} \) is:
\[ \text{proj}_{\vec{u}} \vec{w} = \frac{\vec{w} \cdot \vec{u}}{|\vec{u}|^2} \vec{u} = \vec{w} \cdot \vec{u} \cdot \vec{u} \]

Since these projections are equal:
\[ \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u} \]

### Step 3: Using the Perpendicular Condition
Since \( \vec{v} \) and \( \vec{w} \) are perpendicular, we have:
\[ \vec{v} \cdot \vec{w} = 0 \]

### Step 4: Finding the Magnitude of \( |\vec{u} - \vec{v} + \vec{w}| \)
We need to calculate:
\[ |\vec{u} - \vec{v} + \vec{w}| \]

Using the formula for the magnitude of a vector:
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) - 2(\vec{v} \cdot \vec{w}) \]

Substituting the magnitudes:
- \( |\vec{u}|^2 = 1^2 = 1 \)
- \( |\vec{v}|^2 = 2^2 = 4 \)
- \( |\vec{w}|^2 = 3^2 = 9 \)

So we have:
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = 1 + 4 + 9 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) - 2(0) \]
\[ = 14 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) \]

### Step 5: Simplifying the Expression
Since \( \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u} \), we can denote this common value as \( k \):
\[ \vec{u} \cdot \vec{v} = k \quad \text{and} \quad \vec{u} \cdot \vec{w} = k \]
Thus:
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = 14 - 2k + 2k = 14 \]

### Step 6: Final Calculation
Taking the square root gives:
\[ |\vec{u} - \vec{v} + \vec{w}| = \sqrt{14} \]

### Conclusion
Thus, the answer is:
\[ \boxed{\sqrt{14}} \]