Let ` vec u , vec va n d vec w` be such that `| vec u|=1,| vec v|=2a n d| vec w|=3.` If the projection of ` vec v` along ` vec u` is equal to that of ` vec w` along ` vec u` and vectors ` vec va n d vec w` are perpendicular to each other, then `| vec u- vec v+ vec w|` equals `2` b. `sqrt(7)` c. `sqrt(14)` d. `14`

Options

A. `2`
B. `sqrt(7)`
C. `sqrt(14)`
D. `14`

Answer

To solve the problem step by step, we will analyze the given information and apply the relevant mathematical concepts.

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

We know:
1. The projection of \( \vec{v} \) along \( \vec{u} \) is equal to the projection of \( \vec{w} \) along \( \vec{u} \).
2. Vectors \( \vec{v} \) and \( \vec{w} \) are perpendicular to each other.

### Step 2: Set Up the Projection Equations
The projection of a vector \( \vec{a} \) along another vector \( \vec{b} \) is given by:
\[ \text{Projection of } \vec{a} \text{ along } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

For our vectors:
- Projection of \( \vec{v} \) along \( \vec{u} \):
\[ \text{Proj}_{\vec{u}} \vec{v} = \frac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \vec{v} \cdot \vec{u} \quad (\text{since } |\vec{u}| = 1) \]

- Projection of \( \vec{w} \) along \( \vec{u} \):
\[ \text{Proj}_{\vec{u}} \vec{w} = \frac{\vec{w} \cdot \vec{u}}{|\vec{u}|} = \vec{w} \cdot \vec{u} \quad (\text{since } |\vec{u}| = 1) \]

### Step 3: Set the Projections Equal
Since the projections are equal:
\[ \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u} \]

### Step 4: Use the Perpendicular Condition
Since \( \vec{v} \) and \( \vec{w} \) are perpendicular:
\[ \vec{v} \cdot \vec{w} = 0 \]

### Step 5: Calculate the Magnitude of \( |\vec{u} - \vec{v} + \vec{w}| \)
We need to find:
\[ |\vec{u} - \vec{v} + \vec{w}| \]

Using the formula for the magnitude of a vector:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

Let \( \vec{a} = \vec{u}, \vec{b} = -\vec{v}, \vec{c} = \vec{w} \):
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot (-\vec{v}) + (-\vec{v}) \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]

### Step 6: Substitute the Magnitudes and Dot Products
Substituting the known magnitudes:
- \( |\vec{u}|^2 = 1^2 = 1 \)
- \( |\vec{v}|^2 = 2^2 = 4 \)
- \( |\vec{w}|^2 = 3^2 = 9 \)

Thus:
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = 1 + 4 + 9 + 2(-\vec{u} \cdot \vec{v} - \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]

### Step 7: Simplify the Expression
Since \( \vec{v} \cdot \vec{w} = 0 \) (perpendicularity):
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = 14 + 2(-\vec{u} \cdot \vec{v} + \vec{w} \cdot \vec{u}) \]

### Step 8: Recognize Equal Projections
From \( \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u} \), we can denote \( \vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w} = k \):
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = 14 + 2(-k + k) = 14 \]

### Step 9: Final Calculation
Thus:
\[ |\vec{u} - \vec{v} + \vec{w}|^2 = 14 \implies |\vec{u} - \vec{v} + \vec{w}| = \sqrt{14} \]

### Conclusion
The magnitude \( |\vec{u} - \vec{v} + \vec{w}| \) equals \( \sqrt{14} \).