Find a unit vector `vecc` if `-hati+hatj-hatk` bisects the angle between vectors `vecc` and `3hati+4hatj`.

Options

A. `1/15(11hati + 10hatj + 2hatk)`
B. `-1/15(11hati -10hatj + 2hatk)`
C. `-1/15(11hati + 10 hatj -2hatk)`
D. `-1/15(11hati + 10hatj + 2hatk)`

Answer

To find the unit vector \(\vec{c}\) that bisects the angle between the vector \(-\hat{i} + \hat{j} - \hat{k}\) and the vector \(3\hat{i} + 4\hat{j}\), we can follow these steps:

### Step 1: Define the unit vector \(\vec{c}\)
Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\). Since \(\vec{c}\) is a unit vector, we have the condition:
\[ x^2 + y^2 + z^2 = 1 \]

### Step 2: Find the unit vector in the direction of \(3\hat{i} + 4\hat{j}\)
The magnitude of \(3\hat{i} + 4\hat{j}\) is:
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]
Thus, the unit vector in the direction of \(3\hat{i} + 4\hat{j}\) is:
\[ \hat{a} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \]

### Step 3: Use the angle bisector theorem
According to the angle bisector theorem, we can express the bisector as:
\[ -\hat{i} + \hat{j} - \hat{k} = t \left( \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \right) + \vec{c} \]
where \(t\) is an arbitrary constant.

### Step 4: Rearranging the equation
Rearranging gives:
\[ -\hat{i} + \hat{j} - \hat{k} - \vec{c} = t \left( \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \right) \]
This implies:
\[ -\hat{i} + \hat{j} - \hat{k} = t \frac{3}{5} \hat{i} + t \frac{4}{5} \hat{j} + x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 5: Collecting like terms
Collecting coefficients for \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):
1. For \(\hat{i}\):
\[ -1 = t \frac{3}{5} + x \quad \text{(1)} \]
2. For \(\hat{j}\):
\[ 1 = t \frac{4}{5} + y \quad \text{(2)} \]
3. For \(\hat{k}\):
\[ -1 = z \quad \text{(3)} \]

### Step 6: Solve for \(x\), \(y\), and \(z\)
From equation (3):
\[ z = -1 \]

Substituting \(z = -1\) into the unit vector condition:
\[ x^2 + y^2 + (-1)^2 = 1 \implies x^2 + y^2 + 1 = 1 \implies x^2 + y^2 = 0 \]
This implies:
\[ x = 0, \quad y = 0 \]

### Step 7: Substitute back to find \(t\)
Substituting \(x = 0\) into equation (1):
\[ -1 = t \frac{3}{5} \implies t = -\frac{5}{3} \]

Substituting \(t = -\frac{5}{3}\) into equation (2):
\[ 1 = -\frac{5}{3} \cdot \frac{4}{5} + y \implies 1 = -\frac{4}{3} + y \implies y = 1 + \frac{4}{3} = \frac{7}{3} \]

### Step 8: Final unit vector \(\vec{c}\)
Thus, substituting \(x\), \(y\), and \(z\) back into the unit vector:
\[ \vec{c} = 0\hat{i} + \frac{7}{3}\hat{j} - 1\hat{k} \]
However, we need to normalize this vector to ensure it is a unit vector.

### Step 9: Normalize \(\vec{c}\)
The magnitude of \(\vec{c}\) is:
\[ \sqrt{0^2 + \left(\frac{7}{3}\right)^2 + (-1)^2} = \sqrt{0 + \frac{49}{9} + 1} = \sqrt{\frac{49}{9} + \frac{9}{9}} = \sqrt{\frac{58}{9}} = \frac{\sqrt{58}}{3} \]
Thus, the unit vector \(\vec{c}\) is:
\[ \vec{c} = \frac{0}{\frac{\sqrt{58}}{3}}\hat{i} + \frac{\frac{7}{3}}{\frac{\sqrt{58}}{3}}\hat{j} + \frac{-1}{\frac{\sqrt{58}}{3}}\hat{k} = 0\hat{i} + \frac{7}{\sqrt{58}}\hat{j} - \frac{3}{\sqrt{58}}\hat{k} \]

### Final Answer
\[ \vec{c} = 0\hat{i} + \frac{7}{\sqrt{58}}\hat{j} - \frac{3}{\sqrt{58}}\hat{k} \]