The vector c directed along the bisectors of the angle between the vectors ` a = 7i- 4j - 4k` and `b = -2i - j + 2k` if `|c| = 3sqrt(6)` is given by

Options

A. `i - 7j + 2k`
B. `i + 7j -2k`
C. `-i + 7j -2k`
D. `i-7j-2k`

Answer

To find the vector \( \mathbf{c} \) directed along the bisector of the angle between the vectors \( \mathbf{a} = 7\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} \) and \( \mathbf{b} = -2\mathbf{i} - \mathbf{j} + 2\mathbf{k} \) with the magnitude \( |\mathbf{c}| = 3\sqrt{6} \), we can follow these steps:

### Step 1: Calculate the unit vectors of \( \mathbf{a} \) and \( \mathbf{b} \)

First, we need to find the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \).

\[ |\mathbf{a}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9 \]

\[ |\mathbf{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Now, we can find the unit vectors \( \hat{a} \) and \( \hat{b} \):

\[ \hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{7\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}}{9} \]

\[ \hat{b} = \frac{\mathbf{b}}{|\mathbf{b}|} = \frac{-2\mathbf{i} - \mathbf{j} + 2\mathbf{k}}{3} \]

### Step 2: Express the vector \( \mathbf{c} \)

The vector \( \mathbf{c} \) along the bisector of the angle between \( \mathbf{a} \) and \( \mathbf{b} \) can be expressed as:

\[ \mathbf{c} = \lambda (\hat{a} + \hat{b}) \]

### Step 3: Substitute the unit vectors into the equation

Substituting \( \hat{a} \) and \( \hat{b} \):

\[ \mathbf{c} = \lambda \left( \frac{7\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}}{9} + \frac{-2\mathbf{i} - \mathbf{j} + 2\mathbf{k}}{3} \right) \]

To combine these, we need a common denominator. The common denominator is 9:

\[ \mathbf{c} = \lambda \left( \frac{7\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}}{9} + \frac{-6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}}{9} \right) \]

\[ \mathbf{c} = \lambda \left( \frac{(7 - 6)\mathbf{i} + (-4 - 3)\mathbf{j} + (-4 + 6)\mathbf{k}}{9} \right) \]

\[ \mathbf{c} = \lambda \left( \frac{1\mathbf{i} - 7\mathbf{j} + 2\mathbf{k}}{9} \right) \]

### Step 4: Find the magnitude of \( \mathbf{c} \)

The magnitude of \( \mathbf{c} \) is given as \( |\mathbf{c}| = 3\sqrt{6} \):

\[ |\mathbf{c}| = |\lambda| \cdot \frac{1}{9} \sqrt{1^2 + (-7)^2 + 2^2} \]

Calculating the magnitude inside the square root:

\[ \sqrt{1 + 49 + 4} = \sqrt{54} \]

Thus,

\[ |\mathbf{c}| = |\lambda| \cdot \frac{\sqrt{54}}{9} \]

Setting this equal to \( 3\sqrt{6} \):

\[ |\lambda| \cdot \frac{\sqrt{54}}{9} = 3\sqrt{6} \]

### Step 5: Solve for \( \lambda \)

Cross-multiplying gives:

\[ |\lambda| \cdot \sqrt{54} = 27\sqrt{6} \]

Now, simplifying \( \sqrt{54} = 3\sqrt{6} \):

\[ |\lambda| \cdot 3\sqrt{6} = 27\sqrt{6} \]

Dividing both sides by \( 3\sqrt{6} \):

\[ |\lambda| = 9 \]

### Step 6: Write the final expression for \( \mathbf{c} \)

Substituting \( \lambda \) back into the expression for \( \mathbf{c} \):

\[ \mathbf{c} = 9 \cdot \frac{1\mathbf{i} - 7\mathbf{j} + 2\mathbf{k}}{9} = \mathbf{i} - 7\mathbf{j} + 2\mathbf{k} \]

Thus, the vector \( \mathbf{c} \) directed along the bisector is:

\[ \mathbf{c} = \mathbf{i} - 7\mathbf{j} + 2\mathbf{k} \quad \text{or} \quad \mathbf{c} = -\mathbf{i} + 7\mathbf{j} - 2\mathbf{k} \quad \text{(considering the negative value of } \lambda\text{)} \]