Find a vector in the direction of `5hati-hatj+2hatk`, which has magnitude 8 units.

Answer

To find a vector in the direction of \(5\hat{i} - \hat{j} + 2\hat{k}\) with a magnitude of 8 units, we can follow these steps:

### Step 1: Define the given vector
Let the vector \( \mathbf{A} = 5\hat{i} - \hat{j} + 2\hat{k} \).

### Step 2: Calculate the magnitude of vector \( \mathbf{A} \)
The magnitude of vector \( \mathbf{A} \) is given by the formula:
\[ |\mathbf{A}| = \sqrt{(5)^2 + (-1)^2 + (2)^2} \]
Calculating this:
\[ |\mathbf{A}| = \sqrt{25 + 1 + 4} = \sqrt{30} \]

### Step 3: Find the unit vector in the direction of \( \mathbf{A} \)
The unit vector \( \hat{a} \) in the direction of \( \mathbf{A} \) is calculated as:
\[ \hat{a} = \frac{\mathbf{A}}{|\mathbf{A}|} = \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}} \]

### Step 4: Scale the unit vector to have a magnitude of 8
To find a vector in the direction of \( \mathbf{A} \) with a magnitude of 8, we multiply the unit vector \( \hat{a} \) by 8:
\[ \mathbf{B} = 8 \hat{a} = 8 \left(\frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}}\right) \]
Calculating this:
\[ \mathbf{B} = \frac{40\hat{i} - 8\hat{j} + 16\hat{k}}{\sqrt{30}} \]

### Final Result
Thus, the required vector \( \mathbf{B} \) in the direction of \( 5\hat{i} - \hat{j} + 2\hat{k} \) with a magnitude of 8 units is:
\[ \mathbf{B} = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k} \]