Find a vector in the direction of :
`vec(a)=hat(i)-2hat(j)+2hat(k)`, which has magnitude 15 units

Answer

To find a vector in the direction of \( \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \) with a magnitude of 15 units, we can follow these steps:

### Step 1: Find the magnitude of vector \( \vec{a} \)

The magnitude of a vector \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) is given by the formula:

\[ |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

For \( \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \):
- \( a_1 = 1 \)
- \( a_2 = -2 \)
- \( a_3 = 2 \)

Calculating the magnitude:

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

### Step 2: Find the unit vector in the direction of \( \vec{a} \)

The unit vector \( \hat{a} \) in the direction of \( \vec{a} \) is given by:

\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \]

Substituting the values we found:

\[ \hat{a} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} = \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \]

### Step 3: Find the vector with magnitude 15 in the direction of \( \vec{a} \)

To find a vector \( \vec{x} \) in the direction of \( \vec{a} \) with a magnitude of 15, we can multiply the unit vector \( \hat{a} \) by 15:

\[ \vec{x} = 15 \hat{a} = 15 \left( \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \right) \]

Calculating \( \vec{x} \):

\[ \vec{x} = 15 \cdot \frac{1}{3} \hat{i} - 15 \cdot \frac{2}{3} \hat{j} + 15 \cdot \frac{2}{3} \hat{k} \]
\[ \vec{x} = 5 \hat{i} - 10 \hat{j} + 10 \hat{k} \]

### Final Answer

Thus, the vector in the direction of \( \vec{a} \) with a magnitude of 15 units is:

\[ \vec{x} = 5\hat{i} - 10\hat{j} + 10\hat{k} \]

---