Find a vector in the direction of :
`vec(a)=-2hat(i)+hat(j)+2hat(k)`, which has magnitude 9 units.

Answer

To find a vector in the direction of \(\vec{a} = -2\hat{i} + \hat{j} + 2\hat{k}\) with a magnitude of 9 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:

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

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

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

### 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:

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

### Step 3: Find the vector with magnitude 9 in the direction of \(\vec{a}\)

To find the vector \(\vec{x}\) that has a magnitude of 9 units in the direction of \(\vec{a}\), we can multiply the unit vector \(\hat{a}\) by 9:

\[ \vec{x} = 9 \hat{a} = 9 \left(-\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}\right) \]

Calculating this gives:

\[ \vec{x} = 9 \left(-\frac{2}{3}\right)\hat{i} + 9 \left(\frac{1}{3}\right)\hat{j} + 9 \left(\frac{2}{3}\right)\hat{k} \]

\[ \vec{x} = -6\hat{i} + 3\hat{j} + 6\hat{k} \]

### Final Answer

Thus, the vector in the direction of \(\vec{a}\) with a magnitude of 9 units is:

\[ \vec{x} = -6\hat{i} + 3\hat{j} + 6\hat{k} \]

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