The direction cosines of the vector `(2 hati + 2 hatj -hatk)` are `"___________."`

Answer

To find the direction cosines of the vector \( \vec{v} = 2\hat{i} + 2\hat{j} - \hat{k} \), we can follow these steps:

### Step 1: Identify the components of the vector
The vector \( \vec{v} \) can be expressed in terms of its components:
- \( A = 2 \) (coefficient of \( \hat{i} \))
- \( B = 2 \) (coefficient of \( \hat{j} \))
- \( C = -1 \) (coefficient of \( -\hat{k} \))

### Step 2: Calculate the magnitude of the vector
The magnitude \( |\vec{v}| \) of the vector is given by the formula:
\[ |\vec{v}| = \sqrt{A^2 + B^2 + C^2} \]
Substituting the values:
\[ |\vec{v}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 3: Calculate the direction cosines
The direction cosines \( L, M, N \) are calculated using the formulas:
\[ L = \frac{A}{|\vec{v}|}, \quad M = \frac{B}{|\vec{v}|}, \quad N = \frac{C}{|\vec{v}|} \]
Substituting the values:
\[ L = \frac{2}{3}, \quad M = \frac{2}{3}, \quad N = \frac{-1}{3} \]

### Final Answer
The direction cosines of the vector \( 2\hat{i} + 2\hat{j} - \hat{k} \) are:
\[ \left( \frac{2}{3}, \frac{2}{3}, -\frac{1}{3} \right) \]

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