The projection of a vector on the co-ordinate axis are 2,3 and -6. Find the direction cosines of the vector.

Answer

To find the direction cosines of a vector given its projections on the coordinate axes, we can follow these steps:

### Step 1: Identify the projections
The projections of the vector on the coordinate axes are given as:
- \( x = 2 \)
- \( y = 3 \)
- \( z = -6 \)

### Step 2: Calculate the magnitude of the vector
The magnitude \( |V| \) of the vector can be calculated using the formula:
\[ |V| = \sqrt{x^2 + y^2 + z^2} \]
Substituting the values:
\[ |V| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 3: Calculate the direction cosines
The direction cosines \( l, m, n \) are given by the formulas:
\[ l = \frac{x}{|V|}, \quad m = \frac{y}{|V|}, \quad n = \frac{z}{|V|} \]
Substituting the values:
\[ l = \frac{2}{7}, \quad m = \frac{3}{7}, \quad n = \frac{-6}{7} \]

### Step 4: Write the final answer
The direction cosines of the vector are:
\[ l = \frac{2}{7}, \quad m = \frac{3}{7}, \quad n = \frac{-6}{7} \]

### Summary
Thus, the direction cosines of the vector are:
\[ \left( \frac{2}{7}, \frac{3}{7}, -\frac{6}{7} \right) \]

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