If OP=8 and OP makes angles `45^(@) and 60^(@)` with OX-axis and OY-axis respectively, then OP is equal to

Options

A. `8(sqrt(2)hati+hatj+-hatk)`
B. `4(sqrt(2)hati+hatj+-hatk)`
C. `(1)/(4)(sqrt(2)hati+hatj+-hatk)`
D. `(1)/(8)(sqrt(2)hati+hatj+-hatk)`

Explanation

To solve the problem, we need to find the vector OP given that its length is 8 and it makes angles of 45° with the OX-axis and 60° with the OY-axis.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- The angle OP makes with the OX-axis is 45°.
- The angle OP makes with the OY-axis is 60°.
- We can use these angles to find the components of the vector OP.

2. **Using Trigonometric Functions**:
- The components of the vector OP can be expressed in terms of its magnitude and the angles it makes with the axes.
- Let the vector OP be represented as OP = (x, y, z).
- From the angle with the OX-axis:
\[ \cos(45^\circ) = \frac{x}{|OP|} \implies x = |OP| \cdot \cos(45^\circ) = 8 \cdot \frac{1}{\sqrt{2}} = 4\sqrt{2} \]
- From the angle with the OY-axis:
\[ \cos(60^\circ) = \frac{y}{|OP|} \implies y = |OP| \cdot \cos(60^\circ) = 8 \cdot \frac{1}{2} = 4 \]

3. **Finding the Z-component**:
- Since we are not given any angle with the OZ-axis, we can assume that the angle with the OZ-axis is such that we can find z using the Pythagorean theorem in three dimensions.
- The magnitude of OP is given by:
\[ |OP| = \sqrt{x^2 + y^2 + z^2} \]
- Substituting the known values:
\[ 8 = \sqrt{(4\sqrt{2})^2 + 4^2 + z^2} \]
- Calculating:
\[ 8 = \sqrt{32 + 16 + z^2} \]
\[ 8 = \sqrt{48 + z^2} \]
- Squaring both sides:
\[ 64 = 48 + z^2 \]
\[ z^2 = 64 - 48 = 16 \implies z = 4 \]

4. **Final Vector**:
- Thus, the vector OP can be expressed as:
\[ OP = (4\sqrt{2}, 4, 4) \]

### Conclusion:
The vector OP is equal to \( (4\sqrt{2}, 4, 4) \) and its length is confirmed to be 8.