If `P` is a point in space such that `OP=12` and `vec(OP)` is inclied at angle of `45^(@)` and `60^(@)` with OX and OY respectively, then the position vector of `P` is

Options

A. `6hati+thatj+-6sqrt(2)hatk`
B. `6hati+6sqrt(2)hatj+-6hatk`
C. `6sqrt(2)hati+6sqrt(j)+-6hatk`
D. none of these

Explanation

To find the position vector of point \( P \) in space given the conditions, we can follow these steps:

### Step 1: Understand the problem
We know that the point \( P \) is such that the distance \( OP = 12 \) units, and it makes angles of \( 45^\circ \) with the x-axis and \( 60^\circ \) with the y-axis. We need to find the position vector \( \vec{OP} \).

### Step 2: Define the angles
Let:
- \( \alpha = 45^\circ \) (angle with the x-axis)
- \( \beta = 60^\circ \) (angle with the y-axis)
- \( \gamma \) = angle with the z-axis (unknown)

### Step 3: Use the formula for angles
According to the formula for direction cosines, we have:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

### Step 4: Calculate the cosines
Calculate \( \cos \alpha \) and \( \cos \beta \):
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos 60^\circ = \frac{1}{2} \)

Now substitute these values into the equation:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \gamma = 1 \]
This simplifies to:
\[ \frac{1}{2} + \frac{1}{4} + \cos^2 \gamma = 1 \]

### Step 5: Solve for \( \cos^2 \gamma \)
Combine the fractions:
\[ \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \]
Thus, we have:
\[ \frac{3}{4} + \cos^2 \gamma = 1 \]
Subtract \( \frac{3}{4} \) from both sides:
\[ \cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 6: Find \( \cos \gamma \)
Taking the square root gives:
\[ \cos \gamma = \pm \frac{1}{2} \]

### Step 7: Write the direction vector
The direction vector \( \vec{OP} \) can be expressed as:
\[ \vec{OP} = \cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k} \]
Substituting the values we found:
\[ \vec{OP} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \left(\pm \frac{1}{2}\right) \hat{k} \]

### Step 8: Multiply by the magnitude
Now, multiply the direction vector by the magnitude \( OP = 12 \):
\[ \vec{OP} = 12 \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \left(\pm \frac{1}{2}\right) \hat{k} \right) \]
This gives:
\[ \vec{OP} = 12 \cdot \frac{1}{\sqrt{2}} \hat{i} + 12 \cdot \frac{1}{2} \hat{j} + 12 \cdot \left(\pm \frac{1}{2}\right) \hat{k} \]
Calculating each term:
\[ \vec{OP} = 6\sqrt{2} \hat{i} + 6 \hat{j} + 6\left(\pm 1\right) \hat{k} \]

### Final Answer
Thus, the position vector of point \( P \) is:
\[ \vec{OP} = 6\sqrt{2} \hat{i} + 6 \hat{j} + 6\hat{k} \quad \text{or} \quad \vec{OP} = 6\sqrt{2} \hat{i} + 6 \hat{j} - 6\hat{k} \]

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