A vector equally inclined to the vectors `hat(i)-hat(j)+hat(k) and hat(i)+hat(j)-hat(k)` then the plane containing them is

Options

A. `(hat(i)+hat(j)-hat(k))/(sqrt(3))`
B. `hat(j)-hat(k)`
C. `2hat(i)`
D. `hat(i)`

Answer

To find a vector that is equally inclined to the vectors \(\hat{i} - \hat{j} + \hat{k}\) and \(\hat{i} + \hat{j} - \hat{k}\), we can follow these steps:

### Step 1: Define the Vectors
Let:
- Vector \( \mathbf{A} = \hat{i} - \hat{j} + \hat{k} \)
- Vector \( \mathbf{B} = \hat{i} + \hat{j} - \hat{k} \)

### Step 2: Calculate the Magnitudes of the Vectors
The magnitude of vector \( \mathbf{A} \) is calculated as follows:
\[ |\mathbf{A}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

The magnitude of vector \( \mathbf{B} \) is calculated similarly:
\[ |\mathbf{B}| = \sqrt{(1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Find the Angle Bisector
The angle bisector \( \mathbf{C} \) of vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be found using the formula:
\[ \mathbf{C} = \frac{\mathbf{A}}{|\mathbf{A}|} + \frac{\mathbf{B}}{|\mathbf{B}|} \]

Substituting the values:
\[ \mathbf{C} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} + \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} \]

### Step 4: Simplify the Expression
Combining the two fractions:
\[ \mathbf{C} = \frac{(\hat{i} - \hat{j} + \hat{k}) + (\hat{i} + \hat{j} - \hat{k})}{\sqrt{3}} = \frac{2\hat{i}}{\sqrt{3}} \]

### Step 5: Identify the Direction of the Angle Bisector
The vector \( \mathbf{C} \) simplifies to:
\[ \mathbf{C} = \frac{2}{\sqrt{3}} \hat{i} \]
This indicates that the vector \( \mathbf{C} \) is in the direction of \( \hat{i} \).

### Step 6: Determine the Plane Containing Vectors
The plane containing the vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be described by the normal vector, which can be found using the cross product \( \mathbf{A} \times \mathbf{B} \).

### Step 7: Calculate the Cross Product
Calculating the cross product:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:
\[ = \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(1) - (-1)(1)) \]
\[ = \hat{i}(1 - 1) - \hat{j}(-1 - 1) + \hat{k}(1 + 1) \]
\[ = 0\hat{i} + 2\hat{j} + 2\hat{k} \]

Thus, the normal vector to the plane is \( \mathbf{N} = 2\hat{j} + 2\hat{k} \).

### Conclusion
The plane containing the vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be described by the equation:
\[ 2y + 2z = d \]
where \( d \) is a constant that can be determined based on specific points in the plane.