Let `a=hat(i)+hat(j)+hat(k), b=-hat(i)+hat(j)+hat(k), c=hat(i)-hat(j)+hat(k) and d=hat(i)+hat(j)-hat(k)`. Then, the line of intersection of planes one determined by a, b and other determined by c, d is perpendicular to

Options

A. X-axis
B. Y-axis
C. Both X and Y axes
D. Both y and z-axes

Answer

To solve the problem, we need to find the line of intersection of two planes defined by the vectors \( \mathbf{a}, \mathbf{b} \) and \( \mathbf{c}, \mathbf{d} \). Then, we will determine the direction of this line of intersection and check to which vector it is perpendicular.

### Step 1: Identify the vectors
The given vectors are:
\[ \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{b} = -\hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{c} = \hat{i} - \hat{j} + \hat{k} \]
\[ \mathbf{d} = \hat{i} + \hat{j} - \hat{k} \]

### Step 2: Find the normal vectors of the planes
The normal vector of a plane defined by two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be found using the cross product \( \mathbf{n} = \mathbf{u} \times \mathbf{v} \).

**For the plane defined by \( \mathbf{a} \) and \( \mathbf{b} \):**
\[ \mathbf{n_1} = \mathbf{a} \times \mathbf{b} \]

Calculating \( \mathbf{n_1} \):
\[ \mathbf{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{n_1} = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} \]
\[ = \hat{i}(1 - 1) - \hat{j}(1 + 1) + \hat{k}(1 + 1) \]
\[ = 0\hat{i} - 2\hat{j} + 2\hat{k} = -2\hat{j} + 2\hat{k} \]

Thus,
\[ \mathbf{n_1} = -2\hat{j} + 2\hat{k} \]

**For the plane defined by \( \mathbf{c} \) and \( \mathbf{d} \):**
\[ \mathbf{n_2} = \mathbf{c} \times \mathbf{d} \]

Calculating \( \mathbf{n_2} \):
\[ \mathbf{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{n_2} = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \]
\[ = \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(1) - (-1)(1)) \]
\[ = \hat{i}(1 - 1) - \hat{j}(-1 - 1) + \hat{k}(1 + 1) \]
\[ = 0\hat{i} + 2\hat{j} + 2\hat{k} \]

Thus,
\[ \mathbf{n_2} = 2\hat{j} + 2\hat{k} \]

### Step 3: Find the direction of the line of intersection
The direction of the line of intersection of the two planes can be found by taking the cross product of the normal vectors:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating \( \mathbf{d} \):
\[ \mathbf{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ 0 & 2 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{d} = \hat{i} \begin{vmatrix} -2 & 2 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 2 \\ 0 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & -2 \\ 0 & 2 \end{vmatrix} \]
\[ = \hat{i}((-2)(2) - (2)(0)) - \hat{j}(0 - 0) + \hat{k}(0 - 0) \]
\[ = -4\hat{i} + 0\hat{j} + 0\hat{k} \]

Thus,
\[ \mathbf{d} = -4\hat{i} \]

### Step 4: Determine to which vector the line is perpendicular
The line of intersection is represented by the direction vector \( \mathbf{d} = -4\hat{i} \). We need to check which of the original vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \) is perpendicular to this direction vector. A vector \( \mathbf{v} \) is perpendicular to another vector \( \mathbf{w} \) if their dot product is zero:
\[ \mathbf{v} \cdot \mathbf{w} = 0 \]

Calculating the dot products:
1. \( \mathbf{a} \cdot \mathbf{d} = (1)(-4) + (1)(0) + (1)(0) = -4 \) (not perpendicular)
2. \( \mathbf{b} \cdot \mathbf{d} = (-1)(-4) + (1)(0) + (1)(0) = 4 \) (not perpendicular)
3. \( \mathbf{c} \cdot \mathbf{d} = (1)(-4) + (-1)(0) + (1)(0) = -4 \) (not perpendicular)
4. \( \mathbf{d} \cdot \mathbf{d} = (1)(-4) + (1)(0) + (-1)(0) = -4 \) (not perpendicular)

None of the vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \) are perpendicular to the line of intersection.

### Conclusion:
The line of intersection of the planes determined by \( \mathbf{a}, \mathbf{b} \) and \( \mathbf{c}, \mathbf{d} \) is not perpendicular to any of the given vectors.