Let `veca=hati+hatj+hatk, vecb=-hati+hatj+hatk, vecc=hati-hatj+hatk` and `vecd=hati+hatj-hatk`. Then, the line of intersection of planes one determined by `veca, vecb` and other determined by `vecc, vecd` is perpendicular to

Options

A. `x`-axis
B. y-axis
C. z-axis
D. none of these

Explanation

To solve the problem, we need to find the line of intersection of the two planes determined by the vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{d}\). Then, we will determine which axis this line is perpendicular to.

### Step 1: Identify the vectors
Given:
- \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{b} = -\hat{i} + \hat{j} + \hat{k}\)
- \(\vec{c} = \hat{i} - \hat{j} + \hat{k}\)
- \(\vec{d} = \hat{i} + \hat{j} - \hat{k}\)

### Step 2: Find the normal vectors of the planes
The normal vector of the plane formed by vectors \(\vec{a}\) and \(\vec{b}\) can be found using the cross product:
\[ \vec{n_1} = \vec{a} \times \vec{b} \]

Calculating \(\vec{n_1}\):
\[ \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{vmatrix} \]

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

So, \(\vec{n_1} = 2\hat{k} - 2\hat{j}\).

Now, for the plane formed by \(\vec{c}\) and \(\vec{d}\):
\[ \vec{n_2} = \vec{c} \times \vec{d} \]

Calculating \(\vec{n_2}\):
\[ \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]

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

So, \(\vec{n_2} = 2\hat{i} + 2\hat{j} + 2\hat{k}\).

### Step 3: Find the direction of the line of intersection
The direction of the line of intersection of the two planes is given by the cross product of the normal vectors:
\[ \vec{d} = \vec{n_1} \times \vec{n_2} \]

Calculating \(\vec{d}\):
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ 2 & 2 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{d} = \hat{i} \begin{vmatrix} -2 & 2 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 2 \\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & -2 \\ 2 & 2 \end{vmatrix} \]
\[ = \hat{i}(-4 - 4) - \hat{j}(0 - 4) + \hat{k}(0 + 4) \]
\[ = -8\hat{i} + 4\hat{j} + 4\hat{k} \]

### Step 4: Determine the perpendicularity
To find out which axis the line is perpendicular to, we can analyze the direction vector \(\vec{d} = -8\hat{i} + 4\hat{j} + 4\hat{k}\).

The line is perpendicular to the x-axis if its x-component is zero, which it is not. The same applies to the y-axis and z-axis.

### Conclusion
The line is not perpendicular to any of the axes, so the answer is "none of these."