A vector(d) is equally inclined to three vectors `a=hat(i)-hat(j)+hat(k), b=2hat(i)+hat(j) and c=3hat(j)-2hat(k)`. Let x, y, z be three vectors in the plane a, b:b, c:c, a respectively, then

Options

A. `x*d=14`
B. `y*d=3`
C. `z*d=0`
D. `r*d=0,` where `r=lambdax+muy+deltaz`

Answer

To solve the problem step by step, we need to find the vector \( \mathbf{d} \) that is equally inclined to the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \). We will also determine the relationships between the vectors \( \mathbf{x} \), \( \mathbf{y} \), and \( \mathbf{z} \) in the plane formed by \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \).

### Step 1: Define the vectors
Given the vectors:
- \( \mathbf{a} = \hat{i} - \hat{j} + \hat{k} \)
- \( \mathbf{b} = 2\hat{i} + \hat{j} \)
- \( \mathbf{c} = 3\hat{j} - 2\hat{k} \)

### Step 2: Check if the vectors are coplanar
To check if the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) are coplanar, we can calculate the scalar triple product (box product) using the determinant of the matrix formed by the coefficients of the vectors.

\[ \text{Box}(\mathbf{a}, \mathbf{b}, \mathbf{c}) = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & 0 \\ 0 & 3 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} 1 & 0 \\ 3 & -2 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & 0 \\ 0 & -2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 0 \\ 3 & -2 \end{vmatrix} = (1)(-2) - (0)(3) = -2 \)

2. \( \begin{vmatrix} 2 & 0 \\ 0 & -2 \end{vmatrix} = (2)(-2) - (0)(0) = -4 \)

3. \( \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} = (2)(3) - (1)(0) = 6 \)

Putting it all together:

\[ = 1(-2) + 1(-4) + 1(6) = -2 - 4 + 6 = 0 \]

### Step 4: Conclusion about coplanarity
Since the scalar triple product is zero, the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) are coplanar.

### Step 5: Relationship of vectors \( \mathbf{x}, \mathbf{y}, \mathbf{z} \)
Since \( \mathbf{x}, \mathbf{y}, \mathbf{z} \) are in the plane formed by \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), we can express \( \mathbf{z} \) as a linear combination of \( \mathbf{x} \) and \( \mathbf{y} \):

\[ \mathbf{z} = \lambda \mathbf{x} + \mu \mathbf{y} \]

for some scalars \( \lambda \) and \( \mu \).

### Step 6: Vector \( \mathbf{d} \)
The vector \( \mathbf{d} \) is equally inclined to \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), which implies that \( \mathbf{d} \) is perpendicular to the normal of the plane formed by these vectors. Thus, we have:

\[ \mathbf{d} \cdot \mathbf{a} = \mathbf{d} \cdot \mathbf{b} = \mathbf{d} \cdot \mathbf{c} \]

### Step 7: Perpendicularity condition
Since \( \mathbf{d} \) is perpendicular to the plane formed by \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), we can say:

\[ \mathbf{z} \cdot \mathbf{d} = 0 \]

### Final Result
Thus, the relationships we have established are:
1. \( \mathbf{z} \cdot \mathbf{d} = 0 \) (indicating that \( \mathbf{z} \) is perpendicular to \( \mathbf{d} \)).
2. \( \mathbf{r} = \lambda \mathbf{x} + \mu \mathbf{y} + \delta \mathbf{z} \) (where \( \mathbf{r} \) is any vector in the plane).