If `(x,y,z) ne (0,0,0) and (hati + hatj +3hatk )x + (3 hati - 3 hatj + hatk )y +(-4 hati + 5 hatj ) z` `= a (x hati + y hatj + z hatk ),` then the values of a are

Options

A. 0, -2
B. 2, 0
C. 0, -1
D. 1, 0

Explanation

To solve the given vector equation, we start with the expression:

\[ (\hat{i} + \hat{j} + 3\hat{k})x + (3\hat{i} - 3\hat{j} + \hat{k})y + (-4\hat{i} + 5\hat{j})z = a(x\hat{i} + y\hat{j} + z\hat{k}) \]

### Step 1: Expand the Left-Hand Side
We will expand the left-hand side by distributing \(x\), \(y\), and \(z\) with their respective coefficients.

\[ = (x\hat{i} + x\hat{j} + 3x\hat{k}) + (3y\hat{i} - 3y\hat{j} + y\hat{k}) + (-4z\hat{i} + 5z\hat{j}) \]

### Step 2: Combine Like Terms
Now, we combine the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):

- Coefficient of \(\hat{i}\): \(x + 3y - 4z\)
- Coefficient of \(\hat{j}\): \(x - 3y + 5z\)
- Coefficient of \(\hat{k}\): \(3x + y\)

Thus, we can rewrite the left-hand side as:

\[ (x + 3y - 4z)\hat{i} + (x - 3y + 5z)\hat{j} + (3x + y)\hat{k} \]

### Step 3: Set Up the Equation
Now, we equate the coefficients from both sides of the equation:

\[ (x + 3y - 4z) = a x \]
\[ (x - 3y + 5z) = a y \]
\[ (3x + y) = a z \]

### Step 4: Rearranging the Equations
Rearranging each equation gives us:

1. \(x + 3y - 4z - ax = 0\) \(\Rightarrow (1 - a)x + 3y - 4z = 0\)
2. \(x - 3y + 5z - ay = 0\) \(\Rightarrow x + (-3 + a)y + 5z = 0\)
3. \(3x + y - az = 0\) \(\Rightarrow 3x + y - az = 0\)

### Step 5: Form the Determinant
We can set up a determinant with the coefficients of \(x\), \(y\), and \(z\):

\[ \begin{vmatrix} 1 - a & 3 & -4 \\ 1 & -3 + a & 5 \\ 3 & 1 & -a \end{vmatrix} = 0 \]

### Step 6: Calculate the Determinant
Calculating the determinant:

\[ = (1 - a)\left((-3 + a)(-a) - 5\right) - 3\left(1(-a) - 5(3)\right) - 4\left(1(1) - 3(-3 + a)\right) \]

Expanding this determinant will lead to a cubic equation in \(a\).

### Step 7: Solve the Cubic Equation
After simplifying, we find:

\[ -a^3 - 2a^2 - a = 0 \]

Factoring out \(-a\):

\[ -a(a^2 + 2a + 1) = 0 \]

This gives us:

\[ a = 0 \quad \text{or} \quad a^2 + 2a + 1 = 0 \]

The quadratic can be factored as:

\[ (a + 1)^2 = 0 \Rightarrow a = -1 \]

### Final Values of \(a\)
Thus, the values of \(a\) are:

\[ a = 0 \quad \text{and} \quad a = -1 \]