What are the values of x for which the two vectors `(x^(2)-1)hat(i)+(x+2)hat(j)+x^(2)hat(k) and 2 hat (i) - x hat (j) + 3 hat(k)` are orthogonal ?

Options

A. No real value of `x`
B. `x=(1)/(2) and x=-1`
C. `x=-(1)/(2) and x=1`
D. `x=-1and x=2`

Explanation

To find the values of \( x \) for which the two vectors

\[ \mathbf{A} = (x^2 - 1) \hat{i} + (x + 2) \hat{j} + x^2 \hat{k} \]

and

\[ \mathbf{B} = 2 \hat{i} - x \hat{j} + 3 \hat{k} \]

are orthogonal, we need to set their dot product equal to zero.

### Step 1: Calculate the dot product of the two vectors

The dot product \( \mathbf{A} \cdot \mathbf{B} \) is given by:

\[ \mathbf{A} \cdot \mathbf{B} = (x^2 - 1)(2) + (x + 2)(-x) + (x^2)(3) \]

### Step 2: Expand the dot product

Now we will expand this expression:

1. \( (x^2 - 1)(2) = 2x^2 - 2 \)
2. \( (x + 2)(-x) = -x^2 - 2x \)
3. \( (x^2)(3) = 3x^2 \)

Combining these, we have:

\[ \mathbf{A} \cdot \mathbf{B} = (2x^2 - 2) + (-x^2 - 2x) + (3x^2) \]

### Step 3: Combine like terms

Now, combine the terms:

\[ \mathbf{A} \cdot \mathbf{B} = 2x^2 - 2 - x^2 - 2x + 3x^2 \]

This simplifies to:

\[ (2x^2 - x^2 + 3x^2) - 2 - 2x = 4x^2 - 2x - 2 \]

### Step 4: Set the dot product to zero

For the vectors to be orthogonal, we set the dot product equal to zero:

\[ 4x^2 - 2x - 2 = 0 \]

### Step 5: Simplify the equation

We can simplify this equation by dividing everything by 2:

\[ 2x^2 - x - 1 = 0 \]

### Step 6: Factor the quadratic equation

Now we will factor the quadratic equation:

\[ (2x + 1)(x - 1) = 0 \]

### Step 7: Solve for \( x \)

Setting each factor to zero gives us:

1. \( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \)
2. \( x - 1 = 0 \) → \( x = 1 \)

### Final Solution

Thus, the values of \( x \) for which the two vectors are orthogonal are:

\[ x = -\frac{1}{2} \quad \text{and} \quad x = 1 \]

---