The two vectors `(x^2 - 1)hati +(x+2)hatj + x^2 hatk` and `2hati -xhatj + 3hatk` are orthogonal

Options

A. for no real value of x
B. for x = -1
C. for x = 1/2
D. for x = -1/2 and x = 1

Answer

To determine the values of \( x \) for which the 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 find the dot product of these two vectors and set it equal to zero.

### Step 1: Write the dot product of the 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 the expression:

\[ = 2(x^2 - 1) - x(x + 2) + 3x^2 \]

Expanding each term gives:

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

### Step 3: Combine like terms

Now, we combine the like terms:

\[ = 2x^2 - 2 - x^2 - 2x + 3x^2 \]

Combining these results:

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

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

Since the vectors are 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 all terms by 2:

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

### Step 6: Factor the quadratic equation

Next, we factor the quadratic equation:

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

Grouping the terms:

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

Factoring out \( (x - 1) \):

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

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

Setting each factor equal to zero gives us:

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

### Final Answer

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

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

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