For what value of x, the vector `vecA=3hati+xhatj-4hatk` is perpendicular to be vector `vecB=2hati-2hatj+6hatk`?

Answer

To find the value of \( x \) such that the vector \( \vec{A} = 3\hat{i} + x\hat{j} - 4\hat{k} \) is perpendicular to the vector \( \vec{B} = 2\hat{i} - 2\hat{j} + 6\hat{k} \), we can use the property that two vectors are perpendicular if their dot product is zero.

### Step-by-Step Solution:

1. **Write the dot product formula**:
The dot product of two vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]
where \( A_x, A_y, A_z \) are the components of \( \vec{A} \) and \( B_x, B_y, B_z \) are the components of \( \vec{B} \).

2. **Identify the components**:
For \( \vec{A} = 3\hat{i} + x\hat{j} - 4\hat{k} \):
- \( A_x = 3 \)
- \( A_y = x \)
- \( A_z = -4 \)

For \( \vec{B} = 2\hat{i} - 2\hat{j} + 6\hat{k} \):
- \( B_x = 2 \)
- \( B_y = -2 \)
- \( B_z = 6 \)

3. **Calculate the dot product**:
Substitute the components into the dot product formula:
\[ \vec{A} \cdot \vec{B} = (3)(2) + (x)(-2) + (-4)(6) \]
Simplifying this gives:
\[ \vec{A} \cdot \vec{B} = 6 - 2x - 24 \]
\[ \vec{A} \cdot \vec{B} = -2x - 18 \]

4. **Set the dot product to zero**:
For the vectors to be perpendicular, we set the dot product equal to zero:
\[ -2x - 18 = 0 \]

5. **Solve for \( x \)**:
Rearranging the equation:
\[ -2x = 18 \]
Dividing both sides by -2:
\[ x = -9 \]

### Final Answer:
The value of \( x \) for which the vector \( \vec{A} \) is perpendicular to the vector \( \vec{B} \) is \( x = -9 \).