If `A =2i + 2j + 3k, B =-i+2j+k` and `C=3i+j`, then A +t B is perpendicular to C if t is equal to

Options

A. 8
B. 4
C. 6
D. 2

Answer

To solve the problem, we need to find the value of \( t \) such that the vector \( A + tB \) is perpendicular to the vector \( C \). This means that the dot product of \( A + tB \) and \( C \) must equal zero.

### Step-by-Step Solution:

1. **Identify the vectors**:
- \( A = 2i + 2j + 3k \)
- \( B = -i + 2j + k \)
- \( C = 3i + j \)

2. **Express \( A + tB \)**:
\[ A + tB = (2i + 2j + 3k) + t(-i + 2j + k) \]
Distributing \( t \):
\[ A + tB = (2 - t)i + (2 + 2t)j + (3 + tk) \]

3. **Set up the dot product with \( C \)**:
The dot product \( (A + tB) \cdot C \) must equal zero:
\[ ((2 - t)i + (2 + 2t)j + (3 + tk)) \cdot (3i + j) = 0 \]

4. **Calculate the dot product**:
\[ = (2 - t) \cdot 3 + (2 + 2t) \cdot 1 + (3 + t) \cdot 0 \]
Simplifying:
\[ = 3(2 - t) + (2 + 2t) \]
\[ = 6 - 3t + 2 + 2t \]
\[ = 8 - t \]

5. **Set the dot product equal to zero**:
\[ 8 - t = 0 \]

6. **Solve for \( t \)**:
\[ t = 8 \]

### Final Answer:
The value of \( t \) is \( 8 \).