The dot product of two vectors of magnitudes 3 units and 5 units cannot be :-
(i) -20 (ii) 16 (iii) -10 (br) 14

Options

A. (i,iii)
B. (i,ii)
C. (i, iv)
D. (ii, iii, iv)

Answer

To solve the problem, we need to determine the range of possible values for the dot product of two vectors with given magnitudes.

### Step-by-Step Solution:

1. **Understand the Dot Product Formula**:
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
where \( |\mathbf{A}| \) and \( |\mathbf{B}| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them.

2. **Identify the Magnitudes**:
We have two vectors with magnitudes:
\[ |\mathbf{A}| = 3 \text{ units}, \quad |\mathbf{B}| = 5 \text{ units} \]

3. **Calculate the Maximum and Minimum Values of the Dot Product**:
- The maximum value occurs when \( \cos \theta = 1 \) (i.e., the vectors are in the same direction):
\[ \text{Maximum} = 3 \times 5 \times 1 = 15 \]
- The minimum value occurs when \( \cos \theta = -1 \) (i.e., the vectors are in opposite directions):
\[ \text{Minimum} = 3 \times 5 \times (-1) = -15 \]

4. **Determine the Range of Possible Values**:
Therefore, the dot product can take any value in the range:
\[ -15 \leq \mathbf{A} \cdot \mathbf{B} \leq 15 \]

5. **Evaluate the Given Options**:
Now we check which of the given options falls outside this range:
- (i) -20: This is less than -15 (not possible)
- (ii) 16: This is greater than 15 (not possible)
- (iii) -10: This is within the range (-15 to 15)
- (iv) 14: This is also within the range (-15 to 15)

6. **Conclusion**:
The values that cannot be the dot product of the two vectors are -20 and 16. However, since the question asks for which one cannot be, we can conclude that the answer is:
- The dot product of two vectors of magnitudes 3 units and 5 units cannot be **-20** or **16**.

### Final Answer:
The dot product of two vectors of magnitudes 3 units and 5 units cannot be: **(i) -20** and **(ii) 16**.