The magnitude of two vectors are 16 and 12 units respectively and the magnitude of their scalar product is `98sqrt2` units. The angle between the vectors would be

Options

A. 30°
B. 45°
C. 60°
D. 90°

Answer

To solve the problem step by step, we will use the formula for the scalar (dot) product of two vectors.

### Step 1: Write down the given information
- Magnitude of vector A, |A| = 16 units
- Magnitude of vector B, |B| = 12 units
- Magnitude of the scalar product (dot product) of the two vectors = 98√2 units

### Step 2: Recall the formula for the scalar product
The scalar product (dot product) of two vectors A and B can be expressed as:
\[ A \cdot B = |A| |B| \cos(\theta) \]
where θ is the angle between the two vectors.

### Step 3: Substitute the known values into the formula
Using the values we have:
\[ 98\sqrt{2} = |A| |B| \cos(\theta) \]
Substituting the magnitudes:
\[ 98\sqrt{2} = 16 \times 12 \cos(\theta) \]

### Step 4: Calculate the product of the magnitudes
Calculate \( 16 \times 12 \):
\[ 16 \times 12 = 192 \]

### Step 5: Rewrite the equation
Now, we can rewrite the equation:
\[ 98\sqrt{2} = 192 \cos(\theta) \]

### Step 6: Solve for cos(θ)
To find cos(θ), we rearrange the equation:
\[ \cos(\theta) = \frac{98\sqrt{2}}{192} \]

### Step 7: Simplify the fraction
Now, we simplify the right side:
1. Calculate \( 98\sqrt{2} \):
- \( 98\sqrt{2} \approx 138.59 \) (using \( \sqrt{2} \approx 1.414 \))
2. Divide by 192:
\[ \cos(\theta) \approx \frac{138.59}{192} \approx 0.721 \]

### Step 8: Calculate θ using the inverse cosine
Now, we find θ:
\[ \theta = \cos^{-1}(0.721) \]

### Step 9: Use a calculator to find θ
Using a calculator:
\[ \theta \approx 43.6^\circ \]

### Step 10: Conclusion
Thus, the angle between the two vectors is approximately \( 43.6^\circ \).