Topic: Engineering Mathematics
The sine of the angle between the two vectors a = 3i + j + k and b = 2i - 2j + k is
- A. $\sqrt{74/99}$
- B. $\sqrt{25/99}$
- C. $\sqrt{37/99}$
- D. $\sqrt{5/51}$
Answer: A. $\sqrt{74/99}$
Explanation: First, find the cosine of the angle using the dot product formula: cos(theta) = (a . b) / (|a| * |b|). The dot product is (3*2) + (1*-2) + (1*1) = 5. The magnitudes are |a| = $\sqrt{9 + 1 + 1}$ = $\sqrt{11}$ and |b| = $\sqrt{4 + 4 + 1}$ = $\sqrt{9}$ = 3. Therefore, cos(theta) = 5 / (3 * $\sqrt{11}$) = 5 / $\sqrt{99}$. The sine of the angle is calculated using the identity sin(theta) = $\sqrt{1 - cos²(theta}$), which gives $\sqrt{1 - 25/99}$ = $\sqrt{74/99}$.