Let $\vec{AB}=2\hat{i}+10\hat{j}+11\hat{k}$ and $\vec{AC}=-\hat{i}+2\hat{j}+2\hat{k}$. If $\theta$ is the angle between $\vec{AB}$ and $\vec{AC}$, then $\sin\theta=$ ________.

Question

If $ \vec{A} = 2\hat{i} + 3\hat{j} $ and $ \vec{B} = 4\hat{i} - \hat{j} $, then the dot product $ \vec{A} \cdot \vec{B} $ is:

Answer 1

To find the sine of the angle $\theta$ between the vectors $\vec{AB}$ and $\vec{AC}$, we can use the following vector mathematics concepts:

Formula for the dot product: The dot product of two vectors $\vec{U}$ and $\vec{V}$ is given by: $\vec{U} \cdot \vec{V} = |\vec{U}| |\vec{V}| \cos\theta$
Magnitude of a vector: The magnitude of a vector $\vec{U} = a\hat{i} + b\hat{j} + c\hat{k}$ is found using: $|\vec{U}| = \sqrt{a^2 + b^2 + c^2}$
Relation between sine and cosine: Using the trigonometric identity: $\sin^2\theta + \cos^2\theta = 1$ we can express $\sin\theta$ as: $\sin\theta = \sqrt{1 - \cos^2\theta}$

Let's solve step by step:

Compute the dot product of $\vec{AB}$ and $\vec{AC}$:
- $ \vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k} $
- $ \vec{AC} = -\hat{i} + 2\hat{j} + 2\hat{k} $
- Dot product: $\vec{AB} \cdot \vec{AC} = 2 \times (-1) + 10 \times 2 + 11 \times 2 = -2 + 20 + 22 = 40$
Calculate the magnitudes:
- $|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$
- $|\vec{AC}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
Find $\cos\theta$: $\cos\theta = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|} = \frac{40}{15 \times 3} = \frac{40}{45} = \frac{8}{9}$
Calculate $\sin\theta$ using $\sin^2\theta + \cos^2\theta = 1$:
- $\cos^2\theta = \left(\frac{8}{9}\right)^2 = \frac{64}{81}$
- $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{64}{81} = \frac{17}{81}$
- $\sin\theta = \sqrt{\frac{17}{81}} = \frac{\sqrt{17}}{9}$

Thus, the value of $\sin\theta$ is indeed $\frac{\sqrt{17}}{9}$. This confirms that the correct answer is $\frac{\sqrt{17}}{9}$.

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