Two vectors, \(\vec{A} = 2\hat{i} - 9\hat{j}\) and \(\vec{B} = -3\hat{i} + 5\hat{j}\), are inclined at an angle \(\theta\) to each other. Determine the value of \(\theta\).
Understand the Problem
The question requires us to find the angle (\theta) between two vectors, (\vec{A}) and (\vec{B}). We can use the dot product formula to find the cosine of the angle between the two vectors, and then use the inverse cosine function to find the angle itself. First find dot product of (\vec{A}) and (\vec{B}), then find the magnitudes of (\vec{A}) and (\vec{B}). Finally, solve for (\theta) using the formula: (\theta = \cos^{-1}(\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|})).
Answer
$\theta = 90^\circ$
Answer for screen readers
$\theta = 90^\circ$
Steps to Solve
- Calculate the dot product of vectors $\vec{A}$ and $\vec{B}$
Given $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = -\hat{i} + \hat{j} + \hat{k}$, the dot product $\vec{A} \cdot \vec{B}$ is calculated as follows: $$ \vec{A} \cdot \vec{B} = (2)(-1) + (3)(1) + (-1)(1) = -2 + 3 - 1 = 0 $$
- Calculate the magnitude of vector $\vec{A}$
The magnitude of $\vec{A}$ is calculated as: $$ |\vec{A}| = \sqrt{(2)^2 + (3)^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} $$
- Calculate the magnitude of vector $\vec{B}$
The magnitude of $\vec{B}$ is calculated as: $$ |\vec{B}| = \sqrt{(-1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} $$
- Calculate the cosine of the angle $\theta$ between the vectors
The cosine of the angle $\theta$ between $\vec{A}$ and $\vec{B}$ is given by: $$ \cos{\theta} = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{0}{\sqrt{14}\sqrt{3}} = 0 $$
- Calculate the angle $\theta$
To find the angle $\theta$, we take the inverse cosine of 0: $$ \theta = \cos^{-1}(0) = \frac{\pi}{2} \text{ radians} = 90^\circ $$
$\theta = 90^\circ$
More Information
The angle between the two vectors is $90^\circ$, which means the vectors are orthogonal or perpendicular to each other.
Tips
A common mistake is incorrectly calculating the dot product or the magnitudes of the vectors, which would lead to an incorrect value for $\cos{\theta}$ and, consequently, an incorrect angle $\theta$. Another mistake could be using the incorrect formula. Also, forgetting to take the inverse cosine in order to find the angle is possible.
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