Find the direction cosines of the line connecting the origin to point P (2, -1, 3).
Understand the Problem
The question asks us to find the direction cosines of the line connecting the origin (0,0,0) to the point P (2, -1, 3). To do this, we need to find the vector from the origin to P, normalize this vector, and the components of the normalized vector will be the direction cosines.
Answer
The direction cosines are $\frac{2}{\sqrt{14}}$, $\frac{-1}{\sqrt{14}}$, and $\frac{3}{\sqrt{14}}$.
Answer for screen readers
The direction cosines of the line connecting the origin to the point P(2, -1, 3) are: $$ \cos \alpha = \frac{2}{\sqrt{14}}, \quad \cos \beta = \frac{-1}{\sqrt{14}}, \quad \cos \gamma = \frac{3}{\sqrt{14}} $$
Steps to Solve
- Find the vector $\vec{OP}$
The vector $\vec{OP}$ from the origin $O(0,0,0)$ to the point $P(2, -1, 3)$ is given by: $$ \vec{OP} = \langle 2 - 0, -1 - 0, 3 - 0 \rangle = \langle 2, -1, 3 \rangle $$
- Calculate the magnitude of $\vec{OP}$
The magnitude (or length) of the vector $\vec{OP}$ is: $$ ||\vec{OP}|| = \sqrt{(2)^2 + (-1)^2 + (3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} $$
- Normalize the vector $\vec{OP}$
To find the direction cosines, we need to normalize the vector $\vec{OP}$. This means dividing the vector by its magnitude: $$ \hat{u} = \frac{\vec{OP}}{||\vec{OP}||} = \frac{\langle 2, -1, 3 \rangle}{\sqrt{14}} = \left\langle \frac{2}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right\rangle $$
- Identify the direction cosines
The components of the normalized vector $\hat{u}$ are the direction cosines. Therefore, the direction cosines are: $$ \cos \alpha = \frac{2}{\sqrt{14}}, \quad \cos \beta = \frac{-1}{\sqrt{14}}, \quad \cos \gamma = \frac{3}{\sqrt{14}} $$
The direction cosines of the line connecting the origin to the point P(2, -1, 3) are: $$ \cos \alpha = \frac{2}{\sqrt{14}}, \quad \cos \beta = \frac{-1}{\sqrt{14}}, \quad \cos \gamma = \frac{3}{\sqrt{14}} $$
More Information
Direction cosines are the cosines of the angles between the vector and the coordinate axes. They provide a way to describe the orientation of a vector in space. The sum of the squares of the direction cosines is always equal to 1. In this case, $(\frac{2}{\sqrt{14}})^2 + (\frac{-1}{\sqrt{14}})^2 + (\frac{3}{\sqrt{14}})^2 = \frac{4}{14} + \frac{1}{14} + \frac{9}{14} = \frac{14}{14} = 1$.
Tips
A common mistake is forgetting to normalize the vector. The direction cosines are the components of a unit vector, so dividing by the magnitude is crucial. Another mistake is incorrect calculation of the magnitude of the vector, especially with negative components.
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