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Thread: Projection of a vector onto a line

  1. #1
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    Smile Projection of a vector onto a line

    I need to find the projection of vector $\displaystyle \vec{a} = \sqrt{2}\cdot\vec{i}-3\cdot\vec{j}-5\cdot\vec{k}$ onto the line that makes $\displaystyle \alpha = 45^{\circ}$ angle with $\displaystyle Ox$, $\displaystyle \gamma = 60^{\circ}$ angle with $\displaystyle Oz$ and an acute angle with $\displaystyle Oy$
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  2. #2
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    Quote Originally Posted by sillyme View Post
    I need to find the projection of vector $\displaystyle \vec{a} = \sqrt{2}\cdot\vec{i}-3\cdot\vec{j}-5\cdot\vec{k}$ onto the line that makes $\displaystyle \alpha = 45^{\circ}$ angle with $\displaystyle Ox$, $\displaystyle \gamma = 60^{\circ}$ angle with $\displaystyle Oz$ and an acute angle with $\displaystyle Oy$
    If a line makes angles $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma$ with the x, y, and z axes respectively, then the "direction cosines" satisfy $\displaystyle cos^2(\alpha)+ cos^2(\beta)+ cos^2(\gamma)= 1$ and the vector $\displaystyle cos(\alpha)\vec{i}+ cos(\beta)\vec{j}+ cos(\gamma)\vec{k}$ is a unit vector in the direction of the line.

    Here, $\displaystyle \alpha= 45$ so $\displaystyle cos(\alpha)= \frac{1}{\sqrt{2}}$ and $\displaystyle cos^2(\alpha)= \frac{1}{2}$.

    $\displaystyle \beta= 60$ so $\displaystyle cos(\beta)= \frac{1}{2}$ and $\displaystyle cos^2(\beta)= \frac{1}{4}$.

    That means that $\displaystyle cos^2(\gamma)= 1- 1/2- 1/4= 1/4$ and so $\displaystyle cos(\gamma)= 1/2$

    A unit vector in the direction of the line is $\displaystyle \frac{\sqrt{2}}{2}\vec{i}+ \frac{1}{2}\vec{j}+ \frac{1}{2}\vec{k}$

    Can you find the projection of your given vector onto that?
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  3. #3
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    Yes. Now I can calculate the projection. Thank you HallsofIvy!
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