MHB Projection of $\overrightarrow{c}$ on $\overrightarrow{a}$: Example

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To find the projection of vector $\overrightarrow{c}$ on vector $\overrightarrow{a}$, the formula used is $$\frac{\overrightarrow{c} \cdot \overrightarrow{a}}{||\overrightarrow{a}||^2}\overrightarrow{a}$$. An example with $\overrightarrow{c} =(4, 2, -6)$ and $\overrightarrow{a}=(-2, 2, 2)$ shows the calculation steps leading to the projection result. The calculations confirm that the projection is correctly derived. The discussion concludes with affirmation that the computed projection is indeed accurate.
mathmari
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Hey! :o

To find the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$ do we have to use the formula $$\frac{\overrightarrow{c} \cdot \overrightarrow{a}}{||\overrightarrow{a}||^2}\overrightarrow{a}$$ ?? (Wondering)

For example, if we have $\overrightarrow{c} =(4, 2, -6)$ and $\overrightarrow{a}=(-2, 2, 2)$ :

$$\frac{(4, 2, -6) \cdot (-2, 2, 2)}{||(-2, 2, 2)||^2}(-2, 2, 2)=\frac{-8+4-12}{4+4+4}(-2, 2, 2)=\frac{-16}{12}(-2, 2, 2)=\frac{-4}{3}(-2, 2, 2)$$

Is this the asked projection?? (Wondering)
 
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Yes, that is correct, assuming you are to compute the vector projection. (Yes)
 
Rido12 said:
Yes, that is correct, assuming you are to compute the vector projection. (Yes)

Great! Thank you! (Yes)
 

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