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I'm trying to do homework for my physics class, and it says I should find 'the component of $\vec{a}$ along the direction of $\vec{b}$'. The vectors are:

$\vec{a} = 7.1\hat i + 8.97 \hat j $

$\vec{b} = 5.8\hat i + 2.5\hat j$

I know how to find the $x$ and $y$ components but I've never done this before. How do I do it?

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Compute $\dfrac{\vec a \cdot \vec b}{|b|}$ as the component of $\vec a$ along $\vec b$.

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Using the formula

$$\text{comp}_b a = \frac{a \cdot b}{\vert b \vert}$$

with the given vectors

$$\vec{a} = <7.1, 8.9>$$ $$\vec{b} = <5.8,2.5>$$ we get that

$$a\cdot b = <7.1, 8.9>\cdot <5.8, 2.5> = (7.1\cdot 5.8) + (8.9\cdot 2.5) = 63.43$$

Then $$\vert b\vert = \sqrt{{5.8}^{2} + 2.5^{2}} = \sqrt{33.64 + 6.25}= \sqrt{39.89}$$

Therefore the answer is $$\frac{63.43}{\sqrt{39.89}}$$

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I believe the component of A along B must be a vector. The previous answer gives the length of the component of A along B. Now that must be multiplied by a unit vector in the direction of B. So my answer would be:

$comp_{b}A = \frac {A \cdot B}{|B|}$ multiplying by the vector $\frac {B}{|B|}$ we get $\frac {A \cdot B}{|B||B|} B = \frac {A \cdot B}{B \cdot B} B$

Then to get the component of A perpendicular to B, you subtract that from A.

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  • $\begingroup$ Precise answer ! $\endgroup$
    – Vicrobot
    Jul 14 '20 at 18:26
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Hint: the component of $a$ along $b$ (also known as the scalar projection of $a$ onto $b$) is given by

$$\text{comp}_b a = \frac{a \cdot b}{\vert b \vert}$$

where $a \cdot b$ is the dot product.

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