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I'm still learning how to work with vectors and ran to this question...

Given vector: $V = 3i + 4j$ and for vector $F = 9i + 12j$

a) Find the component of F parallel to V

b) the component of F perpendicular to V

c) The work, W, done by force F through displacement

For question a) i know that to vectors a parallel when $\lambda(V) = F$ but i'm not so sure how use it with equations i'm just used to working with the $(x,y,z)$ vectors

For question b) I think to be perpendicular the projection of v - projection f must be 0.

For question c) i got nothing

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  • $\begingroup$ Are vectors $F$ and $f$ the same? $\endgroup$ – N. F. Taussig Nov 6 '14 at 1:14
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Note that $\textbf{F} = 3\textbf{v}$ so it's clear that $\textbf{F}$ is parallel to $\textbf{v}$ or $\textrm{proj}_\textbf{v} \textbf{F} = \textbf{F}$. It follows that the perpendicular component of $\textbf{F}$ is $\textbf{0}$.

For the last question, work is $W = \int \textbf{F}\cdot d\textbf{r}$ where $\textbf{r}$ is displacement. You didn't specify what the displacement is. $\textbf{v}$ is usually a notation for velocity. If $\textbf{v}$ is displacement then

$$W = \textbf{F} \cdot \textbf{v} = (3\textbf{i}+4\textbf{j})\cdot(9\textbf{i}+12\textbf{j}) = 3\cdot 9 + 4 \cdot 12 = 75 $$

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The component of f parallel to $v$ is the projection of $4$ onto $v$, $\text{proj}_v(f)$. The component of $f$ perpendicular to $v$ is $f-\text{proj}_v(f)$.

Try drawing vectors in $\mathbb{R}^2$ to see this. Draw two vectors $f$ and $v$. Then draw the projection of $f$ onto $v$.

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For a) v = 3i + 4j = <3,4> f = 9i + 12j = <9,12>

f = v

when f = 3v

Hence since vectors are separated by a factor of 3 from each other then they are parallel

For b) By definition two vectors are parallel when the dot product is zero $a . b = x1x2 + y2y3 = 0 $

$$a.b = 0$$ but the dot product is ... $f.v = 3*9 + 4+12 = 27+48 = 75$ which is not zero hence they are no perpendicular

For c) the work is the dot product $w = f.v = |f|*|v|*cos(\theta)$

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