What is the projection of ## onto ##?

Question

What is the projection of #<5,8,3 ># onto #<2,4,-2 >#?

William Audy · Accepted Answer

The vector projection is #< 3,6,-3 >,# the scalar projection is #3sqrt6#.

Given #veca= < 5,8,3 ># and #vecb= < 2,4,-2 >,# we can find #proj_(vecb)veca#, the vector projection of #veca# onto #vecb# using the following formula: #proj_(vecb)veca=((veca*vecb)/(|vecb|))vecb/|vecb|# That is, the dot product of the two vectors divided by the magnitude of #vecb#, multiplied by #vecb# divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide #vecb# by its magnitude in order to obtain a unit vector (vector with magnitude of #1#). You might notice that the first quantity is scalar, as we know that when we take the dot product of two vectors, the resultant is a scalar. Therefore, the scalar projection of #a# onto #b# is #comp_(vecb)veca=(a*b)/(|b|)#, also written #|proj_(vecb)veca|#. Taking the dot product of the two vectors will allow us to get started. #veca*vecb=< 5,8,3 > * < 2,4,-2 ># #=> (5*2)+(8*4)+(3*-2)# #=>10+32-6=36# Then we can find the magnitude of #vecb# by taking the square root of the sum of the squares of each of the components. #|vecb|=sqrt((b_x)^2+(b_y)^2+(b_z)^2)# #|vecb|=sqrt((2)^2+(4)^2+(2)^2)# #=>sqrt(4+16+4)=sqrt(24)# And now we have everything we need to find the vector projection of #veca# onto #vecb#. #proj_(vecb)veca=(36)/sqrt(24)*(< 2,4,-2 >)/sqrt(24)# #=>(36 < 2,4,-2 >)/24# #=3/2< 2,4,-2 ># #=>< 3,6,-3 ># The scalar projection of #veca# onto #vecb# is just the first half of the formula, where #comp_(vecb)veca=(a*b)/(|b|)#. Therefore, the scalar projection is #36/sqrt(24)#. This could be simplified further if desired: #=>36/(2sqrt6)# #=>18/sqrt6# #=>(18sqrt6)/6# #=>3sqrt6# I hope that's useful.

Amelia Adams · Answer

undefined To find the projection of a vector $\vec{a}$ onto another vector $\vec{b}$, you can use the formula: $$\text{proj}_{\vec{b}}(\vec{a}) = \left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|^2}\right)\vec{b}$$ where $\cdot$ denotes the dot product and $\|\vec{b}\|$ is the magnitude of vector $\vec{b}$. Given $\vec{a} = <5,8,3>$ and $\vec{b} = <2,4,-2>$, we can calculate: $$\vec{a} \cdot \vec{b} = 5(2) + 8(4) + 3(-2) = 10 + 32 - 6 = 36$$ $$\|\vec{b}\|^2 = \sqrt{2^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}$$ $$\text{proj}_{\vec{b}}(\vec{a}) = \left(\frac{36}{24}\right)<2,4,-2>$$ $$\text{proj}_{\vec{b}}(\vec{a}) = \frac{3}{2}<2,4,-2>$$ $$\text{proj}_{\vec{b}}(\vec{a}) = <3,6,-3>$$

What is the projection of #&lt;5,8,3 &gt;# onto #&lt;2,4,-2 &gt;#?

What is the projection of #<5,8,3 ># onto #<2,4,-2 >#?