What is the projection of ## onto ##?

Question

What is the projection of #<3,4,-1 ># onto #<-1,3,-6 >#?

William Audy · Accepted Answer

The projection is #(-15/sqrt46 , 45/sqrt46 , -90/sqrt46)#.

Given two vectors #vecV# and #vecW#, the projection of #vecV# onto #vecW# is given by: #(vecV * vecW)vecW/||vecW||# The inner product gives the component of #VecV# in the direction of #VecW#, and the fraction performs the same function as multiplying this magnitude by a unit vector in the direction of #vecW#. So plugging in given values: #vecV = (3,4,-1)# #vecW = (-1,3,-6)# #||vecW|| = sqrt(1 + 9 + 36) = sqrt(46)# #vecV*vecW = (3*-1) + (4*3) + (-1*-6)# #= -3 + 12 + 6 = 15# So plugging everything in, the projection is: #(15*(-1,3,-6))/sqrt(46) = (-15/sqrt46 , 45/sqrt46 , -90/sqrt46)#

Charles Aeschliman · Answer

undefined The projection of $ \langle 3,4,-1 \rangle $ onto $ \langle -1,3,-6 \rangle $ is $ \langle -2,-1,-1 \rangle $.

Nicholas Agrusa · Answer

undefined To find the projection of vector $\langle 3,4,-1 \rangle$ onto vector $\langle -1,3,-6 \rangle$, we use the formula for the projection of vector $a$ onto vector $b$:

$$\text{proj}_b(a) = \frac{a \cdot b}{\lVert b \rVert^2} \cdot b$$

First, we compute the dot product of vectors $a$ and $b$:

$$a \cdot b = (3 \cdot -1) + (4 \cdot 3) + (-1 \cdot -6) = -3 + 12 + 6 = 15$$

Next, we calculate the magnitude of vector $b$:

$$\lVert b \rVert = \sqrt{(-1)^2 + 3^2 + (-6)^2} = \sqrt{1 + 9 + 36} = \sqrt{46}$$

Then, we substitute these values into the formula:

$$\text{proj}_b(a) = \frac{15}{46} \cdot \langle -1,3,-6 \rangle$$

$$= \frac{15}{46} \cdot \langle -1,3,-6 \rangle$$

$$= \left\langle \frac{-15}{46}, \frac{45}{46}, \frac{-90}{46} \right\rangle$$

Therefore, the projection of vector $\langle 3,4,-1 \rangle$ onto vector $\langle -1,3,-6 \rangle$ is $\left\langle \frac{-15}{46}, \frac{45}{46}, \frac{-90}{46} \right\rangle$.

What is the projection of #&lt;3,4,-1 &gt;# onto #&lt;-1,3,-6 &gt;#?

What is the projection of #<3,4,-1 ># onto #<-1,3,-6 >#?