What is the projection of $< 2, 4, 3 >$ $<2,4,3 >$ onto $< - 3, 4, - 1 >$ $<-3,4,-1 >$ ?

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Answer 1 · 2017-11-29T07:13:19

The projection is $= \frac{7}{26} < - 3, 4, - 1 >$ $=7/26 <-3,4,-1>$

The vector projection of $\to v$ $vecv$ onto $\to u$ $vecu$ is

$p r o j_{u} v = \frac{\to u . \to v}{{(∣ ∣ ∣ ∣ \to u ∣ ∣ ∣ ∣)}^{2}} \cdot \to u$ $proj_uv= (vec u. vecv ) /(||vecu||)^2*vecu$

Here,

$\to v = < 2, 4, 3 >$ $vecv= < 2,4,3>$ and

$\to u = < - 3, 4, - 1 >$ $vecu = <-3,4,-1>$

The dot product is

$\to u . \to v = < 2, 4, 3 > . < - 3, 4, 1 \to$ $vecu.vecv =<2,4,3> . <-3,4,1->$

$= (2 \cdot - 3) + (4 \cdot 4) + (3 \cdot - 1) = - 6 + 16 - 3 = 7$ $=(2*-3)+(4*4)+(3*-1) =-6+16-3=7$

The magnitude of $\to u$ $vecu$ is

$= | | < - 3, 4, - 1 > | | = \sqrt{{(- 3)}^{2} + {(4)}^{2} + {(- 1)}^{2}}$ $=||<-3,4,-1 >|| = sqrt((-3)^2+(4)^2+(-1)^2)$

$= \sqrt{9 + 16 + 1} = \sqrt{26}$ $=sqrt(9+16+1)=sqrt26$

The vector projection is

$p r o j_{u} v = \frac{7}{26} \cdot < - 3, 4, - 1 >$ $proj_uv=7/26* <-3,4,-1>$

What is the projection of <2,4,3><2,4,3 > onto <−3,4,−1><-3,4,-1 >?