What is the projection of $< - 4, 8, 9 >$ $<-4,8,9 >$ onto $< 8, 1, - 1 >$ $<8,1,-1 >$ ?

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Answer 1 · 2017-12-05T18:09:07

The projection is $= < - 4, - \frac{1}{2}, \frac{1}{2} >$ $=<-4,-1/2, 1/2>$

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,

$\to v = < - 4, 8, 9 >$ $vecv= < -4,8,9>$ and

$\to u = < 8, 1, - 1 >$ $vecu = <8,1,-1>$

The dot product is

$\to u . \to v = < - 4, 8, 9 > . < 8, 1, - 1 >$ $vecu.vecv =<-4,8,9> . <8,1,-1>$

$= (- 4 \cdot 8) + (8 \cdot 1) + (9 \cdot - 1) = - 32 + 8 - 9 = - 33$ $=(-4*8)+(8*1)+(9*-1) =-32+8-9=-33$

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

$= | | < 8, 1, - 1 > | | = \sqrt{{(8)}^{2} + {(1)}^{2} + {(- 1)}^{2}}$ $=||<8,1,-1 >|| = sqrt((8)^2+(1)^2+(-1)^2)$

$= \sqrt{64 + 1 + 1} = \sqrt{66}$ $=sqrt(64+1+1)=sqrt66$

The vector projection is

$p r o j_{u} v = - \frac{33}{66} \cdot < 8, 1, - 1 > = < - 4, - \frac{1}{2}, \frac{1}{2} >$ $proj_uv=-33/66* <8,1,-1> = <-4,-1/2, 1/2>$

What is the projection of <−4,8,9><-4,8,9 > onto <8,1,−1><8,1,-1 >?