What is the projection of $< - 5, 2, 8 >$ $<-5,2,8>$ onto $< 4, - 6, 3 >$ $<4,-6,3 >$ ?

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What is the projection of $< - 5, 2, 8 >$ $<-5,2,8>$ onto $< 4, - 6, 3 >$ $<4,-6,3 >$ ?

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Answer 1 · 2018-01-26T10:55:13

The projection is $= - \frac{8}{\sqrt{61}} < 4, - 6, 3 >$ $=-8/sqrt61<4, -6,3>$

Explanation:

The vector projection of $\to b$ $vecb$ onto $\to a$ $veca$ is

$p r o j_{\to a} \to b = \frac{\to a . \to b}{{(∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣)}^{2}} \to a$ $proj_(veca)vecb=(veca.vecb)/(||veca||)^2veca$

$\to a = < 4, - 6, 3 >$ $veca=<4,-6,3>$

$\to b = < - 5, 2, 8 >$ $vecb= <-5, 2,8>$

The dot product is

$\to a . \to b = < 4, - 6, 3 > . < - 5, 2, 8 >$ $veca.vecb =<4,-6,3>. <-5,2,8>$

$= (4) \cdot (- 5) + (- 6) \cdot (2) + (3) \cdot (8) = - 20 - 12 + 24 = - 8$ $= (4)*(-5)+(-6) *(2)+(3)*(8)=-20-12+24=-8$

The modulus of $\to a$ $veca$ is

$= ∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = | | < 4, - 6, 3 > | | = \sqrt{{(4)}^{2} + {(- 6)}^{2} + {(3)}^{2}} = \sqrt{61}$ $=||veca||=||<4,-6,3>|| =sqrt((4)^2+(-6)^2+(3)^2)=sqrt61$

Therefore,

$p r o j_{\to a} \to b = - \frac{8}{\sqrt{61}} < 4, - 6, 3 >$ $proj_(veca)vecb=-8/sqrt61<4, -6,3>$

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What is the projection of $< - 5, 2, 8 >$ $<-5,2,8>$ onto $< 4, - 6, 3 >$ $<4,-6,3 >$ ?

1 Answer

Explanation:

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What is the projection of <−5,2,8><-5,2,8> onto <4,−6,3><4,-6,3 >?

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