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

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

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Answer 1 · 2018-01-11T18:03:33

The projection is $= \frac{76}{61} < 4, - 6, 3 >$ $=76/61<4, -6,3>$

Explanation:

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

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

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

$\to b = < 4, - 6, 8 >$ $vecb= <4, -6,8>$

The dot product is

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

$= 4 \cdot 4 + (- 6) \cdot (- 6) + 3 \cdot 8 = 16 + 36 + 24 = 76$ $= 4*4+(-6)*(-6)+3*8=16+36+24=76$

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,

$p r o j_{\to a} \to b = \frac{76}{61} < 4, - 6, 3 >$ $proj_(veca)vecb=76/61<4, -6,3>$

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

1 Answer

Explanation:

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