What is the projection of $< 5, - 2, 3 >$ $<5,-2,3 >$ onto $< 1, - 6, - 3 >$ $<1,-6,-3 >$ ?

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What is the projection of $< 5, - 2, 3 >$ $<5,-2,3 >$ onto $< 1, - 6, - 3 >$ $<1,-6,-3 >$ ?

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Answer 1 · 2018-01-12T07:32:33

The projection is $= \frac{4}{23} < 1, - 6, - 3 >$ $=4/23<1, -6,-3>$

Explanation:

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

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

$\to a = < 1, - 6, - 3 >$ $veca=<1,-6,-3>$

$\to b = < 5, - 2, 3 >$ $vecb= <5, -2,3>$

The dot product is

$\to a . \to b = < 1, - 6, - 3 > . < 5, - 2, 3 >$ $veca.vecb =<1,-6,-3>. <5,-2,3>$

$= 1 \cdot 5 + (- 6) \cdot (- 2) + (- 3) \cdot 3 = 5 + 12 - 9 = 8$ $= 1*5+(-6)*(-2)+(-3)*3=5+12-9=8$

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

$= ∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = | | < 1, - 6, - 3 > | | = \sqrt{1^{2} + {(- 6)}^{2} + {(- 3)}^{2}} = \sqrt{46}$ $=||veca||=||<1,-6,-3>|| =sqrt(1^2+(-6)^2+(-3)^2)=sqrt46$

Therefore,

$p r o j_{\to a} \to b = \frac{8}{46} < 1, - 6, - 3 > = \frac{4}{23} < 1, - 6, - 3 >$ $proj_(veca)vecb=8/46<1, -6,-3> = 4/23<1, -6,-3>$

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What is the projection of $< 5, - 2, 3 >$ $<5,-2,3 >$ onto $< 1, - 6, - 3 >$ $<1,-6,-3 >$ ?

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Explanation:

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