What is the projection of $< 4, - 6, 3 >$ $<4,-6,3 >$ onto $< 1, 5, 2 >$ $<1,5,2 >$ ?

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Answer 1 · 2018-06-04T08:11:13

The vector projection is $= < - \frac{2}{3}, - \frac{10}{3}, - \frac{4}{3} >$ $=<-2/3,-10/3,-4/3>$

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

$p r o j_{\to u} (\to v) = \frac{< \to u, \to v >}{< \to u, \to u >} \to u$ $proj_(vecu)(vecv)= (< vecu, vecv >)/ (< vecu, vecu >) vecu$

$\to u = < 1, 5, 2 >$ $vecu = <1, 5, 2>$

$\to v = < 4, - 6, 3 >$ $vecv= <4, -6,3>$

The dot product is

$< \to u, \to v > = < 1, 5, 2 > . < 4, - 6, 3 >$ $< vecu, vecv > = <1, 5, 2> .<4, -6,3>$

$= (1 \times 4) + (5 \times - 6) + (2 \times 3)$ $=(1xx4)+(5xx-6)+(2xx3)$

$= 4 - 30 + 6$ $=4-30+6$

$= - 20$ $=-20$

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

$< \to u, \to u > = | | < 1, 5, 2 > | | = \sqrt{1^{2} + {(5)}^{2} + 2^{2}}$ $< vecu, vecu > = ||<1, 5, 2>|| =sqrt(1^2+(5)^2+2^2)$

$= \sqrt{1 + 25 + 4}$ $=sqrt(1+25+4)$

$= \sqrt{30}$ $=sqrt30$

Therefore, the vector projection is

$p r o j_{\to u} (\to v) = - \frac{20}{30} < 1, 5, 2 >$ $proj_(vecu)(vecv)=-20/30<1, 5, 2>$

$= < - \frac{2}{3}, - \frac{10}{3}, - \frac{4}{3} >$ $= <-2/3,-10/3,-4/3>$

What is the projection of <4,-6,3 ><4,−6,3><4,-6,3 > onto <1,5,2 ><1,5,2><1,5,2 >?