What is the projection of $(i - 2 j + 3 k)$ $(i -2j + 3k)$ onto $(2 i + j + 2 k)$ $( 2i+j+2k)$ ?

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Answer 1 · 2018-07-08T08:58:21

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

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

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

The vectors are

$\to a = < 2, 1, 1 >$ $veca= <2,1,1 >$

and

$\to b = < 1, - 2, 3 >$ $vecb= <1,-2,3>$

The dot product is

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

$= (2 \times 1) + (1 \times (- 2)) + 1 \times 3 = 2 - 2 + 3 = 3$ $= (2xx1)+(1xx(-2))+ 1xx3=2-2+3=3$

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

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = < 2, 1, 1 > = \sqrt{2^{2} + 1^{2} + 1^{2}} = \sqrt{6}$ $||veca|| = <2,1,1> = sqrt(2^2+1^2+1^2)=sqrt6$

Therefore,

$p r o j_{\to a} \to b = \frac{3}{6} \cdot < 2, 1, 1 >$ $proj_(veca)vecb=(3)/(6)*<2,1,1 >$

$= < 1, \frac{1}{2}, \frac{1}{2} >$ $= <1,1/2,1/2>$

What is the projection of (i -2j + 3k)(i−2j+3k)(i -2j + 3k) onto ( 2i+j+2k)(2i+j+2k) ( 2i+j+2k)?