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

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Answer 1 · 2018-05-08T06:19:13

The projection is $= - \frac{2}{3} \to i - \frac{4}{9} \to j + \frac{2}{3} \to k$ $=-2/3veci-4/9vecj+2/3veck$

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

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

Here

$\to a = < 3, 2, - 3 >$ $veca= <3,2,-3>$

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

The dot product is

$\to a . \to b = < 3, 2, - 3 > . < - 1, 1, 1 > = - 3 + 2 - 3 = - 4$ $veca.vecb = <3,2,-3>. <-1,1,1> = -3+2-3=-4$

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

$∣ ∣ \to a ∣ ∣ = | < 3, 2, - 3 > | = \sqrt{9 + 4 + 9} = \sqrt{18}$ $|veca|=|<3,2, -3>| = sqrt(9+4+9)=sqrt18$

Therefore,

$p r o j_{\to a} \to b = - \frac{4}{18} < 3, 2, - 3 >$ $proj_(veca)vecb=-4/18 <3,2,-3>$

$= - \frac{2}{9} < 3, 2, - 3 >$ $=-2/9 <3,2,-3>$

$= < - \frac{2}{3}, - \frac{4}{9}, \frac{2}{3} >$ $= <-2/3, -4/9, 2/3>$

$= - \frac{2}{3} \to i - \frac{4}{9} \to j + \frac{2}{3} \to k$ $=-2/3veci-4/9vecj+2/3veck$

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