What is the projection of $< - 3, 1, 3 >$ $< -3 , 1 ,3 >$ onto $< 8, 5, - 3 >$ $< 8 ,5 ,-3>$ ?

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Answer 1 · 2017-01-28T10:23:58

The vector projection is $= - \frac{2}{7} < 8, 5, - 3 >$ $=-2/7<8,5,-3>$
The scalar projection is $= - \frac{4}{\sqrt{2}}$ $=-4/sqrt2$

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

$= \frac{\to a . \to b}{{∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣}^{2}} \to a$ $=(veca.vecb)/(||veca||^2)veca$

The dot product is $\to a . \to b = < - 3, 1, 3 > . < 8, 5, - 3 >$ $veca.vecb=<-3,1,3>.<8,5,-3>$

$= - 24 + 5 - 9 = - 28$ $=-24+5-9=-28$

The modulus of $\to a$ $veca$ is $∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = | | < 8, 5, - 3 > | |$ $||veca||=||<8,5,-3>||$

$= \sqrt{64 + 25 + 9} = \sqrt{98}$ $=sqrt(64+25+9)=sqrt98$

The vector projection is $= - \frac{28}{98} < 8, 5, - 3 >$ $=-28/98<8,5,-3>$

$= - \frac{2}{7} < 8, 5, - 3 >$ $=-2/7<8,5,-3>$

The scalar projection is $= \frac{\to a . \to b}{∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣}$ $=(veca.vecb)/(||veca||)$

$= - \frac{28}{\sqrt{98}} = - \frac{4}{\sqrt{2}}$ $=-28/sqrt98=-4/sqrt2$

What is the projection of < -3 , 1 ,3 ><−3,1,3>< -3 , 1 ,3 > onto < 8 ,5 ,-3><8,5,−3>< 8 ,5 ,-3>?