What is the projection of $< 7, - 8, 3 >$ $<7,-8,3 >$ onto $< 5, - 6, 1 >$ $<5,-6,1 >$ ?

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Answer 1 · 2017-02-19T07:17:35

The vector projection is $= \frac{43}{31} < 5, - 6, 1 >$ $=43/31<5,-6,1>$
The scalar projection is $= \frac{86}{\sqrt{62}}$ $=86/sqrt62$

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

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

The dot product is

$\to a . \to b = < 5, - 6, 1 > \cdot < 7, - 8, 3 >$ $veca.vecb=<5,-6,1>*<7,-8,3>$

$= 5 \cdot 7 + (- 6 \cdot - 8) + 3 \cdot 1$ $=5*7+(-6*-8)+3*1$

$= 35 + 48 + 3$ $=35+48+3$

$= 86$ $=86$

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

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = | | < 5, - 6, 1 > | |$ $||veca||=||<5,-6,1>||$

$= \sqrt{25 + 36 + 1}$ $=sqrt(25+36+1)$

$= \sqrt{62}$ $=sqrt62$

The vector projection is

$= \frac{86}{62} \cdot < 5, - 6, 1 >$ $=86/62*<5,-6,1>$

$= \frac{43}{31} < 5, - 6, 1 >$ $=43/31<5,-6,1>$

The scalar projection is

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

$= \frac{86}{\sqrt{62}}$ $=86/sqrt62$

What is the projection of <7,-8,3 ><7,−8,3><7,-8,3 > onto <5,-6,1 ><5,−6,1><5,-6,1 >?