Question #0b785

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Question #0b785

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Answer 1 · 2017-04-14T11:40:13

The vector projection is $= \frac{13}{97} < 7, - 8, - 9 >$ $=13/97<7,-8,-9>$
The scalar projection is $= \frac{26}{\sqrt{194}}$ $=26/sqrt194$

Explanation:

The vector projection of $\to A$ $vecA$ over $\to B$ $vecB$ is

$= \frac{\to A . \to B}{{∣ ∣ ∣ ∣ ∣ ∣ \to B ∣ ∣ ∣ ∣ ∣ ∣}^{2}} \to B$ $=(vecA.vecB)/(||vecB||^2)vecB$

$\to A = < 2, - 6, 4 >$ $vecA=<2,-6,4>$

$\to B = < 7, - 8, - 9 >$ $vecB=<7,-8,-9>$

The dot product is

$\to A . \to B = < 2, - 6, 4 > . < 7, - 8, - 9 > = (2 \cdot 7 + 6 \cdot 8 - 9 \cdot 4) = 14 + 48 - 36 = 26$ $vecA.vecB=<2,-6,4>.<7,-8,-9> =(2*7+6*8-9*4)=14+48-36=26$

The modulus of $\to B$ $vecB$ is

$= ∣ ∣ ∣ ∣ ∣ ∣ \to B ∣ ∣ ∣ ∣ ∣ ∣ = | | < 7, - 8, - 9 > | | = \sqrt{7^{2} + 8^{2} + 9^{2}} = \sqrt{194}$ $=||vecB||=||<7,-8,-9>||=sqrt(7^2+8^2+9^2)=sqrt194$

The vector projection is

$= \frac{26}{194} < 7, - 8, - 9 >$ $=26/194<7,-8,-9>$

The scalar projection is

$= \frac{\to A . \to B}{∣ ∣ ∣ ∣ ∣ ∣ \to B ∣ ∣ ∣ ∣ ∣ ∣} = \frac{26}{\sqrt{194}}$ $=(vecA.vecB)/(||vecB||)=26/sqrt194$

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Question #0b785

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