What is the dot product of $< - 4, 0, - 2 >$ $<-4,0,-2 >$ and $< 3, - 9, 1 >$ $<3,-9,1>$ ?

Question

1634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-01T14:37:25

The answer is $= - 14$ $=-14$

The dot product of $2$ $2$ vectors

$\to x = < a, b, c >$ $vecx= < a,b,c >$

and

$\to y = < d, e, f >$ $vecy = < d,e,f >$

is

$\to x . \to y = < a, b, c > . < d, e, f > = a d + b e + c f$ $vecx . vecy = < a,b, c> . < d, e, f > =ad+be+cf$

Therefore,

$< - 4, 0, - 2 > . < 3, - 9, 1 > = (- 4) \cdot (3) + (0) \cdot (- 9) + (- 2) \cdot (1)$ $< -4,0,-2 > . < 3, -9, 1> = (-4)*(3)+(0)*(-9)+(-2)*(1)$

$= - 12 + 0 - 2$ $=-12+0-2$

$= - 14$ $=-14$

What is the dot product of <−4,0,−2><-4,0,-2 > and <3,−9,1><3,-9,1>?