What is the angle between $< 9, 1, 4 >$ $<9,1,4>$ and $< 6, - 1, 4 >$ $<6,-1,4 >$ ?

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Answer 1 · 2018-08-02T06:27:33

$θ = {cos}^{- 1} (\frac{69}{7 \sqrt{106}})$ $theta=cos^-1(69/(7sqrt106))$

Let , $\to a = < 9, 1, 4 > and \to b = < 6, - 1, 4 >$ $veca=<9,1,4> and vecb=<6,-1,4>$ be the two vectors.

$:.|veca|=sqrt((9)^2+(1)^2+(4)^2)=sqrt(81+1+16)=sqrt98=7sqrt2$

$and|vecb|=sqrt((6)^2+(-1)^2+(4)^2)=sqrt(36+1+16)=sqrt53$

Dot product : $a*b=9xx6+1xx(-1)+4xx4$

$:.a*b=54-1+16=69$

Now the angle between $veca and vecb$ is:

$theta=cos^-1((a*b)/(|veca||vecb|))=cos^-1(69/(7sqrt2sqrt53))$

$:. theta=cos^-1(69/(7sqrt106))=$ $(16.78)^circ$

What is the angle between <9,1,4> <9,1,4><9,1,4> and <6,-1,4 ><6,−1,4><6,-1,4 >?