Calculating the Angle Between Two Vectors

[needhelp83]

Calculate the angle between the vectors:
A = 6.8i + 4.5j + 6.2k
B = 8.2i + 2.3j – 7.0k


A*B= AB cos
A*B=AxBx + AyBy + AzBz
A*B= (6.8)(8.2) + (4.5)(2.3) + (6.2)(-7.0)=22.71
22.71 cos = 87.48 degrees

Correct?

 

[radou]

Calculate the angle between the vectors:
A = 6.8i + 4.5j + 6.2k
B = 8.2i + 2.3j – 7.0k


A*B= AB cos
A*B=AxBx + AyBy + AzBz
A*B= (6.8)(8.2) + (4.5)(2.3) + (6.2)(-7.0)=22.71
22.71 cos = 87.48 degrees

Correct?

What exactly did you write here? A.B=|A||B|cos(A,B) implies cos(A,B) = (A.B) / (|A||B|).

 

[needhelp83]

What exactly did you write here? A.B=|A||B|cos(A,B) implies cos(A,B) = (A.B) / (|A||B|).

I am not too sure what you mean by the A.B and |A||B|

 

[Ja4Coltrane]

A.B means A "dot" B
This is a dot product and conceptually means to multiply two vectors only in the same direction as each other (for example, if you push on a cart downward, and it goes forward because of the angle of your force, the dot product would be the magnitude (that's what the lines mean) of the force in the direction of the distance (that's why Cos@) times the magnitude of the distance) or you could do magnitude of distance in direction of force times magnitude of force.

 

[OlderDan]

Calculate the angle between the vectors:
A = 6.8i + 4.5j + 6.2k
B = 8.2i + 2.3j – 7.0k


A*B= AB cosθ
A*B=AxBx + AyBy + AzBz
A*B= (6.8)(8.2) + (4.5)(2.3) + (6.2)(-7.0)=22.71
22.71 cos = 87.48 degrees

Correct?

The 22.71 part is good, but where did you calculate the AB in AB cosθ?

 

[radou]

I am not too sure what you mean by the A.B and |A||B|

By |A||B| I meant what you denoted as AB - the product of the absolute values of the vectors.

 

[needhelp83]

A=6.8+4.5+6.2=17.5
B=8.2+2.3-7.0=3.5

AB=61.25

22.71/61.25=cos theta=68.28 degrees

I think it should be 78.40 degrees because I found this dot product calculator, but I am not sure how they go to this answer

 

[Doc Al]

A=6.8+4.5+6.2=17.5
B=8.2+2.3-7.0=3.5

The components of a vector are perpendicular--you can't just add them to find the magnitude!

How you find the magnitude of a vector from its components?

 

[OlderDan]

A=6.8+4.5+6.2=17.5
B=8.2+2.3-7.0=3.5

AB=61.25

22.71/61.25=cos theta=68.28 degrees

I think it should be 78.40 degrees because I found this dot product calculator, but I am not sure how they go to this answer

That is not correct for AB
A = sqrt(A*A)
B = sqrt(B*B)
where I have used your symbol * for the dot product.

What you did for A*B was perfect. Do it again for A*A and B*B To find A and B and their product.

 

[needhelp83]

A=sqrt(AxAx + AyAy + AzAz)=sqrt((6.8)(6.8) + (4.5)(4.5) + (6.2)(6.2))=10.24

B=sqrt(BxBx + ByBy + BzBz)=sqrt((8.2)(8.2) + (2.3)(2.3) + (-7.0)(-7.0))=11.02

AB=(10.24)(11.02)=112.85

22.71/112.85=0.20 cos theta= 78.46 degrees

Think I finally got it. Correct?

 

[radou]

A=sqrt(AxAx + AyAy + AzAz)=sqrt((6.8)(6.8) + (4.5)(4.5) + (6.2)(6.2))=10.24

B=sqrt(BxBx + ByBy + BzBz)=sqrt((8.2)(8.2) + (2.3)(2.3) + (-7.0)(-7.0))=11.02

AB=(10.24)(11.02)=112.85

22.71/112.85=0.20 cos theta= 78.46 degrees

Think I finally got it. Correct?

Looks good.