+0

# Find the cross product of the following vector pairs. Then use the cross product to find the angle between them. a. a = 3i − 2j + 5k and b =

0
708
9

Find the cross product of the following vectors. Then use the cross product to find the angle between them.

a = 3i − 2j + 5k and b = 7i + 4j − 8k

Guest Jun 14, 2014

#1
+27246
+8

Alan  Jun 14, 2014
#1
+27246
+8

Alan  Jun 14, 2014
#2
0

sorry, im a little confused. did you just turn a and b into and v?

Guest Jun 14, 2014
#3
+27246
+5

I used u and v as general vectors.  a and b are your specific vectors.  That is they are specific examples of u and v.  But if you prefer to think of it as turning a and b into u and v ok!  Whatever works for you!

In fact, why don't you rewrite my u and v in terms of a and b (including their components - so u1 becomes a1 etc.)?

Alan  Jun 14, 2014
#4
0

thanks for the help

Guest Jun 14, 2014
#5
+27246
0

You're welcome.

Alan  Jun 14, 2014
#6
0

sorry, think i did something wrong..my a × b= -4,11,4.

so i'll call it y. y= a3 b1 - a1 b3.

a = 3 − 2 + 5 and b = 7 + 4 − 8

y=(5)(7)-(3)(8)
= 35-24
=11

I noticed however you got 59..where did i go wrong?

Guest Jun 14, 2014
#7
0

nevermind. i figured it out. two negatives is a positive. Thanks again for the help.

Guest Jun 14, 2014
#8
0

i have a new issue now.

p = −2i − 3j and q =4i +7j. (Note: these vectors are 3D).

Does that mean there HAS to be a third?

Guest Jun 14, 2014
#9
+27246
+5

Yes, there has to be a third component; but it is zero!

So p = −2i − 3j + 0k and q =4i +7j + 0k.  It measn p and q are both in the i,j plane; and their cross-product will be in the k direction.

Alan  Jun 15, 2014