Question

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.)

a=j+k , b=i+2j-3k

Answer 1

I got arccos=(1/2sqrt7) It's suppose to be a negative one instead though... I can't find the problem unless the book is wrong.

Answer 2

wait a min I think I know what I did wrong...

Answer 3

Is i=0?

Answer 4

What do you mean?
for the first vector, \(\large\vec a\) the i-component is zero...

Answer 5

Okay then I know what I did wrong... I am dyslexic. Lol

Answer 6

Well... just to get you started, the cosine of the angle between two vectors is given by
\[\Large \cos(\theta) = \frac{\vec u \cdot \vec v}{||\vec u || \ ||\vec v ||}\]

Answer 7

i tend to stack, then multiply down the cols, and add the results for a dot product

a=0+j+k ,
b=i+2j-3k
-----------
2 - 3 = -1 right?

Answer 8

Whatever works :D

Answer 9

\[\Large ||\vec a || = \sqrt 2\]\[\Large \left|\left|\vec b\right|\right|=\sqrt {14}\]

Answer 10

i thought is was |b| = sqrt(2*7) :)

Answer 11

Oh, silly me, how could I have missed that? D:
<slashes wrists>

Answer 12

Okay thanks guys. I was looking at j as i and made j=0 so that must have screwed me up somewhere...

Answer 13

It'd be nice if you show your answer...

Answer 14

Magnitude of a=sqrt2 and magnitude b=sqrt14

Answer 15

No, I mean your final answer as to the angle between the two vectors.

Answer 16

ohh Haha

Answer 17

theta=arc cos(-1/2sqrt17) =100.893* which is about 101*

Answer 18

...trying to visualise in my head...

Answer 19

That does seem about right :D

Answer 20

since a*b=-1 the angle is going to be greater than 90*

Answer 21

Okay, it's correct.
Well done ^_^

Answer 22

If it was 1 then the angle would be less than 90*

Answer 23

cos is negative in q2 in q3 yes, and positive in q1 and q4

Answer 24

cos(90) = 0

Answer 25

I don't know these shortcuts.. D:

Answer 26

well if youd spend less time slashing, and more time nose grinding ...