Question

A question on vectors

Answer 1

Vectors are
A=i+j-2k
B= 3i +3j-6k

(A) Parallel
(B)Antiparralel
(C)Perpendicular
(D)at an acute angle with each other


Are they parralel

Answer 2

Indeed, they are. Do you see why?

Answer 3

I mean take out three common from the B vector , you get A . The ratio is the same

Answer 4

vector(A)=k*vector(B)
here k=3

Answer 5

That is correct. Remember that vectors are parallel (if we include antiparallel as a different case) iff there is some scalar \(k>0\) such that:
\[
\mathbf{a}=k\mathbf{b}
\]

Answer 6

But i have another question, actually

Answer 7

Go for it

Answer 8

Let's break this question into more parts , although it is not asking
How do we know vectors are anti parallel

Answer 9

We know vectors are antiparallel if there is some scalar \(k<0\) (!!) such that \(\mathbf{a}=k\mathbf{b}\).

Answer 10

Oh , that's news to to me hehe , learn't something new
So if the scalar (k) is less than zero they are antiparralel right?

Answer 11

Yes, though some people consider antiparallel and parallel to be just 'parallel' so be careful because it depends on the literature.

Answer 12

oh, How to know if they

Answer 13

perpendicular

Answer 14

That requires the definition of the dot product, have you studied it, yet? I can introduce it, if not.

Answer 15

Yeah i know all of that stuff

Answer 16

dot and scalar product or cross product

Answer 17

Yeah, so the scalar product allows you to know if they're perpendicular as \(\mathbf{a\perp b}\) iff \(\mathbf{a\cdot b}=0\).

Answer 18

And how to know the the last option

Answer 19

so here in "perpendicular" the theta must be 90

Answer 20

ab = a.b.cos90

Answer 21

great!!

Answer 22

Well, since:
\[
a\cdot b=|a||b|\cos\theta
\]Rearranging gives us:
\[
\frac{a\cdot b}{|a||b|}=\cos\theta
\]Hence it is acute when \(\cos \theta > \cos \frac{\pi}{2} > 0\), so:
\[
\frac{a\cdot b}{|a||b|}>0
\]

Answer 23

Yep, I believe you were right on track!

Answer 24

I am reading this waiot

Answer 25

i mean wait

Answer 26

Messed up, I meant:
\[
\cos \theta > \cos \frac{\pi}{2} = 0
\]

Answer 27

yeah , didn't get the part , the once you wrote just know

Answer 28

Would you like me to explain a step?

Answer 29

If you don't mind

Answer 30

Sure, which one?

Answer 31

From ,
Hence it is..............

Answer 32

@LolWolf r u there

Answer 33

Yeah, sorry, responding to something

Answer 34

oh oh np :)

Answer 35

Right, so we want \(0\le \theta<\frac{\pi}{2}\), right? So, we can apply the cosine function to all sides, giving us:
\[
\cos0\ge\cos\theta>\cos\frac{\pi}{2}
\]Actually getting each value, then:
\[
1\ge\cos\theta>0
\]Since \(\cos\theta\) is always \(\le 1\), then:
\[
\cos\theta >0
\]Is our only requirement.

Answer 36

Great thanks for the efforts you put in , I aprreciate your help.

Answer 37

Sure thing, that's what I'm here for!

Answer 38

sorry , i had internet problem

Why do signs change when we multiply cos throughout

Answer 39

@LolWolf

Answer 40

Since it's a decreasing function from \(0\) to \(\frac{\pi}{2}\), the inequality reverses.

Answer 41

i see