P={ a|a=(1, m)} Q={ b|b=(1-n,1+n)} P and Q are sets of vectors, How to find P∩Q? P={ a|a=(1, m)} Q={ b|b=(1-n,1+n)} P and Q are sets of vectors, P∩Q How P∩Q?

P={ a|a=(1, m)} Q={ b|b=(1-n,1+n)} P and Q are sets of vectors, How to find P∩Q? P={ a|a=(1, m)} Q={ b|b=(1-n,1+n)} P and Q are sets of vectors, P∩Q How P∩Q?

Let 1-n=1 give n=0,
So 1+n=1, then m=1,
So P∩Q={(1,1)}.

The known vector ab is a nonzero vector,|a-b|=|a||b| Sorry I didn't finish. What condition should vector a vector b satisfy

A and b are mutually opposite vectors

If vector AB is a unit vector, is vector BA a unit vector? Is the unit vector direction certain? Thank you, sir.

0

Find the unit vector collinear with vector a=(6,8).

The modulus of a is the root number (6^2+8^2)=10
The unit vector is the modulus of a/a
Yes (3/5,4/5)
Same direction, size 1

The modulus of a is the root number (6^2+8^2)=10
The unit vector is the module of a/a
Yes (3/5,4/5)
Same direction, size 1

If vector a=(6,-8), then the unit vector collinear with vector a is

(6/10,-8/10)=(0.6,-0.8)

The unit vector coordinates perpendicular to vector a=(-6,8) are (8,6)。 (-8,-6) Why not

Let b=(x, y)
Then: x^2+y^2=1
A⊥b, i.e.:
A·b=(-6,8)·(x, y)=-6x+8y=0
I.e.3x=4y
Therefore:(3x/4)^2+x^2=1
I.e. x^2=16/25
I.e. x =4/5 or -4/5
I.e. y =3/5 or -3/5
Therefore: b=(4/5,3/5) or b=(-4/5,-3/5)
There are two answers. No problem.