If vector a and vector b are both unit vectors and the angle between vector a and vector b is 60°, then |vector a+vector b|=?

If vector a and vector b are both unit vectors and the angle between vector a and vector b is 60°, then |vector a+vector b|=?

0

Given that vectors a, b are unit vectors, if their included angle is 60°, what is |a-3b| equal to?

Ab=|a b|*cos60=1/2.
|A-3b|^2=a^2+9b^2-6ab=1+9-6*1/2=7,
| A-3b |=√7.

Given that the vectors a, b are unit vectors and their included angle is 60°, then The modulus of (3 vector a+vector b) is A.4 B. Root number 13 C. Root number 10 D. Root number 7

Can be converted into two sides for 3 and 1, included angle of 120°, find the third side of the problem how to turn you should know!
According to the cosine theorem c^2=a^2+b^2-2abcosθ angle θ is 120°, substitute each value into the formula to obtain the modulus of B root 13

Help. The problem with the vector,, if a0 is a unit vector in the a direction, then a0=_____? If a =3/4e, b =-2/3e, then a is _____?

Easy:
A0=a/|a|

If a0 is a unit vector in the a direction, then a0=

A0=(1/|a|) a

Let vector a0, b0 be two unit vectors on vector a, b, respectively, and the included angle of vector a, b is 60°, try to find the included angle of vector m=2 vector a0-vector b0 and vector n=-2 vector a0+3b0

A0*b0=|a0 b0 cos60°=1/2 Find the module of the vector and m of n:|m|^2=m*m=(2a0-b0)*(2a0-b0)=4|a0|^2 b0|^2-4(a0*b0)=4+1-4×1/2=3, so |m||=√3||n|^2=n*n=(-2a0+3b0)*(-2a0+3b0)=4|a0|^2+9||b0|^2-12(a0*b0)=4...