What is the modulus of a + b + c if the vectors a, b, c are two equal angles and the modulus of a is equal to 1, the modulus of b is equal to 1, the modulus of c is equal to 3? Just explain how it's equal to 2,5. What is the modulus of a + b + c if the two vectors a, b, c form equal angles and the modulus of a is equal to 1, the modulus of b is equal to 1, the modulus of c is equal to 3? Just explain how it's equal to 2,5.

What is the modulus of a + b + c if the vectors a, b, c are two equal angles and the modulus of a is equal to 1, the modulus of b is equal to 1, the modulus of c is equal to 3? Just explain how it's equal to 2,5. What is the modulus of a + b + c if the two vectors a, b, c form equal angles and the modulus of a is equal to 1, the modulus of b is equal to 1, the modulus of c is equal to 3? Just explain how it's equal to 2,5.

Ia+b+cI2=a2+b2+c2+2ab+2bc+2ca (1) When a, b, c are equal to 0°, Ia+b+cI2=a2+b2+c2+2ab+2bc+2ca=1+1+9+2+6+6=25 Ia+b+cI=5(2) When a, b, c are equal to 0°,

Ia+b+cI2=a2+b2+c2+2ab+2bc+2ca (1) When a, b, c are equal to 0°, Ia+b+cI2=a2+b2+c2+2ab+2bc+2ca=1+1+9+2+6+6=25 Ia+b+cI=5(2) when a, b, c are equal to 0°.

Ia+b+cI2=a2+b2+c2+2ab+2bc+2ca (1) When the angles formed by a, b and c are equal to 0°, Ia+b+cI2=a2+b2+c2+2ab+2bc+2ca=1+1+9+2+6+6=25 Ia+b+cI=5(2) When a, b by a, b and c are equal to 0°.

Is equal to the angle of the vector a=(7,1), b=(1,-7). What happens when the module is a vector of 1 Is equal to the angle of vectors a=(7,1), b=(1,-7). What happens when the module is a vector of 1

Let c=(x, y) and vector a=(7,1), b=(1,-7) be equal, and the modulus is 1. Then x2+y2=1(1) is obtained by cos=ac/|a||c|, cos=bc/|b||c|, and |c|=1 as ac/|a|=bc/|b|and |a|=|b|=2√5, so ac=bc is (a-b) c=0 and 6x+8y=0(...

Are the norm of the opposite vector equal? Are the moduli of the opposite equal?

The modules of the opposite vector (i.e. the length of the vector) are equal

Find the vector whose angle is equal to the vector a=(7/2,1/2), b=(1/2,7/2) and whose modulus length is 1 Let c=(m, n), |C |=√(m^2+n^2)=1, Set the angle between vectors a and c as θ Cosθ=a·c/(|a||c|=(7m/2+n/2)/[√(49/4+1/4)*1]=√2(7m/2+n/2)/5, Cosθ=b·c (/|b||c|)=(m/2+7n/2)/√[(1/4+49/4)*1]=√2(m/2+7n/2)/5, √2(7M/2+n/2)/5=√2(m/2+7n/2)/5, M=n, M^2+n^2=1, M=2/2, N=2/2, M, n shall be the same number Then vector c=(√2/2,√2/2), c=(-√2/2,-√2/2), I can't understand the last step. Why should m, n be the same?

Because m=n, it should be the same number.
It's recommended that you understand this and do it after m=n so you do n' t get confused
M^2+n^2=1,
2M^2=1,
M=2/2
N=m=√2/2 or n=m=-√2/2
Therefore, vector c=(√2/2,√2/2), c=(-√2/2,-√2/2),

If |a|=2,|b|=3,|a-b|=√7, what is the angle between vector a and b

|A-b|2=|a|2 b|2-2|a||b|cos@
=4+9-2*2*3Cos@
=(√7)²
Cos@=1/2
@=60° Or write π/3
Step are simple,

|A-b|2=|a|2 b|2-2|a||b|cos@
=4+9-2*2*3Cos@
=(√7)²
Cos@=1/2
@=60° Or written as π/3
Step are simple,

Known vector A =(2,3), B=(-1,2) if m A+n B and A-2 B collinear, then m N equals () A.-1 2 B.1 2 C.-2 D.2

M

A+n

B=(2m-n,3m+2n),

A-2

B=(4,-1), m

A+n

B and

A-2

B collinear,
(2M-n)(-1)-4(3m+2n)=0,-14m=7n, then m
N=-1
2,
Therefore, A.