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Matlab solution of higher applied mathematics problems
(a+3b) is perpendicular to (7a-5b)

You can get:

(a+3b)。 (7a-5b)=0

Simplify to get:

7aa+ 16ab- 15bb=0,

According to the definition of vector: a.b=|a||b|cos included angle.

When two vectors are equal, the included angle is 0, so: aa = | a | 2, bb = | b | 2.

The angle between vector a and vector b is 60 degrees, so: ab = | a || b | cos 60 =1/2 | a |||| b |,

Substituting the above equation, we can get:

7|a|^2+8|a||b|- 15|b|^2=0,

So: (7|a|+ 15|b|)(|a|-|b|)=0.

Discard the negative root and get: |a|=|b|. ( 1)

(Vector a-4 Vector B) (Vector 7 Vector a-2 Vector B)

= 7 | a | 2+8 | b | 2-30ab, plus (1)

= 15|a|^2-30|a||a|cos60

= 15|a|^2- 15|a|^2

=0, get the certificate.