If the magnitude of the cross product of two vectors is one-half the dot product of the same vectors, what is the angle between the two vectors?

26.57 degrees

Explanation:

Let the two vectors be a and b

The dot product of two vectors is expressed as;

a.b = |a||b|cosθ

The cross product is expressed as;

a×b = |a||b|sinθ

If the magnitude of the cross product of two vectors is one-half the dot product of the same vectors, then;

|a×b| = 1/2|a.b|

|a||b|sinθ = 1/2|a||b|cosθ

sinθ = 1/2cosθ

2sinθ = cosθ

sinθ/cosθ = 1/2

tanθ = 1/2

θ = arctan(1/2)

θ = 26.57 degrees

Hence  the angle between the two vectors is 26.57 degrees

cant without the picture! ?

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y