Answer:
Value of angle between vector a and b is [tex]56.30^{\circ}[/tex].
Explanation:
Vectors a and b have scalar product 6.00
Let [tex]\theta[/tex] be the angle between a and b.
[tex]\vec{a}.\vec{b} = 6[/tex]
ab cos [tex]\theta[/tex] = 6 ...(1)
Vectors a and b have magnitude of vector product 9.00
[tex]\vec{a} \times\vec{b} = 9[/tex]
ab sin [tex]\theta[/tex] = 9 ...(2)
Dividing equation (2) by (1) we get
[tex]\frac{ab sin \theta}{ab cos \theta} = \frac{9}{6}[/tex]
tan [tex]\theta[/tex] = 1.5
[tex]\theta = tan ^{-1} (1.5)[/tex]
[tex]\theta[/tex] = [tex]56.30^{\circ}[/tex]
Thus, value of angle between vector a and b is [tex]56.30^{\circ}[/tex].
The angle between two vectors will be "56.30°".
According to the question,
Vectors a and b have scalar product = 6.00
Magnitude = +9.00
Let,
Angle between "a" and "b" = [tex]\Theta[/tex]
then,
[tex]\vec{a}. \vec{b} = 6[/tex]
ab Cosθ = 6 ...(Equation 1)
and,
[tex]\vec{a}\times \vec{b} = 9[/tex]
ab Sinθ = 9 ...(Equation 2)
By dividing both equations, we get
→ [tex]\frac{ab \ Sin\Theta}{ab \ Cos\Theta}=\frac{9}{6}[/tex]
[tex]tan \Theta = 1.5[/tex]
[tex]\Theta = tan^{-1} (1.5)[/tex]
[tex]=56.30^{\circ}[/tex]
Thus the above answer is correct.
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