Respuesta :
Answer:
[tex]\text{Vector product}=172.66\ m^2[/tex]
Explanation:
Given that,
The magnitude of vector A, [tex]|A|=13\ m[/tex]
The magnitude of vector B, [tex]|B|=16\ m[/tex]
Scalar product of A and B, [tex]A{\cdot} B=116\ m^2[/tex]
The formula for the scalar product is given by :
[tex]A{\cdot} B=|A||B|\cos\theta[/tex]
Where, [tex]\theta[/tex] is the angle between A and B.
[tex]\cos\theta=\dfrac{116}{13\times 16}\\\\\theta=\cos^{-1}\left(0.5576\right)\\\\\theta=56.11^{\circ}[/tex]
The formula for the vector product is given by :
[tex]A\times B=|A||B|\sin\theta\\\\=13\times 16\times \sin56.11\\\\=172.66\ m^2[/tex]
So, the vector product between these two vectors is [tex]172.66\ m^2[/tex].
The magnitude of the vector product is 172.74 m2.
How do you calculate the magnitude of the vector product?
Given that the magnitude of vector A is 13 and the magnitude of vector B is 16. The scalar product of vectors A and B are 116 m2.
Let us consider that [tex]\theta[/tex] is the angle between vector A and vector B, then the scalar product can be given as below.
[tex]A.B = |A||B|cos\theta[/tex]
Substituting the values in the above equation, we get the angle between vectors A and B.
[tex]116 = 13 \times 16 \times cos\theta[/tex]
[tex]cos \theta = 0.557[/tex]
[tex]\theta = cos^{-1} 0.557[/tex]
[tex]\theta = 56.15^\circ[/tex]
Now the magnitude of the vector product is given below.
[tex]A \times B = |A||B|sin\theta[/tex]
[tex]A\times B = 13\times 16\times sin 56.15^\circ[/tex]
[tex]A\times B = 172.74 \;\rm m^2[/tex]
Hence we can conclude that the magnitude of the vector product is 172.74 m2.
TO know more about the vector quantities, follow the link given below.
https://brainly.com/question/21797532.
