Answer to Question #168599 in Physics for Sheila Ann Gudoy

Question

Vector A has magnitude 6.00 m and vector has magnitude 3.00 m. The vector product between these two vectors has magnitude 12,0 square meters. What are the two possible values for the scalar product of these two vectors?

Answer 1

According to the definition, the vector product gives us the magnitude of the vector and its direction:

$A\times B=||A|||B||sin\theta.$

Let's find the angle $\theta$ between the two vectors:

$\theta=sin^{-1}(\dfrac{A\times B}{||A|||B||})=sin^{-1}(\dfrac{12\ m^2}{6\ m\cdot3\ m})=41.8^{\circ}.$

If our vectors located in the second quadrant, for example, the angle between them will be:

$\theta=180^{\circ}-41.8^{\circ}=138.2^{\circ}.$

Then, we can find the two possible values for the scalar product of these two vectors:

$A\cdot B=||6\ m|||3\ m||cos41.8^{\circ}=13.42\ m^2,$ $A\cdot B=||6\ m|||3\ m||cos138.2^{\circ}=-13.42\ m^2.$