Question

Evaluate the cross products$\overrightarrow{A}\times \overrightarrow{B}$ and $\overrightarrow{C}\times \overrightarrow{D}$. (part )

Short answer

The cross product is $21$ and into the page. (part a)

The cross product is$24$and out of the page.(part b)

Step-by-step solution

Step :1 Introduction  (part  a )

The cross-product of two vectors is also a vector. The direction of the vector is derived from the right-hand rule. The expression for the magnitude of cross product is:

$\overrightarrow{A}\times \overrightarrow{B}=AB\sin \alpha$

Where $A$and $B$are the magnitude of the vectors and$\alpha$ is the angle between them.

Step :2 Explanation(part a)

The expression for the cross product is:

$\overrightarrow{A}\times \overrightarrow{B}=AB\sin \alpha$

Substitute 6 for $A,4$for $B$and$45°$for $\alpha$in the equation,

$\overrightarrow{A}\times \overrightarrow{B}=AB\sin \alpha$

$\begin{array}{r}=\left(6\right)\left(4\right)\sin {45}^{\circ }\\ \end{array}$

$=21.21$

The direction of the vector is into the page calculated form right hand rule.

Step :3 Concept introduction (part b )

A vector is formed by the cross product of two vectors. The right-hand rule is used to determine the vector's direction. The magnitude of the cross product is expressed as:

$\overrightarrow{A}\times \overrightarrow{B}=AB\sin \alpha$

Where $A$and $B$are the magnitude of the vectors and $\alpha$is the angle between them.

Step :4 Explanation (part b )

The expression for the magnitude of cross product is:

$\overrightarrow{C}\times \overrightarrow{D}=CD\sin \alpha$

Substitute $6$for $C,4$for $D$and $90°$for $\alpha$ in the equation,

$\overrightarrow{C}\times \overrightarrow{D}=CD\sin \alpha$

$=\left(6\right)\left(4\right)\sin 90°$

$=24$

The direction of the vector is out of the page calculated form right hand rule.