Question

A vector$\overrightarrow{\mathbf{a}}$ of magnitude 10units and another vector$\overrightarrow{\mathbf{b}}$of magnitude 6.0units differ in directions by$\mathbf{60}\mathbf{°}$. Find (a) the scalar product of the two vectors and (b) the magnitude of the vector product $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}$.

Short answer

a) The scalar product of two vectors,$\overrightarrow{a}.\overrightarrow{b}$,is 30 units

b) The magnitude of the vector product $\left|\overrightarrow{a},\overrightarrow{b}\right|$is 52 units.

Step-by-step solution

Given

The magnitude of the vector $\overrightarrow{a}$ is 10 units.

The magnitude of the vector $\overrightarrow{b}$is 6.0 units

The angle between $\overrightarrow{a}$and$\overrightarrow{b}$ is$\theta =60°$

Understanding the concept

Use the formula of the scalar product of two vectors and vector product of two vectors.

Formula:

$\overrightarrow{a}.\overrightarrow{b}=ab\cos \theta$…………..(i)

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=ab\sin \theta$ …………..(ii)

(a) Calculate the scalar product of two vectors

The dot product of vectors$\overrightarrow{a}$and$\overrightarrow{b}$can be calculated using equation (i)

$$\begin{gathered} \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\mathrm{abcos\theta } \\ =10\times 6\times 0.5 \\ =30\mathrm{units} \end{gathered}$$

Therefore, the dot product of vectors $\overrightarrow{a}$and$\overrightarrow{b}$ is 30 units.

(b) Calculate the magnitude of vector product a→×b→

Use the equation (ii) to calculate the cross product ofvectors$\overrightarrow{a}$ and $\overrightarrow{b}$.

$$\begin{gathered} \left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|=\mathrm{absin}60 \\ =52\mathrm{units} \end{gathered}$$

Therefore, the cross product of vectors $\overrightarrow{a}$and$\overrightarrow{b}$ is 52 units.