91Ó°ÊÓ

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

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

The angle between $\stackrel{→}{a}$and$\stackrel{→}{b}$ is$\mathrm{θ}=60°$

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

Formula:

$\stackrel{→}{a}.\stackrel{→}{b}=ab\cos \mathrm{θ}$…â¶Ä¦â¶Ä¦â¶Ä¦..(¾±)

$\left|\stackrel{→}{a}×\stackrel{→}{b}\right|=ab\sin \mathrm{θ}$ …â¶Ä¦â¶Ä¦â¶Ä¦..(¾±¾±)

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

$$\begin{gathered} \stackrel{→}{\mathrm{a}}.\stackrel{→}{\mathrm{b}}=\mathrm{²¹²ú³¦´Ç²õθ} \\ =10×6×0.5 \\ =30\mathrm{units} \end{gathered}$$

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

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

$$\begin{gathered} \left|\stackrel{→}{\mathrm{a}}×\stackrel{→}{\mathrm{b}}\right|=\mathrm{absin}60 \\ =52\mathrm{units} \end{gathered}$$

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