Question

In Fig. 3-38 , the magnitude of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{3}$, the magnitude of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{is}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{4}$, and $\mathit{\varphi }\mathbf{=}\mathbf{46}\mathbf{°}$. Find the area of the triangle contained between the two vectors and the thin diagonal line.

Short answer

Area of a triangle is8.4

Step-by-step solution

To understand the concept

The magnitude of the vector product is the area of parallelogram spanned by these two vectors. Therefore, vector product is used to compute the area of a triangle.

Here the area of a triangle is given by,

$$\begin{gathered} A=\frac{1}{2}\left|\stackrel{\rightharpoonup }{\mathrm{a}}\times \stackrel{\rightharpoonup }{\mathrm{b}}\right|=\frac{1}{2}\mathrm{ab}\mathrm{sin\varphi }\left(1\right) \\ \mathrm{Given}, \\ \stackrel{\rightharpoonup }{\mathrm{a}}=4.3 \\ \stackrel{\rightharpoonup }{\mathrm{b}}=5.4 \\ \mathrm{The}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{vector}\stackrel{\rightharpoonup }{\mathrm{a}}\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{b}}\mathrm{is}\mathrm{\varphi }=46° \end{gathered}$$

To find the area of a triangle

Substituting the values of $\stackrel{\rightharpoonup }{a},\stackrel{\rightharpoonup }{b}$, and$\varphi$ in equation (i), the area of a triangle is given by

$\mathrm{A}=\frac{1}{2}\mathrm{absin\varphi }=\frac{1}{2}\left(4.3\right)\left(5.4\right)\sin \left(46\right)$

Thus area of triangle$A=\overline{)8.4}$