Lecture Four: Applications on Determinants

1. Area of triangle

Formula:

$$\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$$

2. Area of Parallelogram

Formula:

$$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$$

3. Area of parallelepiped

Formula:

$$\left|\begin{array}{ccc}\uparrow & \uparrow & \uparrow \\ \overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\\ \downarrow & \downarrow & \downarrow \end{array}\right|$$

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IMPORTANT

If the determinant is zero then, points/vectors are collinear (lie on the same line), and the area/volume is zero.

Examples

1. Find the area of $\mathrm{△}abc$ where $a=(1,2),b=(3,-2),c=(4,-1)$

$$\begin{array}{c}\text{area of}\mathrm{△}abc=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 3& -2& 1\\ 4& -1& 1\end{array}\right|\\ \\ =1\left|\begin{array}{cc}-2& 1\\ -1& 1\end{array}\right|-2\left|\begin{array}{cc}3& 1\\ 4& 1\end{array}\right|+1\left|\begin{array}{cc}3& -2\\ 4& -1\end{array}\right|\\ \\ =\frac{1}{2}\cdot 1(-2+1)-2(3-4)+1(-3+8)\\ \\ =\frac{1}{2}\cdot 6=3c{m}^2\end{array}$$

TIP

Remember to apply the Sign rule when calculating the determinant.

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2. Find the Area of the Parallelogram $abcd$ where $a=(5,3),b=(2,4),c=(1,6)$

$$\begin{array}{c}\text{area of}abc=\left|\begin{array}{ccc}5& 3& 1\\ 2& 4& 1\\ 1& 6& 1\end{array}\right|\\ \\ =5\left|\begin{array}{cc}4& 1\\ 6& 1\end{array}\right|-3\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|+1\left|\begin{array}{cc}2& 4\\ 1& 6\end{array}\right|\\ \\ =5(4-6)-3(2-1)+1(12-4)=-5c{m}^2\end{array}$$

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3. Find $x$ in These Cases:

Given:

$$\overrightarrow{u}=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right),\overrightarrow{v}=\left(\begin{array}{c}-2\\ -1\\ 4\end{array}\right),\overrightarrow{w}=\left(\begin{array}{c}8\\ 10\\ x\end{array}\right)$$

$I$. The 3-vectors are colinear ($V=0$)

For vectors $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ to be colinear the determinant must be zero.

$$\begin{array}{c}\left|\begin{array}{ccc}4& -2& 8\\ 5& -1& 10\\ 6& 4& x\end{array}\right|=0\\ \\ \Rightarrow 4\left|\begin{array}{cc}-1& 10\\ 4& x\end{array}\right|+2\left|\begin{array}{cc}5& 10\\ 6& x\end{array}\right|+8\left|\begin{array}{cc}5& -1\\ 6& 4\end{array}\right|\\ \\ \Rightarrow 4(-x-40)+2(5x-60)+8(20+6)=0\\ \\ \Rightarrow -4x-160+10x-120+208\\ \\ \Rightarrow 6x-72=0,6x=72\\ \\ \Rightarrow x=12\end{array}$$

$II$. The volume of the parallelepiped is $18c{m}^3$

$$\begin{array}{c}\left|\begin{array}{ccc}4& -2& 8\\ 5& -1& 10\\ 6& 4& x\end{array}\right|=18\\ \\ \Rightarrow 4\left|\begin{array}{cc}-1& 10\\ 4& x\end{array}\right|+2\left|\begin{array}{cc}5& 10\\ 6& x\end{array}\right|+8\left|\begin{array}{cc}5& -1\\ 6& 4\end{array}\right|\\ \\ \Rightarrow 4(-x-40)+2(5x-60)+8(20+6)=18\\ \\ \Rightarrow -4x-160+10x-120+208\\ \\ \Rightarrow 6x-72=18,6x=90\\ \\ \Rightarrow x=15\end{array}$$