Question

determine the area of the triangle with vertices A(1, 1,2) B (2,4,6) C(0,5,5)​

Answer 1

Coordinate Geometry (3D)

Formula to find area of a triangle:

Let the three vertices of the triangle be $A\:(x_{1},\:y_{1},\:z_{1})$, $B\:(x_{2},\:y_{2},\:z_{2})$, $C\:(x_{3},\:y_{3},\:z_{3})$ respectively and let the area of the triangle be $\Delta$.

Then,

$\Delta^{2}=\frac{1}{4}\left|\begin{array}{ccc}y_{1}&z_{1}&1\\y_{2}&z_{2}&1\\y_{3}&z_{3}&1\end{array}\right|^{2}$

$+\frac{1}{4}\left|\begin{array}{ccc}z_{1}&x_{1}&1\\z_{2}&x_{2}&1\\z_{3}&x_{3}&1\end{array}\right|^{2}+\frac{1}{4}\left|\begin{array}{ccc}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right|^{2}$

Solution:

The vertices of the triangle $\Delta ABC$ are $A\:(1,\:1,\:2)$, $B\:(2,\:4,\:6)$ and $C\:(0,\:5,\:5)$.

So we have:

$\quad\quad x_{1}=1,\: y_{1}=1,\: z_{1}=2$

$\quad\quad x_{2}=2,\: y_{2}=4,\: z_{2}=6$

$\quad\quad x_{3}=0,\:y_{3}=5,\:z_{3}=5$

$\therefore \left|\begin{array}{ccc}y_{1}&z_{1}&1\\y_{2}&z_{2}&1\\y_{3}&z_{3}&1\end{array}\right|=\left|\begin{array}{ccc}1&2&1\\4&6&1\\5&5&1\end{array}\right|$

$=1(6-5)-2(4-5)+1(20-30)$

$\quad\quad$(Expanding along $R_{1}$)

$=1+2-10$

$=-7$,

$\quad$$\left|\begin{array}{ccc}z_{1}&x_{1}&1\\z_{2}&x_{2}&1\\z_{3}&x_{3}&1\end{array}\right|=\left|\begin{array}{ccc}2&1&1\\6&2&1\\5&0&1\end{array}\right|$

$=2(2-0)-1(6-5)+1(0-10)$

$\quad\quad$(Expanding along $R_{1}$)

$=4-1-10$

$=-7$ and

$\quad$$\left|\begin{array}{ccc}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right|=\left|\begin{array}{ccc}1&1&1\\2&4&1\\0&5&1\end{array}\right|$

$=1(4-5)-1(2-0)+1(10-0)$

$\quad\quad$(Expanding along $R_{1}$)

$=-1-2+10$

$=7$

Putting the values in the above formula to find area of a triangle in three dimensions, we get

$\quad\Delta^{2}=\frac{1}{4}\:(-7)^{2}+\frac{1}{4}\:(-7)^{2}+\frac{1}{4}\:(7)^{2}$

$\quad\quad\:=\frac{1}{4}\:(49+49+49)$

$\quad\quad\:=\frac{1}{4}\times 147$

$\quad\quad\:=36.75$

where $\Delta$ is the area of the given triangle

$\Rightarrow \Delta^{2}=36.75$

$\Rightarrow \Delta=6.062$

Answer: Therefore, the area of the given triangle is 6.062 sq. units.

Answer 2

The area of the triangle is 6.06 square units.

Step-by-step explanation:

The vertices of the triangle are:

A (1, 1, 2); B (2, 4, 6); C (0, 5, 5)​

Now,

$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}$ = $(2\hat{i}+4 \hat{j}+6 \hat{k}) - (\hat{i}+1 \hat{j}+2 \hat{k})$ = $\hat{i}+3 \hat{j}+4 \hat{k}$

$\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}$ = $(0\hat{i}+5 \hat{j}+5 \hat{k}) - (2\hat{i}+4 \hat{j}+6 \hat{k})$ = $-2\hat{i}+ \hat{j}-1 \hat{k}$

Now,

$\overrightarrow{A B} \times \overrightarrow{A C} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & 3 & 4 \\-2 & 1 & -1\end{array}\right|$

∴ $\overrightarrow{A B} \times \overrightarrow{A C}$ = $-7 \hat{i}-7 \hat{j}+7 \hat{k}$

Now,

$|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-7)^{2}+(-7)^{2}+(7)^{2}}$

$|\overrightarrow{A B} \times \overrightarrow{A C}|$ = √(49 + 49 + 49)

∴ $|\overrightarrow{A B} \times \overrightarrow{A C}|$ = 12.12

The area of the triangle is given by the formula:

$A =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$

On substituting the values, we get,

A = 1/2 × 12.12

∴ A = 6.06 square units