Question

Find the area of a triangle having the points A(1,1,1),B(1,2,3)and C(2,3,1) as it's vertices using vectors

Answer 1

Vector algebra

We are given a simple question and in the question, we are given that the sides of a triangle at the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) are its vertices. With this information, we are asked to find the area of the triangle using vectors.

Let's first draw the diagram to solve the problem.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A (1, 1, 1) }\put(0.5,-0.3){ \bf B(1, 2, 3) }\put(5.2,-0.3){ \bf C (2, 3, 1) }\end{picture}$

[Check the attachment if any case you're not able to see the diagram]

The basic formula of a area of traingle is equal to half the product of its base and height. i.e.

$\implies \boxed{Area_{(Triangle)} = \dfrac{1}{2} \times \big|\overrightarrow{AB} \times \overrightarrow{AC}\big|}$

Now, consider the side of AB:

$\implies \overrightarrow{AB} = (1 - 1) \hat{i} + (2 - 1)\hat{j} + (3 - 1)\hat{k} \\ \\ \implies \overrightarrow{AB} = 0\hat{i} + 1\hat{j} + 2\hat{k}$

Similarly, consider the side of AC:

$\implies \overrightarrow{AC} = (2 - 1) \hat{i} + (3 - 1)\hat{j} + (1 - 1)\hat{k} \\ \\ \implies \overrightarrow{AC} = 1\hat{i} + 2\hat{j} + 0\hat{k}$

Now, using the determinat method to find the product of both vectors.

$\implies \overrightarrow{AB} \times \overrightarrow{AC} = \begin{array}{|ccc|} \hat{i} & \hat{j} & \hat{k} \\ 0 &1&2 \\ 1&2&0\end{array}$

$\implies \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}\big[(1 \times 0) - (2 \times 2) \big] - \hat{j}\big[(0 \times 0) - (2 \times 1) \big] + \hat{k}\big[(0 \times 2) - (1 \times 1) \big]$

$\implies \overrightarrow{AB} \times \overrightarrow{AC} =\hat{i}\big[(0 - 4) \big] - \hat{j}\big[0 - 2 \big] + \hat{k}\big[0 - 1 \big] \\ \\ \implies \overrightarrow{AB} \times \overrightarrow{AC} = - 4\hat{i}- ( - 2)\hat{j}+ ( - 1)\hat{k} \\ \\ \implies\overrightarrow{AB} \times \overrightarrow{AC} = - 4\hat{i} + 2\hat{j} - 1\hat{k}$

Now, magnitude of both vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ will be:

$\implies \big|\overrightarrow{AB} \times \overrightarrow{AC}\big| = \sqrt{( - 4)^{2} + ( 2)^{2} + ( -1)^{2}} \\ \\ \implies \big|\overrightarrow{AB} \times \overrightarrow{AC}\big| = \sqrt{16 + 4 + 1} \\ \\ \implies \big|\overrightarrow{AB} \times \overrightarrow{AC}\big| = \sqrt{21}$

Now, by using the formula of area of traingle and solving the equation, we get:

$\implies Area_{(Triangle)} = \dfrac{1}{2} \times \big|\overrightarrow{AB} \times \overrightarrow{AC}\big| \\ \\ \implies Area_{(Triangle)} = \dfrac{1}{2} \times \sqrt{21} \\ \\ \implies \boxed{Area_{(Triangle)} = \dfrac{ \sqrt{21} }{2}}$

Hence, the area of traingle is √21/2 sq units.

Answer 2

Answer:

Question :-

• Find the area of a triangle having the points A(1,1,1),B(1,2,3)and C(2,3,1) as it's vertices using vectors.

Find Out :-

• Find the area of a triangle.

Solution :-

As we know that :

☣ Area of a Triangle :-

$\mapsto \red{ \boxed{\sf{Area\: Of\: Triangle = \dfrac{1}{2} \times \big|\overrightarrow{AB} \times \overrightarrow{AC}\big|}}}$

☣ In case of AB :-

$\leadsto \sf \overrightarrow{AB} = (1 - 1) \hat{i} + (2 - 1)\hat{j} + (3 - 1)\hat{k}$

$\leadsto \sf \overrightarrow{AB} = (0) \hat{i} + (1)\hat{j} + (2)\hat{k}$

$\leadsto \bf \overrightarrow{AB} = 0\hat{i} + 1\hat{j} + 2\hat{k}$

☣ In case of AC :-

$\leadsto \sf \overrightarrow{AC} = (2 - 1) \hat{i} + (3 - 1)\hat{j} + (1 - 1)\hat{k}$

$\leadsto \sf \overrightarrow{AC} = (1) \hat{i} + (2)\hat{j} + (0)\hat{k}$

$\leadsto \bf \overrightarrow{AC} = 1\hat{i} + 2\hat{j} + 0\hat{k}$

☣ Product of both vectors :-

$\small \leadsto \sf \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}\big\{(1 \times 0) - (2 \times 2) \big\} - \hat{j}\big\{(0 \times 0) - (2 \times 1) \big\} + \hat{k}\big\{(0 \times 2) - (1 \times 1) \big\}$

$\small \leadsto \sf \overrightarrow{AB} \times \overrightarrow{AC} =\hat{i}\big\{(0 - 4) \big\} - \hat{j}\big\{0 - 2 \big\} + \hat{k}\big\{0 - 1 \big\}$

$\small \leadsto \bf \overrightarrow{AB} \times \overrightarrow{AC} = - 4\hat{i} + 2\hat{j} - 1\hat{k}$

☣ Magnitude of both vectors :-

$\leadsto \sf \overrightarrow{AB} \times \overrightarrow{AC} = \sqrt{( - 4)^{2} + ( 2)^{2} + ( -1)^{2}}$

$\leadsto \sf \overrightarrow{AB} \times \overrightarrow{AC} = \sqrt{( - 4)(- 4) + ( 2)(2) + ( - 1)(- 1)}$

$\leadsto \sf \overrightarrow{AB} \times \overrightarrow{AC} = \sqrt{16 + 4 + 1}$

$\leadsto \sf \overrightarrow{AB} \times \overrightarrow{AC} = \sqrt{16 + 5}$

$\leadsto \bf\big|\overrightarrow{AB} \times \overrightarrow{AC}\big| = \sqrt{21}$

☣ Area of traingle :-

$\leadsto \sf Area\: Of\: Triangle_{(ABC)} = \dfrac{1}{2} \times \big|\overrightarrow{AB} \times \overrightarrow{AC}\big|$

$\leadsto \sf Area\: Of\: Triangle_{(ABC)} = \dfrac{1}{2} \times \sqrt{21}$

$\leadsto \sf Area\: Of\: Triangle_{(ABC)} = \dfrac{1 \times \sqrt{21}}{2}$

$\leadsto {\small{\bold{\purple{\underline{Area\: Of\: Triangle_{(ABC)} = \dfrac{ \sqrt{21}}{2}}}}}}$

Henceforth, the area of traingle is √21/2 sq units.

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