Question

find the area of triangle whose vertices are A(1, 2,3) B(2, 3,1) C((3, 1,2)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given the vertices of triangle ABC,

• Coordinates of vertex A = (1, 2, 3)

• Coordinates of vertex B = (2, 3, 1)

• Coordinates of vertex C = (3, 1, 2)

Now,

• We have to find the area of triangle ABC.

We use here

• Vector method to find the area of triangle as it is in 3 dimension.

Now,

$\rm :\longmapsto\: \vec{AB} \: = \vec{OB} - \vec{OA}$

$\rm :\longmapsto\:\vec{AB} = 2 \hat{i} + 3\hat{j} + \hat{k} - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\rm :\longmapsto\:\vec{AB} = 2 \hat{i} + 3\hat{j} + \hat{k} - \hat{i} - 2\hat{j} - 3\hat{k}$

$\rm :\longmapsto\:\vec{AB} = \hat{i} + \hat{j} - 2\hat{k}$

Aɢᴀɪɴ,

$\rm :\longmapsto\:\vec{AC} = \vec{OC} - \vec{OA}$

$\rm :\longmapsto\:\vec{AC} = 3\hat{i} + \hat{j} + 2\hat{k} - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\rm :\longmapsto\:\vec{AC} = 3\hat{i} + \hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} - 3\hat{k}$

$\rm :\longmapsto\:\vec{AC} = 2\hat{i} - \hat{j} - \hat{k}$

Now,

↝ Consider,

$\rm :\longmapsto\: \vec{AB} \times \vec{AC}$

$\rm \: = \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\1&1& - 2\\2& - 1& - 1\end{array}\right | \end{gathered}$

$\rm \: = \: \: \: \hat{i}( - 1 - 2) - \hat{j}( - 1 + 4) + \hat{k}( - 1 - 2)$

$\rm \: = \: \: \: - 3\hat{i} - 3\hat{j} - 3\hat{k}$

Therefore,

$\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = \sqrt{ {( - 3)}^{2} + {( - 3)}^{2} + {( - 3)}^{2} }$

$\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = \sqrt{9 + 9 + 9}$

$\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = \sqrt{27}$

$\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = 3\sqrt{3}$

↝ Hence,

$\rm :\longmapsto\:Area_{(\triangle \: ABC)} = \dfrac{1}{2} | \vec{AB} \times \vec{AC} |$

$\rm :\longmapsto\:Area_{(\triangle \: ABC)} = \dfrac{1}{2} \times 3 \sqrt{3}$

$\rm :\longmapsto\:Area_{(\triangle \: ABC)} = \dfrac{3 \sqrt{3} }{2} \: sq. \: units$

Additional Information :-

$\boxed{ \red{ \sf \: Area_{(\parallel \: gram)} = | \vec{a} \times \vec{b} |}}$

$\boxed{ \red{ \sf \: \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}}}$

$\boxed{ \red{ \sf \: \vec{a} \times \vec{a} = 0}}$

$\boxed{ \red{ \sf \: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|sin \theta}}$

$\boxed{ \red{ \sf \: \vec{a} \times \vec{b} = \vec{0} \implies \: \vec{a} \parallel\vec{b}}}$