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

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

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   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,

   [tex]\rm :\longmapsto\: \vec{AB} \: = \vec{OB} – \vec{OA}[/tex]

   [tex]\rm :\longmapsto\:\vec{AB} = 2 \hat{i} + 3\hat{j} + \hat{k} – (\hat{i} + 2\hat{j} + 3\hat{k})[/tex]

   [tex]\rm :\longmapsto\:\vec{AB} = 2 \hat{i} + 3\hat{j} + \hat{k} – \hat{i} – 2\hat{j} – 3\hat{k}[/tex]

   [tex]\rm :\longmapsto\:\vec{AB} = \hat{i} + \hat{j} – 2\hat{k}[/tex]

   Aɢᴀɪɴ,

   [tex]\rm :\longmapsto\:\vec{AC} = \vec{OC} – \vec{OA}[/tex]

   [tex]\rm :\longmapsto\:\vec{AC} = 3\hat{i} + \hat{j} + 2\hat{k} – (\hat{i} + 2\hat{j} + 3\hat{k})[/tex]

   [tex]\rm :\longmapsto\:\vec{AC} = 3\hat{i} + \hat{j} + 2\hat{k} – \hat{i} – 2\hat{j} – 3\hat{k}[/tex]

   [tex]\rm :\longmapsto\:\vec{AC} = 2\hat{i} – \hat{j} – \hat{k}[/tex]

   Now,

   ↝ Consider,

   [tex]\rm :\longmapsto\: \vec{AB} \times \vec{AC}[/tex]

   [tex] \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}[/tex]

   [tex] \rm \: = \: \: \: \hat{i}( – 1 – 2) – \hat{j}( – 1 + 4) + \hat{k}( – 1 – 2)[/tex]

   [tex] \rm \: = \: \: \: – 3\hat{i} – 3\hat{j} – 3\hat{k}[/tex]

   Therefore,

   [tex]\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = \sqrt{ {( – 3)}^{2} + {( – 3)}^{2} + {( – 3)}^{2} } [/tex]

   [tex]\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = \sqrt{9 + 9 + 9} [/tex]

   [tex]\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = \sqrt{27} [/tex]

   [tex]\rm :\longmapsto\: | \vec{AB} \times \vec{AC} | = 3\sqrt{3} [/tex]

   ↝ Hence,

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

   [tex]\rm :\longmapsto\:Area_{(\triangle \: ABC)} = \dfrac{1}{2} \times 3 \sqrt{3} [/tex]

   [tex]\rm :\longmapsto\:Area_{(\triangle \: ABC)} = \dfrac{3 \sqrt{3} }{2} \: sq. \: units[/tex]

   Additional Information :-

   [tex] \boxed{ \red{ \sf \: Area_{(\parallel \: gram)} = | \vec{a} \times \vec{b} |}}[/tex]

   [tex] \boxed{ \red{ \sf \: \vec{a} \times \vec{b} = – \vec{b} \times \vec{a}}}[/tex]

   [tex] \boxed{ \red{ \sf \: \vec{a} \times \vec{a} = 0}}[/tex]

   [tex] \boxed{ \red{ \sf \: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|sin \theta}} [/tex]

   [tex] \boxed{ \red{ \sf \: \vec{a} \times \vec{b} = \vec{0} \implies \: \vec{a} \parallel\vec{b}}} [/tex]

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