Question

Find the area of the triangle whose vertices are (1, – 1, – 3), (4, – 3, 1), (3,– 1, 2).

Solution

Let A (1, – 1, – 3), B (4, – 3, 1), C (3, – 1, 2) be vertices of Δ ABC

AB space equals space square root of left parenthesis 4 minus 1 right parenthesis squared plus left parenthesis negative 3 plus 1 right parenthesis squared plus left parenthesis 1 plus 3 right parenthesis squared end root space equals space square root of 9 plus 4 plus 16 end root space equals space square root of 29 AC space equals space square root of left parenthesis 3 minus 1 right parenthesis squared plus left parenthesis negative 1 plus 1 right parenthesis squared plus left parenthesis 2 plus 3 right parenthesis squared end root space equals space square root of 4 plus 0 plus 25 end root space equals space square root of 29

Direction-ratios of AB are 4 – 1, – 3 + 1, 1 + 3 i.e.. 3, – 2, 4 respectively.
Direction-ratios of AC are 3 – 1, –1 + 1, 2 + 3 i.e., 2, 0, 5 respectively.
$therefore space space space cos space straight A space equals space fraction numerator left parenthesis 3 right parenthesis thin space left parenthesis 2 right parenthesis space plus space left parenthesis negative 2 right parenthesis space left parenthesis 0 right parenthesis space plus space left parenthesis 4 right parenthesis thin space left parenthesis 5 right parenthesis over denominator square root of 9 plus 4 plus 16 end root space square root of 4 plus 0 plus 25 end root end fraction space equals space fraction numerator 6 plus 0 plus 20 over denominator square root of 29 space square root of 29 end fraction space equals space 26 over 29 therefore space space sin space straight A space equals space square root of 1 minus space cos squared straight A end root space equals space square root of 1 minus open parentheses 26 over 29 close parentheses squared end root space equals space square root of 1 minus 676 over 841 end root space equals space square root of 165 over 841 end root space equals space fraction numerator square root of 165 over denominator 29 end fraction$
Area of $increment ABC space equals space 1 half AB. space AC space sin space straight A space equals space 1 half square root of 29. space square root of 29. space fraction numerator square root of 165 over denominator 29 end fraction space equals space 1 half square root of 165 space sq. space units. space$

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Three Dimensional Geometry

Find the area of the triangle whose vertices are (1, – 1, – 3), (4, – 3, 1), (3,– 1, 2).

Some More Questions From Three Dimensional Geometry Chapter

If α, β, γ are the angles which a line makes with the axes, prove that sin²α + sin² β + sin²γ = 2.

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For Daily Free Study Material Join wiredfaculty

Three Dimensional Geometry

Find the area of the triangle whose vertices are (1, – 1, – 3), (4, – 3, 1), (3,– 1, 2).

Some More Questions From Three Dimensional Geometry Chapter

If α, β, γ are the angles which a line makes with the axes, prove that sin2α + sin2 β + sin2γ = 2.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

If α, β, γ are the angles which a line makes with the axes, prove that sin²α + sin² β + sin²γ = 2.