If the vertices of a triangle $$areA1$$,$$2$$,$$1$$,$$B4$$,$$3$$,$$1$$,$$C3$$,$$1$$,$$5$$ then Area of Δ $$ABC42$$×$$cos2A=$$

Question

If the vertices of a triangle $$areA1$$,$$2$$,$$1$$,$$B4$$,$$3$$,$$1$$,$$C3$$,$$1$$,$$5$$ then  Area of Δ $$ABC42$$×$$cos2A=$$

Infinity Learn · Accepted Answer

The vertices of a triangle $$areA1$$,$$2$$,$$1$$,$$B4$$,$$3$$,$$1$$,$$C3$$,$$1$$,$$5The$$ area of the triangle  $$ABCis12AB$$¯×AC¯HereAB¯$$=3i+j$$ and AC¯$$=2i$$−$$j+4kHence$$ the area of the triangle is $$12ijk3102$$−$$14=12i4$$−$$j12+k$$−$$3$$−$$2=124i$$−$$12j$$−$$5kTherefore$$, the area of the triangle is  $$16+144+252=1852And$$ $$cosA=AB$$¯⋅AC¯AB¯AC¯$$=6$$−$$19+14+1+16=51021=542$$            Consider  Area of Δ $$ABC42$$×$$cos2A=1852$$×$$42$$×$$542=18510$$

If the vertices of a triangle are $A (1, 2, 1), B (4, 3, 1), C (3, 1, 5)$ then $\frac{Area of Δ A B C}{42 \times \cos^{2} A} =$

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If the vertices of a triangle areA1,2,1,B4,3,1,C3,1,5 then Area of Δ ABC42×cos2A=

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If the vertices of a triangle are $A (1, 2, 1), B (4, 3, 1), C (3, 1, 5)$ then $\frac{Area of Δ A B C}{42 \times \cos^{2} A} =$