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MATHEMATICS · Single correct · CBSE · Mathematics · Vector Algebra · Vector Algebra
Problem Consider △ A B C \triangle ABC △ A B C with A = ( a ) A = (\mathbf{a}) A = ( a ) , B = ( b ) B = (\mathbf{b}) B = ( b ) and C = ( c ) C = (\mathbf{c}) C = ( c ) . If b ⋅ ( a + c ) = b ⋅ b + a ⋅ c \mathbf{b} \cdot (\mathbf{a} + \mathbf{c}) = \mathbf{b} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} b ⋅ ( a + c ) = b ⋅ b + a ⋅ c , ∣ b − a ∣ = 3 |\mathbf{b} - \mathbf{a}| = 3 ∣ b − a ∣ = 3 and ∣ c − b ∣ = 4 |\mathbf{c} - \mathbf{b}| = 4 ∣ c − b ∣ = 4 , then the angle between the medians A M AM A M and B D BD B D is
(1) π − cos − 1 ( 1 5 13 ) \pi - \cos^{-1}\left(\frac{1}{5\sqrt{13}}\right) π − cos − 1 ( 5 13 1 )
(2) π − cos − 1 ( 1 13 5 ) \pi - \cos^{-1}\left(\frac{1}{13\sqrt{5}}\right) π − cos − 1 ( 13 5 1 )
(3) cos − 1 ( 1 5 13 ) \cos^{-1}\left(\frac{1}{5\sqrt{13}}\right) cos − 1 ( 5 13 1 )
(4) cos − 1 ( 1 13 5 ) \cos^{-1}\left(\frac{1}{13\sqrt{5}}\right) cos − 1 ( 13 5 1 )
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