If x and y are two non-collinear vectors and ABC is a triangle whose side length satisfying 5a-12bx-26b-10cy-12c-13ax-y-0- then which of the following is-are correct -

Given : nbsp; (5a 12b)x+(26b 10c)y+(12c 13a)(x×y)=0, As, x,y, x×y are non coplanar, i.e. linearly independent So, all scalars should be zero ⇒ 5a 12b=26b 10c= ...

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Byju's Answer

Standard XII

Mathematics

Condition for Three Vectors to Form a Triangle

If x and y ar...

Question

If $\to x$ and $\to y$ are two non-collinear vectors and $A B C$ is a triangle whose side length satisfying $(5 a - 12 b) \to x + (26 b - 10 c) \to y + (12 c - 13 a) (\to x \times \to y) = \to 0,$ then which of the following is/are correct ?

△ A B C

is an acute angle triangle

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△ A B C

is right angle triangle

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△ A B C

is an obtuse angle triangle

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∠ B = 90^{\circ}

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Solution

The correct option is B $△ A B C$ is right angle triangle
Given : $(5 a - 12 b) \to x + (26 b - 10 c) \to y + (12 c - 13 a) (\to x \times \to y) = \to 0,$
As, $\to x, \to y, \to x \times \to y$ are non-coplanar, i.e. linearly independent
So, all scalars should be zero
$\Rightarrow 5 a - 12 b = 26 b - 10 c = 12 c - 13 a = 0 \Rightarrow 5 a = 12 b, 13 b = 5 c, 12 c = 13 a \Rightarrow \frac{a}{12} = \frac{b}{5} = \frac{c}{13} = k \Rightarrow a = 12 k, b = 5 k, c = 13 k$
As, $c^{2} = a^{2} + b^{2}$
So, $△ A B C$ is a right angle triangle and $∠ C = 90^{\circ}$

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If →x and →y are two non-collinear vectors and ABC is a triangle whose side length satisfying (5a−12b)→x+(26b−10c)→y+(12c−13a)(→x×→y)=→0, then which of the following is/are correct ?

If $\to x$ and $\to y$ are two non-collinear vectors and $A B C$ is a triangle whose side length satisfying $(5 a - 12 b) \to x + (26 b - 10 c) \to y + (12 c - 13 a) (\to x \times \to y) = \to 0,$ then which of the following is/are correct ?