If x- and y- are two non-parallel vectors and ABC is a triangle with side lengths a- b and c satisfying 20 a -15 b x -15 b -12 c y - 12 c -20 a x - y -0- then triangle ABC isA- an acute-angled triangleB- an obtuse-angled triangleC- a right-angled triangleD- an isosceles triangle

The correct option is C a right angled triangleSince, nbsp;x, y and x×y are linearly independent, we have nbsp; 20a 15b=15b 12c=12c 20a=0 ⇒a3=b4=c5 ⇒ c^2= ...

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Byju's Answer

Standard XII

Mathematics

Condition for Three Vectors to Form a Triangle

If x⃗ and y⃗ ...

Question

If $\to x$ and $\to y$ are two non-parallel vectors and $A B C$ is a triangle with side lengths $a, b$ and $c$ satisfying $(20 a - 15 b) \to x + (15 b - 12 c) \to y + (12 c - 20 a) (\to x \times \to y) = \to 0$ , then triangle $A B C$ is

an acute-angled triangle

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an obtuse-angled triangle

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a right-angled triangle

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an isosceles triangle

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Solution

The correct option is C a right-angled triangle
Since, $\to x, \to y$ and $\to x \times \to y$ are linearly independent, we have
$20 a - 15 b = 15 b - 12 c = 12 c - 20 a = 0$
$\Rightarrow \frac{a}{3} = \frac{b}{4} = \frac{c}{5}$
$\Rightarrow c^{2} = a^{2} + b^{2}$
Hence, $△ A B C$ is right angled.

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Similar questions

Q. If

\to x

and

\to y

are two non-parallel vectors and

A B C

is a triangle with side lengths

a, b

and

c

satisfying

(20 a - 15 b) \to x + (15 b - 12 c) \to y + (12 c - 20 a) (\to x \times \to y) = \to 0

, then triangle

A B C

Q. If

\to x

and

\to y

are two non-collinear vectors and

A B C

is a triangle with sides

a, b, c

satisfying

(20 a - 15 b) \to x + (15 b - 12 c) \to y + (12 c - 20 a)

(\to x \times \to y) = \to 0

, then the triangle

A B C

Question 7

Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is

a) Obtuse-angled triangle
b) Acute angled triangle
c) Right angled triangle
d) An isosceles right triangle.

Q. In

△ A B C, {sin}^{2} A + {sin}^{2} B + {sin}^{2} C = 2,

then the triangle is always

$△$ ABC is drawn with the length of side AB being 6 cm. If $∠$ A is 65 $^{\circ}$ and $∠$ B is 65 $^{\circ}$ , then $△$ ABC is

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If →x and →y are two non-parallel vectors and ABC is a triangle with side lengths a, b and c satisfying (20a−15b)→x+(15b−12c)→y+(12c−20a)(→x×→y)=→0, then triangle ABC is

The correct option is C a right-angled triangleSince, →x, →y and →x×→y are linearly independent, we have 20a−15b=15b−12c=12c−20a=0 ⇒a3=b4=c5 ⇒c2=a2+b2 Hence, △ABC is right angled.

If $\to x$ and $\to y$ are two non-parallel vectors and $A B C$ is a triangle with side lengths $a, b$ and $c$ satisfying $(20 a - 15 b) \to x + (15 b - 12 c) \to y + (12 c - 20 a) (\to x \times \to y) = \to 0$ , then triangle $A B C$ is