If the position vectors of the vertices of a triangle be 6i + 4j + 5k, 4i + 5j + 6k and 5i + 6j + 4k then the triangle is

Right angled triangle

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Equilateral triangle

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Isosceles triangle

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None of these

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Solution

The correct option is B Equilateral triangle
We have,

Let the vertices of triangle is

$- - \to O A = (6 \to i + 4 \to j + 5 \to k)$

$- - \to O B = (4 \to i + 5 \to j + 6 \to k)$

$- - \to O C = (5 \to i + 6 \to j + 4 \to k)$

Now,

$- - \to A B = - - \to O B - - - \to O A$

$- - \to A B = (4 \to i + 5 \to j + 6 \to k) - (6 \to i + 4 \to j + 5 \to k)$

$- - \to A B = 4 \to i + 5 \to j + 6 \to k - 6 \to i - 4 \to j - 5 \to k$

$- - \to A B = - 2 \to i + \to j + \to k$

$- - \to B C = - - \to O C - - - \to O B$

$- - \to B C = (5 \to i + 6 \to j + 4 \to k) - (4 \to i + 5 \to j + 6 \to k)$

$- - \to B C = 5 \to i + 6 \to j + 4 \to k - 4 \to i - 5 \to j - 6 \to k$

$- - \to B C = \to i + \to j - 2 \to k$

$- - \to C A = - - \to O A - - - \to O C$

$- - \to C A = (6 \to i + 4 \to j + 5 \to k) - (5 \to i + 6 \to j + 4 \to k)$

$- - \to C A = 6 \to i + 4 \to j + 5 \to k - 5 \to i - 6 \to j - 4 \to k$

$- - \to C A = \to i - 2 \to j + \to k$

Now,

$| A B | = \sqrt{{(- 2)}^{2} + 1^{2} + 1^{2}} = \sqrt{6}$

$| B C | = \sqrt{1^{2} + 1^{2} + {(- 2)}^{2}} = \sqrt{6}$

$| C A | = \sqrt{1^{2} + {(- 2)}^{2} + 1^{2}} = \sqrt{6}$

Then,

$A B = B C = C A = \sqrt{6}$
Hence, It triangle is equilateral triangle.

Suggest Corrections

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