Show that the points $7 ¯ j + 10 ¯ ¯ ¯ k, - ¯ i + 6 ¯ j + 6 ¯ ¯ ¯ k, - 4 ¯ i + 9 ¯ j + 6 ¯ ¯ ¯ k$ form a right angled isosceles triangle.

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Solution

$7^j + 10^k a_{1}, -^i + 6^j + 6^k a_{2}, - 4^i + a^j + 6^k a_{3}$

Distance between $a_{1} a_{2}$

$\sqrt{(- 1 - 0)^{2} +) 6 - 7)^{2} + (6 - 10)^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18}$

Distance between $a_{2} a_{3}$

$\sqrt{(- 4 + 1)^{2} + (9 - 6)^{2} + (6 - 6)^{2}} = \sqrt{9 + 9} = \sqrt{18}$

Distance between $a_{1} a_{3}$

$\sqrt{(- 4 - 0)^{2} + (9 - y)^{2} + (6 - 10)^{2}} = \sqrt{16 + 4 + 16} = \sqrt{36}$

$a_{1} a_{2} = a_{2} a_{3} \Rightarrow$ it is isosceles triangle.

$(a_{1} a_{2})^{2} + (a_{2} a_{3})^{2} = (a_{1} a_{3})^{2}$

$\Rightarrow 18 + 18 = 36$

$\Rightarrow 36 = 36 \Rightarrow$ it is right angled triangle.

$∴$ the given points form a right angled isosceles triangle.

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