Find the lengths of the sides of the triangle and also determine the type of a triangle:L \left(\right. 3 , \textrm{ } - 2 , \textrm{ } - 3 \left.\right) , \textrm{ } M \left(\right. 7 , \textrm{ } 0 , \textrm{ } 1 \left.\right) , \textrm{ } N \textrm{ } \left(\right. 1 , \textrm{ } 2 , \textrm{ } 1 \left.\right)

Engineering

Mathematics

Distance Formulae and Its Application

Area of a Triangle

Section Formulae and Centres of a Triangle

Question

Find the lengths of the sides of the triangle and also determine the type
of a triangle:

L (3, - 2, - 3), M (7, 0, 1), N (1, 2, 1)

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The position vectors

\overset{ˉ}{a}, \overset{ˉ}{b}, \overset{ˉ}{c}

of the points

L, M, N

are

\overset{ˉ}{a} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}, \overset{ˉ}{b} = 7 \hat{i} + \hat{k}, \overset{ˉ}{c} = \hat{i} + 2 \hat{j} + \hat{k}

\overline{L M} = \overset{ˉ}{b} - \overset{ˉ}{a} = (7 \hat{i} + \hat{k}) - (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) = 4 \hat{i} + 2 \hat{j} + 4 \hat{k}

\overline{M N} = \overset{ˉ}{c} - \overset{ˉ}{b} = (\hat{i} + 2 \hat{j} + \hat{k}) - (7 \hat{i} + \hat{k}) = - 6 \hat{i} + 2 \hat{k}

\overline{N L} = \overset{ˉ}{a} - \overset{ˉ}{c} = (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k}) = 2 \hat{i} - 4 \hat{j} - 4 \hat{k}

∴ l (L M) = ∣ \overline{A B}, ∣ = \sqrt{4^{2} + 2^{2} + 4^{2}} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6

∴ l (M N) = ∣ \overline{M N}, ∣ = \sqrt{(- 6)^{2} + 2^{2}} = \sqrt{36 + 4} = \sqrt{40} = \sqrt{10 \times 4} = 2 \sqrt{10}

∴ l (N L) = ∣ \overline{N L}, ∣ = \sqrt{(2)^{2} + (- 4)^{2} + (- 4)^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6

∴ l (L M) = 6, l (M N) = 2 \sqrt{10}, l (N L) = 6

∴ △ L M N

is isosceles.