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Prove that the vectors 2hati-hatj+hatk, hati-3hatj-5hatk, 3hati-4hatj-4hatk are the sides of a right angled triangle. |
Answer» Solution :Let A,B and C be the points whose POSITION vectors are `2hati-hatj-hatk, hati-3hatj-5hatk` and `3hati-4hatj-4hatk` respectively. Then `vec(AB) = (hati-3hatj-5hatk)-(2hati-hatj+hatk) = -hati-2hatj-6hatk` `vec(BC) = (3hati-4hatj-4hatk)-(hati-3hatj+5hatk) =2hati-hatj+hatk`. `vec(CA) = (2hati-hatj+hatk)-(3hati-4hatj-4hatk) = -hati+3hatj+5hatk` Then AB = `|vec(AB)| = sqrt(1+4+36) = sqrt(41)` BC = `sqrt(4+1+1) = sqrt6` CA = `sqrt(1+9+25) = sqrt(35)`. Now `BC^2+CA^2 = AB^2` So `triangle ABC` is a right ANGLED triangle where `ANGLEABC = 90^@` (Proved). |
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