1.

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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