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Question
Show that the points A, B and C with position vectors `veca = 3hati - 4hatj - 4hatk`, `vecb = 2hati - hatj + hatk` and `vecc = hati - 3hatj - 5hatk`, respectively form the vertices of a right angled triangle.
Solution
Here, `vec(AB) = vecb - veca`
`= (2hati - hatj + hatk) - (3hati - 4hatj - 4hatk)`
`= -hati + 3hatj + 5hatk`
`vec (BC) = vecc - vecb = (hati - 3hatj - 5hatk) - (2hati + hatj + hatk)`
`= hati - 2hatj - 6hatk`
`vec(CA) = veca - vecc = (3hati - 4hatj - 4hatk) - (hati - 3hatj - 5hatk)`
`= 2hati - hatj + hatk`
`|vec(AB)|^2 = (-1)^2 + 3^2 + 5^2`
= 1 + 9 + 25
= 35
`|vec(BC)|^2 = (-1)^2 + (-2)^2 + (-6)^2`
= 1 + 4 + 36
= 41
and `|vec(CA)|^2 = 2^2 + (-1)^2 + 1^2`
= 4 + 1 + 1
= 6
So, `|vec(BC)|^2 = |vec(AB)|^2 + |vec(CA)|^2`
Hence, the triangle is a right angles triangle.
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