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Question
Show that the vectors `2hati - hatj + hatk, hati - 3hatj - 5hatk` and `3hati - 4hatj - 4hatk` from the vertices of a right angled triangle.
Solution
Let the given vectors `2hati - hatj + hatk, hati - 3hatj - 5hatk and 3hati - 4hatj - 4hatk` be represented by A, B, C respectively,
Position vector of A `vec(OA) = 2hati - hatj + hatk`
Position vector of B `vec(OB) = hati - 3hatj - 5hatk`
Position vector of C `vec(OC) = 3hati - 4hatj - 4hatk`
Now `vec(AB) = vec(OB) - vec(OA)`
= `(hat1 - 3hatj - 5hatk) - (2hat1 - hatj + hatk)`
= `-hat1 - 2hatj + 6hatk`
`|vec(AB)| = sqrt((-1)^2 + (-2)^2 + (-6)^2) `
`= sqrt(1 + 4 + 36) `
`= sqrt41`
`vec(BC) = vec(OC) - vec(OB) = (3hat1 - 4hatj - 4hatk) - (hat1 - 3hatj - 5hatk) = 2hati - hatj + hatk`
`|vec(BC)| = sqrt((2)^2 + (-1)^2 + (1)^2) `
`= sqrt6`
`vec(CA) = vec(OA) - vec(OC) = (2hati - hatj + hatk) - (3hat1 - 4hatj - 4hatk) = -hati + 3hatj + 5hatk`
`|vec(CA)| = sqrt((-1)^2 + (3)^2 + (5)^2) `
`= sqrt35`
For right angled ΔABC where ∠C = 90°, then
`|vec(AB)|^2 = |vec(BC)|^2 + |vec(CA)|^2`
41 = 6 + 35 = 41
Therefore, a right-angled triangle is constructed from the given vectors.
Hence, proved
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