Question

Show that the vectors `2hati - hatj + hatk, hati - 3hatj - 5hatk`  and `3hati - 4hatj - 4hatk` from the vertices of a right angled triangle.

Answer 1

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