Question

Show that the vectors `-hat"i" - 2hat"j" - 6hat"k", 2hat"i" - hat"j" + hat"k"` and find `-hat"i" + 3hat"j" + 5hat"k"` form a right angled triangle

Answer 1

Let `vec"AB"| = -hat"i" - 2hat"j" - 6hat"k"`

`vec"BC" = 2hat"i" - hat"j" + hat"k"`

and `vec"CA" = -hat"i" - 2hat"j" - 5hat"k"`

`|vec"AB"| = |-hat"i" - 2hat"j" - 6hat"k"|`

= `sqrt((-1)^2 + (-2)^2 + (-6)^2`

AB = `sqrt(1 + 4 + 36)`

= `sqrt(41)`

`|vec"BC"| = |2hat"i" - hat"j" + hat"k"|`

= `sqrt(2^2 + (-1)^2 + 1^2)`

BC = `sqrt(4 + 1 + 1)`

= `sqrt(6)`

`|vec"CA"| = |-hat"i" + 3hat"j" + 5hat"k"|`

= `sqrt((-1)^2 + 3^2 + 5^2)`

CA = `sqrt(1 + 9 + 25)`

= `sqrt(35)`

AB ≠ BC + CA

∴ The given vectors form a triangle, Also

AB² = 41, BC² = 6, CA² = 35

AB² = BC² + CA²

∴ ∆ABC is a right angled triangle.