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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
Solution
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.
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