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Question
Show that the points A(1, 1, 1), B(1, 2, 3) and C(2, – 1, 1) are vertices of an isosceles triangle
Solution
The given vertices of a triangle are
A(1, 1, 1) ,B(1, 2, 3) and C (2, – 1, 1)
Position vector of A is `vec"OA" = hat"i" + hat"j" + hat"k"`
Position vector of B is `vec"OB" = hat"i" + 2hat"j" + 3hat"k"`
Position vector of C is `vec"OC" = 2hat"i" + hat"j" + hat"k"`
`vec"AB" = vec"OB" - vec"OA"`
= `(hat"i" + 2hat"j" + 3hat"k") - (hat"i" + hat"j" + hat"k")`
= `hat"i" + 2hat"j" + 3hat"k" - hat"i" - hat"j" + hat"k"`
`vec"AB" = hat"j" + 2hat"k"`
`vec"BC" = vec"OC" - vec"OB"`
= `(2hat"i" - hat"j" + hat"k") -(hat"i" + 2hat"j" + 3hat"k")`
= `2hat"i" - hat"j" + hat"k" - hat"i" - 2hat"j" - 3hat"k"`
`vec"BC" = hat"i" - 3hat"j" - 2hat"k"`
`vec"CA" = vec"OA" - vec"OC"`
= `(hat"i" + hat"j" + hat"k") - (2hat"i" - hat"j" + hat"k")`
= `hat"i" + hat"j" + hat"k" - 2hat"i" + hat"j" - hat"k"`
`vec"CA" = vec"OA" = vec"OC"`
= `(hat"i" + hat"j" + hat"k") - (2hat"i" - hat"j" + hat"k")`
= `hat"i" + hat"j" + hat"k" - 2hat"i" + hat"j" - hat"k"`
`vec"CA" = -hat"i" + 2hat"j"`
`|vec"AB"| = |hat"j" + 2hat"k"|`
AB = `sqrt(1^2 + 2^2)`
= `sqrt(1 + 4)`
= `sqrt(5)`
`|vec"BC"| = |hat"i" - 3hat"j" - 2hat"k"|`
= `sqrt(1^2 + (-3)^2 + (-2)^2`
= `sqrt(1 + 9 + 4)`
= `sqrt(14)`
`|vec"CA"| = |-hat"i" + 2hat"j"|`
CA = `sqrt((-1)^2 + 2^2`
= `sqrt(1 + 4)`
= `sqrt(5)`
AB + CA = `sqrt(5) + sqrt(5)`
= `2sqrt(5)`
BC = `sqrt(14)`
AB + CA ≠ BC
∴ A, B, C are not collinear.
Hence A, B, C form a triangle.
AB = CA = `sqrt(5)`
∴ ∆ABC is an isosceles triangle.
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