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Question
If points A, B and C have position vectors `2hati, hatj` and `2hatk` respectively, then show that ΔABC is an isosceles triangle.
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Solution
Using distance formula
AB = `sqrt((0 - 2)^2 + (1 - 0)^2 + (0 - 0)^2`
= `sqrt(4 + 1)`
AB = `sqrt(5)` units
BC = `sqrt((0 - 0)^2 + (0 - 1)^2 + (2 - 0)^2`
= `sqrt(1 + 4)`
= `sqrt(5)`
BC = `sqrt(5)` units
AC = `sqrt((0 - 2)^2 + (0 - 0)^2 + (2 - 0)^2`
= `sqrt(4 + 4)`
= `2sqrt(2)`
∵ AB = BC = `sqrt(5) ≠ AC`
`\implies` ΔABC is an isosceles triangle.
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