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If points A, B and C have position vectors 2i^,j^ and 2k^ respectively, then show that ΔABC is an isosceles triangle.

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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.

Sum
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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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2022-2023 (March) Outside Delhi Set 3

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