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Question
The position vectors of the vertices of a triangle are `hat"i" + 2hat"j" + 3hat"k", 3hat"i" - 4hat"j" + 5hat"k"` and `-2hat"i" + 3hat"j" - 7hat"k"`. Find the perimeter of the triangle
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Solution
Let A, B, C be the vertices of the triangle ABC.
Now, `vec"AB" = vec"OB" - vec"OA"`
= `(3hat"i" - 4hat"j" + 5hat"k") - (hat"i" + 2hat"j" + 3hat"k")`
= `3hat"i" - 4hat"j" + 5hat"k" - hat"i" - 2hat"i" - 3hat"k"`
= `2hat"i" - 6hat"j" + 2hat"k"`
|vec"B"| = sqrt(4 + 36 + 4)`
= `sqrt(44)`
= AB
BC = `vec"OC" - vec"OB"`
= `(-2hat"i" + 3hat"j" - 7hat"k") - (3hat"i" - 4hat"j" + 5hat"k")`
= `-2hat"i" + 3hat"j" - 7hat"k" - 3hat"i" + 4hat"j" + 4hat"j" - 5hat"k"`
= `-5hat"i" + 7hat"j" - 12hat"k"`
`|vec"BC"| = sqrt(25 + 49 + 144)`
= `sqrt(218)`
= BC
`vec"AC"= vec"OC" - vec"OA"`
= `(-2hat"i" + 3hat"j" - 7hat"k") - (hat"i" + 2hat"j" + 3hat"k")`
= `-2hat"i" + 3hat"j" - 7hat"k" - hat"i" - 2hat"j" - 3hat"k"`
= `-3hat"i" + hat"j" - 10hat"k"`
`|vec"AC"| = sqrt(9 + 1 +100)`
= `sqrt(110)`
= AC
Premiter of ΔABC = AB + BC + AC
=`sqrt(44) + sqrt(218) + sqrt(110)`.
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