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Question

Find by vector method the perimeter of the triangle whose vectors are (1,1,9),(3,1,5) and (0,5,1).

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Solution

AB=(1,1,9)(3,1,5)=(4,2,4)|AB|=42+22+42=6BC=(3,1,5)(0,5,1)=(3,6,4)|BC|=9+36+16=61CA=(0,5,1)(1,1,9)=(1,4,8)|CA|=12+42+82=64+16+1=9
Perimeter=|AB|+|BC|+|CA|
=6+61+9=15+7.81=22.81

1186589_1222466_ans_c9373b8d6ccb4ad0afdda857edbf2246.png

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