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The points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are |
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Answer» collinear Now, `AB=PV" of "B - PV " of "A` `vec(AB)=(2hati+6hatj+3hatk)-(1hati+2hatj+7hatk)` `=hati+4hatj-4hatk` `implies |AB|=sqrt((1)^(2)+(4)^(2)+(-4)^(2))` `=sqrt(1+16+16)=sqrt(33)` `BC=PV" of "C-PV" of "B` `=(3hati+10hatj-1hatk)-(2hati+6hatj+3hatk)` `=hati+4hatj-4hatk` `implies |BC|=sqrt((1)^(2)+(4)^(2)+(-4)^(2))` `=sqrt(1+16+16)=sqrt(33)` `AC=PV" of "C-PV" of "A` `vec(AC)=(3hati+10hatj-1hatk)-(1hati+2hatj+7hatk)` `=2hati+8hatj-8hatk` `implies |AC|=sqrt(2^(2)+8^(2)+(-8)^(2))=sqrt(4+64+64)` `=sqrt(132)=2sqrt(33)=sqrt(33)+sqrt(33)` `THEREFORE |AC|=|AB|+|BC|` Hence, the given points A, B and C are collinear. |
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