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If the position vectors of the vertices of Delta ABC are 3hati+hatj+2hatk, hati-2hatj+7hatk and -2hati+3hatj+5hatk, then the DeltaABC is |
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Answer» right angled and isosceles `A(3hati+hatj+2hatk),B(hati-2hatj+7hatk)andC(-2hati+3hatj+5hatk)`. Now, `AB=(hati-2hatj+7hatk)-(3hati+hatj+2hatk)` `=-2hati-3hatj+5hatk` `THEREFORE |AB|=sqrt(4+9+25)=sqrt(38)` Similarly, `|BC|=|CA|=sqrt(38)` `therefore |AB|=|BC|=|CA|=sqrt(38)` HENCE, triangle is an equilateral triangle. |
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