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
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
A. right angled and isosceles
B. right angled, but not isosceles
C. isosceles but not right angled
D. equilateral
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Correct Answer - D
Given vertices are
`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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