# Calculate product of vectors `bar(AB)* bar (AC), ` given area of triangle ABC= `sqrt3` , AB=AC=CB?

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### 1 Answer

Notice that the problem provides the information that all sides of triangle have equal lengths, hence, the triangle is equilateral, having the measure of all internal angles of `60^o` .

Since the measures of internal angles are known, you may use the following formula for area of triangle such that:

`A = (AB*AC*sin hatA)/2`

The problem provides the value of area, hence, you need to substitute sqrt3 for A and `60^o` for hatA such that:

`sqrt3 = (AB^2*sin 60^o)/2 `

`sqrt3 = (AB^2*sqrt3/2)/2`

`1 = (AB^2)/4 =gt AB^2 = 4 =gt AB =2`

You should evaluate the dot product of vectors `bar(AB)` and `bar(AC)` such that:

`bar(AB)*bar(AC) = |bar(AB)|*|bar(AC)|*cos(hat(bar(AB)bar(AC)))`

Since the angle between the vectors `bar(AB)` and `bar(AC)` is of `60^o` and the magnitudes of the vectors are equal to the lengths of the sides, yields:

`bar(AB)*bar(AC) = 2*2*cos 60^o`

`bar(AB)*bar(AC) = 4*(1/2) =gt bar(AB)*bar(AC) = 2`

**Hence, evaluating the dot product of vectors bar(AB) and bar(AC) under given conditions, yields `bar(AB)*bar(AC) = 2` .**

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