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If  \(\vec a , \vec b, \vec c\) determine the vertices of a triangle, show that \(\frac{1}{2} [\vec b \times \vec c + \vec c \times \vec a + \vec a \times \vec b]\) gives the vector area of the triangle. Hence deduce the condition that the three points a, b, c are collinear. Also find the unit vector normal to the plane of the triangle.

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Best answer

We know that,

Vector area of ∆ABC is given as,

Thus, shown.

We know that, two vectors are collinear if they lie on the same line or parallel lines.

For \(\vec a, \vec b \,and\, \vec c\) to be collinear, area of the ∆ABC should be equal to 0.

⇒ Area of ∆ABC = 0

Thus, this is the required condition for \(\vec a, \vec b \,and\, \vec c\) to be collinear.

Now, we need to find the unit vector normal to the plane of the triangle.

Let \(\vec n\) be the unit vector normal to the plane of the triangle.

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