Find the volume of parallelopiped

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utkarshakash
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Homework Statement


Let a,b and c are the three vectors such that |a|=|b|=|c| =2 and angle between a and b is ∏/3, b and c is ∏/3 and a and c is ∏/3
If a,b and c represents adjacent edges of paralleopiped then find its volume.

Homework Equations



The Attempt at a Solution



Volume of parallelopied = [itex]\left( \vec{a} \times \vec{b} \right) . \vec{c} \\<br /> |\vec{a} \times \vec{b}| |\vec{c}| cos \theta \\<br /> 4 \sqrt{3} cos \theta[/itex]

But I don't know what is the angle between a x b and c.
 
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haruspex said:
You could approach it as a geometry question (have you worked out what shape the origin and the endpoints of the three vectors give you?). It's not obvious to me how to solve it with a purely vectorial approach.

Since it is a parallelopiped it is obvious that the given vectors are non-coplanar and the angle between each of them is ∏/3. If we consider only the origin and endpoints it will look like a tetrahedron.
 
utkarshakash said:
Since it is a parallelopiped it is obvious that the given vectors are non-coplanar and the angle between each of them is ∏/3. If we consider only the origin and endpoints it will look like a tetrahedron.

Are you sure that those vectors are not not coplanar? :devil:

The three side-edges of the tetrahedron are of the same length and each pair enclose the same angle. You project the tetrahedron onto the plane of the base. What angle do the pairs of edges enclose in the projection?

ehild
 
utkarshakash said:
Since it is a parallelopiped it is obvious that the given vectors are non-coplanar and the angle between each of them is ∏/3. If we consider only the origin and endpoints it will look like a tetrahedron.

Right, so courtesy of Pythagoras you can figure out the height. From there you can compute the volume directly, or out of interest relate it to the angle between axb and c.
 
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