Find the volume of parallelopiped

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Homework Help Overview

The problem involves finding the volume of a parallelepiped defined by three vectors a, b, and c, each with a magnitude of 2 and angles of π/3 between each pair. The original poster attempts to apply the formula for the volume of a parallelepiped but is uncertain about the angle between the cross product of two vectors and the third vector.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants suggest considering the geometric interpretation of the vectors and their endpoints, questioning the coplanarity of the vectors and exploring the shape formed by them.

Discussion Status

Participants are exploring different approaches to the problem, including geometric interpretations and vectorial methods. There is a focus on understanding the relationships between the vectors and the implications of their angles. No explicit consensus has been reached, but various lines of reasoning are being discussed.

Contextual Notes

There is a mention of the vectors being non-coplanar and the challenge of determining the angle between the cross product of two vectors and the third vector. The discussion also references the shape formed by the vectors, likening it to a tetrahedron.

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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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.
 
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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