# Please explain volume of parallelepiped

1. Aug 27, 2011

### maff is tuff

1. The problem statement, all variables and given/known data

Ok so my book says that the volume of a parallelepiped is:

V= |b cross c||a||cos(theta)| = |a dot (b cross c)|, where a, b, and c are vectors

I get the "|b cross c||a||cos(theta)|" part because I can see the geometry but I don't get how they get from that to |a dot (b cross c)|.

2. Relevant equations

3. The attempt at a solution

I see how |a||cos(theta)| looks kinda familiar so I tried to figure out using that.

So I know a dot b = |a||b|cos(theta), so

|a|cos(theta) = (a dot b)/|b| but then if I plug that quantity in it doesn't seem to work.

So can anyone help me see why |b cross c||a||cos(theta)| = |a dot (b cross c)| ?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 27, 2011

### LCKurtz

Think of a, b, and c with their tails together. b x c is a vector perpendicular to the plane of b and c and |b x c| is numerically equal to the area of the parallelogram formed by b and c. To get the volume of the parallelepiped formed with this parallelogram as a base and a as the third side you need to multiply the area of the base by the height of the parallelepiped. But the height of the parallelepiped is the component of a perpendicular to the plane of b and c. If theta is the angle between b x c and a, then this height is h = |a|cos(theta). So the volume is V = h|bxc| = |a|cos(theta)|bxc| = a dot (bxc). Does that help?

3. Aug 27, 2011

### maff is tuff

I already understood how the volume is |bxc||a||cos(theta)| but I don't understand how that equals |a dot(bxc)|. Can you explain how |bxc||a||cos(theta)|= |a dot(bxc)|? Thanks.

4. Aug 27, 2011

### LCKurtz

Oh, that's all you are asking. That is just the basic property of dot product:

$$\vec A \cdot \vec B = |\vec A||\vec B|\cos\theta$$

applied to a and bxc. Here θ is the angle between a and bxc.

5. Aug 30, 2011

### maff is tuff

Ok so I'm still not seeing it.

I want to show that |bxc||a||cos(theta)| = |a dot (bxc)|

ok my first step: I know that a dot b = |a||b|cos(theta) so I can solve for |a|cos(theta)

So now I have |bxc|(a dot b)/||b| = |a dot (bxc)|

So I plug it in and get:

|bxc|[(a dot b)/|b|] and now I am stuck.

6. Aug 30, 2011

### LCKurtz

Just do what I suggested above. Replace B by BxC in the identity.