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I am presented with the following problem:

I have a parallelepiped with adjacent edges $$\vec{u} = [3,2,1]\\ \vec{v} = [2,3,1]\\ \vec{w} = [1,3,3]$$

a) Find volume
b) find area of face determined by $\vec{u}$ and $\vec{w}$
c) find angle between $\vec{u}$ and face determined by $\vec{v}$ and $\vec{w}$.

So for (a), I just used a simple equation:

$$V = ||\vec{u} \cdot (\vec{v} \times \vec{w}) ||,$$ which gave me $V=11 \space \text{cubic units}$.

For (b), i found what $\vec{u} \times \vec{w}$ was, which is $[-7,-8,7]$. But from here, I'm not exactly sure how to find the area?

For part (c), what I did was find $\vec{v} \times \vec{w}$, then did $\vec{u} \times [ \vec{v} \times \vec{w} ]$.. which gave me a vector $[17, -3, -21]$. Then I solved for the angle by rearranging the equation

$$||\vec{u} \times \vec{w}|| = ||\vec{u}||||\vec{w}||sin(\theta)$$

Is that correct?

1 Answers 1

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For b), the formula is $\|u \times w\|$. For c), find the angle between $u$ and $v \times w$ or $-v \times w$, whichever is acute, and subtract it from $90^\circ$ (draw a picture).

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    I found the angle, which was 60.271 degrees... so my solution is simply 90-60 = 30 degrees?2012-06-10
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    Assuming you found the angle correctly, that is right. You should probably not give your answer as $30^\circ$, though, since that is a "special angle" and people might think it's an exact answer. I would use more digits. By the way, using the dot product to compute angles between vectors is better than using the cross product. For one thing, dot products are easier to compute. The other reason is that, even if you know $\sin \theta$, unless $\sin \theta$ is $1$, there are going to be two angles between $0^\circ$ and $180^\circ$ with that $\sin$.2012-06-10
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    Well to find the correct angle, I think it is just v x w and not -v x w since they give the hint "relate the angle between the vector u and the vector normal to the face"...2012-06-10
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    There are two directions that a vector normal to the face may point. If you want your answer to be between $0^\circ$ and $90^\circ$, which is natural, you should consider both directions.2012-06-10
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    So should I provide a solution with both v x w and -v x w?2012-06-10
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    No, you should only provide one solution. If the angle between $u$ and $v \times w$ is acute, subtract it from $90^\circ$. Otherwise, subtract $90^\circ$ from the angle. Again, I recommend drawing a picture.2012-06-10
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    Drawing a picture of the parallelepiped?2012-06-10
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    One more comment. No, draw the plane spanned by $v$ and $w$ head-on so it looks like a line. Sketch $v \times w$ perpendicular to the plane. Sketch $u$ in the case where the angle between $u$ and $v \times w$ is acute and in the case where it is obtuse. Figure out the angle between $u$ and the plane in each case.2012-06-10