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I know how to find area of parallelogram and height if the vectors are adjacent to each other but not sure what to do with diagonal vectors.

Here is my attempt so far.

Given diagonal vector is $\vec a=(1,3,-7)$ and diagonal vector $\vec b=(-3,5,-11)$, cross product of two diagonal is $(2, -32, 14)$ and the magnitude of cross product is $6\sqrt{34}$.

I know the area of parallelogram with two diagonal is $\frac 12|\vec a\times \vec b|$.

I can find height of parallelogram by dividing area of parallelogram ÷ magnitude of base.

How do I find base?

Am I in right track?

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  • $\begingroup$ Have you drawn a diagram? $\endgroup$ – gen-z ready to perish May 17 '18 at 7:01
  • $\begingroup$ Yes. Drew a parallelogram with sides being vector u and w and diagonal being vector a and b. In my diagram, vector u is a base $\endgroup$ – Jasoyuji May 17 '18 at 7:09
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HINT: If the edges of the parallelogram are given by vectors $\vec v$ and $\vec w$, then you have $\vec a = \vec v + \vec w$ and $\vec b = \vec v - \vec w$. What is $\vec a\times\vec b$ in terms of $\vec v\times\vec w$?

Then you need to recover $\vec v$ (or $\vec w$) from $\vec a$ and $\vec b$. How can you do that algebraically?

By the way, I don't know how you know which vector should be the base, hence which amount should be the height.

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  • $\begingroup$ Sorry for not answering with the correct format but I'm really lost. Is cross product of vector a and b same as (Vector v + vector w) × (Vector v - Vector w). Don't know where I am going with this $\endgroup$ – Jasoyuji May 17 '18 at 6:58
  • $\begingroup$ vector a =(1,3,-7)= vector v + vector w and vector b=(-3,5,-11)=Vector w-Vector v. Then vector w =(-1, 4, -9) and vector v=(4,-2, -18). Then find magnitude of vector v then area of parallelogram ÷ magnitude of v. $\endgroup$ – Jasoyuji May 17 '18 at 7:40

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