Cross products and orthogonal planes/vectors

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May 18th 2009, 09:31 AM
cubrikal

Cross products and orthogonal planes/vectors

http://img297.imageshack.us/img297/7...ssproductq.jpg

I know c) is not necessarily true. And I think if d) were true then $(u \times v) \cdot (u \times w) \neq 0$

But aren't a) and b) both true? a) follows from the definition of orthogonal planes because of perpendicular normals, and I can't think of a counterexample for b); somehow though, my thinking is flawed because the question is multiple choice.
May 18th 2009, 10:47 AM
HallsofIvy

Quote:

Originally Posted by cubrikal

http://img297.imageshack.us/img297/7...ssproductq.jpg

I know c) is not necessarily true. And I think if d) were true then $(u \times v) \cdot (u \times w) \neq 0$

But aren't a) and b) both true? a) follows from the definition of orthogonal planes because of perpendicular normals, and I can't think of a counterexample for b); somehow though, my thinking is flawed because the question is multiple choice.

No, b is not necessarily true. For example, if u= <1, 0, 0>, v= <0, 0, 1>, and w= <1, 1, 0>, then $u\times v$ = <0, -1, 0> and $u\times w$ = <0, 0, -1>, both non-zero, so that $(u\times v)\cdot(u\times v)= 0$ but u is not perpendicular to w.
May 18th 2009, 11:02 AM
Danny

Quote:

Originally Posted by cubrikal

http://img297.imageshack.us/img297/7...ssproductq.jpg

I know c) is not necessarily true. And I think if d) were true then $(u \times v) \cdot (u \times w) \neq 0$

But aren't a) and b) both true? a) follows from the definition of orthogonal planes because of perpendicular normals, and I can't think of a counterexample for b); somehow though, my thinking is flawed because the question is multiple choice.

I thinks there's lots of counter examples for (b)

Consider the vectors $u = <1,1,1>, v = <1,2,3>$ and $w = <a,b,c>$ to be determined.

Now $u \times v = <1,-2,1>$ and $u \times w = <a-b,a-c,b-a>$ so that $(u \times w) \cdot (u \times w) = 3c - 3a = 0 \; \Rightarrow\; c = a$ so $w = <a,b,a>$ noting that $a \ne b$ as this would make $u \times w = 0$ .

Now $v \cdot w = 4a + b$ and there a lots of choices for $a$ and $b$ such that this isnt' zero.