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Thread: Cross products and orthogonal planes/vectors

  1. #1
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    Cross products and orthogonal planes/vectors




    I know c) is not necessarily true. And I think if d) were true then $\displaystyle (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.
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    Quote Originally Posted by cubrikal View Post



    I know c) is not necessarily true. And I think if d) were true then $\displaystyle (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 $\displaystyle u\times v$= <0, -1, 0> and $\displaystyle u\times w$= <0, 0, -1>, both non-zero, so that $\displaystyle (u\times v)\cdot(u\times v)= 0$ but u is not perpendicular to w.
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    Quote Originally Posted by cubrikal View Post



    I know c) is not necessarily true. And I think if d) were true then $\displaystyle (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 $\displaystyle u = <1,1,1>, v = <1,2,3>$ and $\displaystyle w = <a,b,c>$ to be determined.

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

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