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anonymous

  • 5 years ago

Find all vector < 1, a, b> orthogonal to <4, -8, 2>. This is what I got so far, but I'm not sure whether my process is correct or not. let u = < 1, a, b> and v = <4, -8, 2>. u*v = < 1, a, b> *<4, -8, 2>= 4-8a+2b set 4-8a+2b= 0

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  1. anonymous
    • 5 years ago
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    Your process if correct assuming you're using the dot product definition :)

  2. anonymous
    • 5 years ago
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    Yes by dot product definition. u*v=0 which orthogonal, but how do I do it? I stuck with solve a and b. For some reason, I keep getting 0 = 0.

  3. anonymous
    • 5 years ago
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    Oh, well, you have one equation in two unknowns, so technically, you have an infinite number of solutions. You can see it clearer if you say, set a=t. Then\[4-8a+2t \rightarrow b=4t-2\]Then the set of all vectors orthogonal to <4,-8,2> is\[\left\{ <1,a,b>|t \in \mathbb{R}, a=t, b=4t-2 \right\}\]That is,\[<1,t,4t-2>\]where t is any real number. You could have chosen b=t and solved for a, or even just solve b for a or a for b. For example, since \[4-8a+2b=0 \rightarrow b=4a-2\]and you could just write,\[<1,a,4a-2>\]where a is real. The point is, when you more unknowns than equations, you will have an infinite number of solutions. Check with a couple of numerical examples if you like.

  4. anonymous
    • 5 years ago
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    Example: Find two vectors orthogonal to <4,-8,2>. Taking your set of possible vectors, <1,a,4a-2>, let a=0 for one, and let a=1 for the other (you can pick any real number, I just picked these for ease). Then two possible orthogonal vectors are: <1,0,-2> and <1,1,2>. We can check: <4,-8,2>.<1,0,-2> = 4 + 0 - 4 = 0 ... good. <4,-8,2>.<1,1,2> = 4 - 8 + 4 = 0 ... good. This is how it works. Hope it helps :)

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