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Given $a = \langle 1,-1,2\rangle$ and $b = \langle 2,1,0\rangle$ . Find $t$ such that the vector $c = \langle 5,t-1,2\rangle$ is perpendicular to $a \times b$.

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What have you tried on this? –  Patrick Li Sep 30 '12 at 0:27
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up vote 1 down vote accepted

That is, you're looking for $t$ such that $c\cdot(a\times b)=0$. You need first, therefore, find $a\times b$, then take the dot product of the result with $c$, and solve the resulting equation for $t$.

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so if a ⋅ b = 0 then that means they are parallel if a x b = non-zero then that means they are perpendicular? –  40Plot Sep 30 '12 at 0:31
    
No. If $a\cdot b=0$, then $a$ and $b$ are perpendicular. If $a\times b\neq 0$, then $a$ and $b$ aren't parallel. That isn't what you're looking for, though. You want $c$ and $(a\times b)$ to be perpendicular, which is equivalent to $c\cdot(a\times b)=0.$ That's the equation you need to work out and then solve for $t$. –  Cameron Buie Sep 30 '12 at 0:34
    
just a question, if i want to test if vector a is parallel to vector c or not then I'd just do a test to see if a x c = 0 . if its not then they are not parallel. If it is = 0 then a || c –  40Plot Sep 30 '12 at 0:45
    
Yes, that's correct. In general, if vectors $v,w\in\Bbb R^3$ are non-zero, then $v\times w=0$ if and only if $v,w$ are parallel. –  Cameron Buie Sep 30 '12 at 0:48
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Hint: note that $a\times b$ is perpendicular to all linear combinations of $a$ and $b$.

More: Find a linear combination of $a$ and $b$ that matches $\langle5,t-1,2\rangle$ in the first and last places:

$a+2b=(5,1,2)$; therefore, $(5,1,2)$ is perpendicular to $a\times b$; that is, $t=2$.

Alternatively: Expand and solve $(5,t-1,2)\cdot(1,-1,2)\times(2,1,0)=0$. $$ \begin{align} 0 &=(5,t-1,2)\cdot(1,-1,2)\times(2,1,0)\\ &=(5,t-1,2)\cdot(-2,4,3)\\ &=4t-8 \end{align} $$ which also gives $t=2$.

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