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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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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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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