Determine the value of $a$ such that the area of the parallelogram determined by $u = (2,1,-1)$ and $v = (1,-1,a)$ is $\sqrt{62}$.
So I did the cross product via determinant and ended up with $(a-1,-2a-1,-3)$. How do I take it from here?
Area of a parallelogram given two vectors
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$\begingroup$
linear-algebra
cross-product
3 Answers
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The magnitude of this cross product equals the area of the parallelogram.
In your question, we thus get, $$\sqrt {(a-1)^2+(-2a-1)^2+(-3)^2} =\sqrt {62} $$ $$\Rightarrow \sqrt {5a^2+2a+11} =\sqrt {62} $$ We can solve this equation to get the answer. Hope it helps.
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The norm of the cross product is the desired area :
$$\mathcal{A}=\sqrt{(a-1)^2+(2a+1)^2+9}=\sqrt{5a^2+2a+11}$$
So you must have $5a^2+2a-51=0$. Hence $a=3$ or $a=-\frac{17}{5}$
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If $\theta$ is the angle between $u$ and $v$ then |$u$ x $v$| = $uvsin(\theta)$ is the area of the parallelogram. Here, u is the length of the $base$, $vsin\theta $ is the $height$ and $area$ is simply $(base)*(height)$. The area is of course the $62^.5$. Solve the equation |$u$ x $v$|= $62^.5$. Note that you already have calculated $u$ x $v$.