How to determine the value of $x$ such that the parallelogram has a given area?
Determine the value(s) of $x$ such that the area of the parallelogram formed by the vectors $a = (x+1, 1, -2)$ and $b = (x, 3, 0)$ is$\sqrt{41}$.
My work (using cross product)
\begin{align} |a \times b| &= ((1)(0) - (-2)(3), (-2)(x) - (x+1)(0), (x+1)(3) - (1)(x))\\ |a \times b| &= (6, -2x, 3x + 3 - x)\\ |a \times b| &= (6, -2x, 2x +3)\\ \sqrt{41} &= (6, -2x, 2x + 3) \end{align}
I don't know what to do next and how I should isolate for $x$. If anyone can help me out, I'd appreciate it.
$\endgroup$ 11 Answer
$\begingroup$You have to be careful with your equality signs. What you already showed is that
$$a \times b = (6, -2x, 2x + 3).$$
In order to get the area to be $\sqrt{41}$, we need
$$\| a \times b \|^2 = 6^2 + (-2x)^2 + (2x + 3)^2 = 41.$$
This is a quadratic equation, which you can solve for $x$.
Spoiler:
$\endgroup$$x_1 = -1/2, \quad x_2 = -1$.
