I'm stuck with this question. Consider the plane $E$ with cartesian equation $x - 2y + az = 6$, and the line $L$ with vector form $(x, y, z) = (1, -1, 0) + t(-3, -a, a)$. For what values of $a$, if any do $L$ and $E$ intersect?

I find it hard to visualise what these 2 equations look like and what their intersection would be like. I'm also not really sure where to start with these questions in general. Thanks.

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You have a line given with a point $\vec p$ and a direction $\vec r$ as

$$L=\{\vec p+t\vec r\in\Bbb R^3 \mid t\in\Bbb R\} \tag 1$$

where $\vec r$ is a function of $a$.

The plane $E$ is given by a vector $\vec n$ that's normal to $E$. Given a point $\vec q\in E$, all points $\vec v\in E$ satisfy $$(\vec v-\vec q)\cdot \vec n = 0\tag2$$ because the line between any two points in the plane is perpendicular to the normal $\vec n$. Now $\vec q\cdot\vec n = d$ is just some real number that's related to the distance between $E$ and the origin. Hence, $E$ can be expressed as

$$E = \{\vec v\in\Bbb R^3 \mid \vec v\cdot\vec n=d\}\tag 3$$

If we intersect $L$ with $E$, then all points in $L\cap E$ must satisty the contition in $(3)$, so insert the term $(1)$ that represents points in $L$ into $(3)$:

$$L\cap E = \{ \vec p+t\vec r \in\Bbb R^3 \mid t\in\Bbb R, (\vec p+t\vec r)\vec n = d\} \tag 4$$

This gives a determinig equation for $t$:

$$t = \frac{d-\vec p\cdot\vec n}{\vec r\cdot \vec n}\tag 5$$

Now you have to drop in values and determine for which values of $a$ there are solutions, and how many solutions there are. In your specific case:

$$\begin{align} d &= 6 \\ \vec p &= (1,-1,0) \\ \vec r &= (-3,-a,a) \\ \vec n &= \cdots \end{align}$$

As your $\vec n$ also depends on $a$, you'll get an equation in $a$ that's quadratic or linear (depending on how you resolve that typo in the specification of $E$).


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