I'm currently studying linear algebra and I'm struggling with a few questions. I would appreciate any help or guidance you can provide.
Question 1: I need to show that the line \begin{align} x-2=\frac{y+3}{2}=\frac{z-1}{4} \end{align} is parallel to the plane $2y-z=1$. Additionally, I need to find the distance between the line and the plane.
Here is what I have attempted so far:
To show that the line is parallel to the plane, I can show that the direction vector of the line is orthogonal to the normal vector of the plane. The direction vector of the line is $\langle 1, 2, 4 \rangle$. The normal vector of the plane is $\langle 0, 2, -1 \rangle$. To check if these vectors are orthogonal, I can take their dot product: \begin{align} \langle 1, 2, 4 \rangle \cdot \langle 0, 2, -1 \rangle = 0 \end{align} Since the dot product is zero, the two vectors are orthogonal and therefore the line is parallel to the plane.
To find the distance between the line and the plane, I can find the projection of the vector between a point on the line and a point on the plane onto the normal vector of the plane. However, I'm not sure how to proceed from here.
Any help would be greatly appreciated! Thank you in advance.
The line $r$ is given as the intersection of two planes:
$$ x-2=\frac{y+3}{2}=\frac{z-1}{4} \quad \quad \Rightarrow \quad \quad \begin{cases} x-2=\frac{y+3}{2} \\ x-2=\frac{z-1}{4} \\ \end{cases} \quad \quad \Rightarrow \quad \quad \begin{cases} y = 2\,x - 7 \\ z = 4\,x - 7 \end{cases} $$
so it can be parameterized like this:
$$ (x,y,z) = (0,-7,-7) + (1,2,4)u \quad \quad \text{with} \; u \in \mathbb{R}. $$
Since the assigned plane $\pi$ has Cartesian equation:
$$ 0\,x+2\,y-z-1=0 $$
noting that:
$$ (1,2,4) \cdot (0,2,-1) = 0 $$
it follows that $r \parallel \pi$, as you rightly wrote and reiterated. Good!
Therefore, a line $s$ passing through $r$ and perpendicular to $r$ can be parameterized as:
$$ (x,y,z) = \underbrace{(0,-7,-7)}_{P} + (0,2,-1)v \quad \quad \text{with} \; v \in \mathbb{R} $$
which intersected with $\pi$ leads to:
$$ 2\left(-7+2\,v\right) - \left(-7-v\right) - 1 = 0 \quad \quad \Leftrightarrow \quad \quad v = \frac{8}{5} $$
i.e. the intersection point is $Q\equiv\left(0,-\frac{19}{5},-\frac{43}{5}\right)$ and the required distance is equal to:
$$ \text{dist}(r,\pi) = ||P-Q|| = \sqrt{\left(0-0\right)^2+\left(-7+\frac{19}{5}\right)^2+\left(-7+\frac{43}{5}\right)^2} = \frac{8}{\sqrt{5}}. $$