QuestionLet $Q(1,2,3)$ be a point in $\mathbb{R}^{3}, P_{1}: x-y+z=-1$ be a plane in $\mathbb{R}^{3}$. (a) Find the parametric equations of the plane $P_{1}$. (b) Find the general equation of the plane $P_{2}$ through $Q$ that is parallel to $P_{1}$ (c) Find the distance between the planes $P_{1}$ and $P_{2}$.

Studdy Solution
To find the distance between the planes $P_{1}$ and $P_{2}$, use the formula for the distance between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$:
$\text{Distance} = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}$
- For $P_{1}: x - y + z = -1$, $d_1 = -1$. - For $P_{2}: x - y + z = 2$, $d_2 = 2$.
- The distance is: $ \frac{|2 - (-1)|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
The distance between the planes $P_{1}$ and $P_{2}$ is:
$$\boxed{\sqrt{3}}$$