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I have to give the equations of the planes that go through the point $(3,4,5)$ that are parallel to the lines with direction $(0,2,1)$ and that are on a distance 2 from the point $(1,1,1)$.

So if you say that the other vector of this plane is given by $(a, b, c)$ then I came to the conclusion that the equations of these planes are given by

$\alpha \leftrightarrow (2c-b)x + ay - 2az = -6a-3b+6c$

but now I don't know how I can use the distance to calculate the specific equations because you can't really derive a relation between $a,b$ and $c$.

Is there something I'm missing or an easier approach to solve this problem?

Mee98
  • 1,133

2 Answers2

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HINT

Let consider the plane equation $ax+by+cz+1=0$ (recall indeed that we can assume wlog $d=1$) and the given conditions

  • $3a+4b+5c+1=0$

  • $0\cdot a+2\cdot b+1\cdot c=0$

  • $\frac{|1\cdot a+1\cdot b+1\cdot c+1|}{\sqrt{a^2+b^2+c^2}}=2$

from which we can find $a,b,c$.

user
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Working this problem from a perspective-geometric point of view, the first two conditions define a pencil of planes that contain the points $[3:4:5:1]$ and $[0:2:1:0]$ (the point at infinity on the given set of parallel lines). These planes are null vectors of the matrix $$\begin{bmatrix}3&4&5&1\\0&2&1&0\end{bmatrix}.$$ This space is spanned by $[1:0:0:-3]$ and $[2:1:-1:0]$, so the planes containing the two points are linear combinations of these basis vectors: $$\mathbf\pi(\lambda,\mu) = \lambda[1:0:0:-3]+\mu[2:1:-2:0] = [\lambda+2\mu:\mu:-2\mu:-3\lambda],$$ i.e., every plane that satisfies the first two conditions has an equation of the form $$(\lambda+2\mu)x + \mu y - 2\mu z - 3\lambda = 0$$ for $\lambda$ and $\mu$ not both zero. Using the usual formula for the distance between a point and plane and squaring to eliminate the square root will give you a homogeneous quadratic equation in $\lambda$ and $\mu$, which in this case easily factors to give two linearly independent solutions.

amd
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