MCV4U: Intersection of Three Planes

How It Works

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

1. Easy

   Describe the intersection of $x + 2y - z = 3$, $2x + 4y - 2z = 1$ and $x + 2y - z = -5$.

   Show solution

   1. The normals are $[1, 2, -1]$, $[2, 4, -2]$ and $[1, 2, -1]$. All parallel.
   2. Divide the second equation by $2$: $x + 2y - z = \frac{1}{2}$. The three left sides match and the constants $3$, $\frac{1}{2}$, $-5$ differ, so the planes are distinct.

   Answer: Three parallel planes: no intersection.

2. Medium

   Find the intersection of $x + y + z = 6$, $2x - y + z = 3$ and $x + 2y - z = 2$.

   Show solution

   1. Classify: $\vec{n_2} \times \vec{n_3} = [2, -1, 1] \times [1, 2, -1] = [-1, 3, 5]$, and $[1, 1, 1] \cdot [-1, 3, 5] = 7 \ne 0$. One point.
   2. Add the first and third equations: $2x + 3y = 8$. Add the second and third: $3x + y = 5$.
   3. From $3x + y = 5$, $y = 5 - 3x$. Then $2x + 15 - 9x = 8$, so $x = 1$ and $y = 2$.
   4. From the first equation, $z = 6 - 1 - 2 = 3$. Check the second: $2 - 2 + 3 = 3$. Check the third: $1 + 4 - 3 = 2$.

   Answer: The planes meet at the point $(1, 2, 3)$.

3. Medium

   Find the intersection of $x + y - z = 2$, $2x - y + z = 1$ and $x - 2y + 2z = -1$.

   Show solution

   1. Classify: $\vec{n_2} \times \vec{n_3} = [2, -1, 1] \times [1, -2, 2] = [0, -3, -3]$, and $[1, 1, -1] \cdot [0, -3, -3] = 0$. Not a single point.
   2. Subtract the first equation from the second: $x - 2y + 2z = -1$. That is exactly the third equation, so it adds nothing: a line.
   3. Add the first two: $3x = 3$, so $x = 1$. Then the first gives $y - z = 1$.
   4. Let $z = t$. Then $y = 1 + t$.

   Answer: The line $x = 1$, $y = 1 + t$, $z = t$, or $\vec{r} = [1, 1, 0] + t[0, 1, 1]$.

4. Test level

   Describe the intersection of $x + y + z = 1$, $x - y + 2z = 3$ and $2x + 3z = 5$.

   Show solution

   1. No two normals are parallel: $[1, 1, 1]$, $[1, -1, 2]$, $[2, 0, 3]$.
   2. But $\vec{n_3} = \vec{n_1} + \vec{n_2}$, so the normals lie in one plane and the triple product is $0$.
   3. Add the first two equations: $2x + 3z = 4$. The third says $2x + 3z = 5$. Both cannot be true.

   Answer: No intersection. The planes meet in pairs along three parallel lines: a triangular prism.

5. Test level

   The planes $x + 2y + z = 4$, $2x - y + 3z = 1$ and $3x + y + kz = m$ are given. For which values of $k$ and $m$ do they meet at a point, along a line, or not at all?

   Show solution

   1. $\vec{n_1} \times \vec{n_2} = [1, 2, 1] \times [2, -1, 3] = [7, -1, -5]$.
   2. Triple product: $[7, -1, -5] \cdot [3, 1, k] = 21 - 1 - 5k = 20 - 5k$, which is zero only when $k = 4$.
   3. So for $k \ne 4$, any $m$, the planes meet at one point.
   4. For $k = 4$, $\vec{n_3} = [3, 1, 4] = \vec{n_1} + \vec{n_2}$. Adding the first two equations gives $3x + y + 4z = 5$.
   5. If $m = 5$, the third equation is that sum, so the planes share a line. If $m \ne 5$, the equations contradict each other, and since no two normals are parallel, the planes form a triangular prism.

   Answer: Point: $k \ne 4$. Line: $k = 4$ and $m = 5$. No intersection (triangular prism): $k = 4$ and $m \ne 5$.

Learn It on Martian Lab

Martian Lab teaches this in the Intersections of Three Planes lesson of its MCV4U Lines and Planes unit. Watch it, then take the Lines and Planes practice tests to see if it stuck.

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