Lines and planes in space
Tonight's idea
A line is a point plus a direction. If it passes through \(A\) along \(\mathbf{d}\), its points are \(\mathbf{r} = \mathbf{a} + t\,\mathbf{d}\), where \(\mathbf{a}\) is the position vector of \(A\) (the arrow from the origin to \(A\)) and \(\mathbf{r} = (x, y, z)\) is a general point.
A plane holds two directions, so describe it by the one direction it does not hold: a normal \(\mathbf{n}\), perpendicular to the plane. \(R\) lies in the plane through \(A\) exactly when \(R - A\) is perpendicular to \(\mathbf{n}\), and the dot product tests that:
\[\mathbf{n}\cdot(\mathbf{r} - \mathbf{a}) = 0 \quad\Longleftrightarrow\quad n_1 x + n_2 y + n_3 z = d,\]
with \(d = \mathbf{n}\cdot\mathbf{a}\). So \(ax + by + cz = d\) is a plane with normal \((a, b, c)\).
From two directions \(\mathbf{u}\), \(\mathbf{v}\) in the plane, the cross product builds a normal:
\[\mathbf{u}\times\mathbf{v} = (u_2 v_3 - u_3 v_2,\; u_3 v_1 - u_1 v_3,\; u_1 v_2 - u_2 v_1).\]
Dot it with \(\mathbf{u}\) and the six terms cancel in pairs (exercise 2), so it is perpendicular to \(\mathbf{u}\), and likewise to \(\mathbf{v}\).
Take any point \(A\) in the plane. The distance from \(P\) to the plane is the shadow of \(P - A\) on \(\mathbf{n}\):
\(\text{distance} = \dfrac{\lvert\mathbf{n}\cdot(\mathbf{p} - \mathbf{a})\rvert}{\lvert\mathbf{n}\rvert} = \dfrac{\lvert n_1 p_1 + n_2 p_2 + n_3 p_3 - d\rvert}{\lvert\mathbf{n}\rvert}.\)
Watch one: the plane through \((1,0,0)\), \((0,2,0)\), \((0,0,3)\)
Call the points \(A\), \(B\), \(C\). Two directions in the plane are \(\overrightarrow{AB} = (-1, 2, 0)\) and \(\overrightarrow{AC} = (-1, 0, 3)\). Cross them:
\[\mathbf{n} = (2\cdot 3 - 0\cdot 0,\; 0\cdot(-1) - (-1)\cdot 3,\; (-1)\cdot 0 - 2\cdot(-1)) = (6, 3, 2).\]
Then \(d = \mathbf{n}\cdot\mathbf{a} = 6\), so the plane is \(6x + 3y + 2z = 6\). The normal has length \(\sqrt{36 + 9 + 4} = 7\), so the origin is \(\lvert 0 - 6\rvert/7 = 6/7\) from the plane and \(P = (2, 2, 2)\) is \(\lvert 12 + 6 + 4 - 6\rvert/7 = 16/7\) away.
Your turn
- A line passes through \((1, 2, 3)\) in the direction \((2, -1, 1)\). Write it in coordinates. Which of \((5, 0, 5)\) and \((3, 1, 5)\) lie on it?
- Let \(\mathbf{u} = (1, 2, 3)\) and \(\mathbf{v} = (4, 5, 6)\). (a) Compute \(\mathbf{u}\times\mathbf{v}\) and check its dot product with each of \(\mathbf{u}\), \(\mathbf{v}\) is zero. (b) Show \(\mathbf{u}\cdot(\mathbf{u}\times\mathbf{v}) = 0\) for every \(\mathbf{u}\), \(\mathbf{v}\).
- Find the plane through \((1, 1, 1)\), \((3, 2, 1)\) and \((1, 2, 3)\), and the distance from \((4, 0, 1)\) to it.
- (a) Why are \(2x - y + 2z = 3\) and \(2x - y + 2z = 9\) parallel? Find the distance between them. (b) Find the point of the plane \(x + 2y + 2z = 9\) closest to the origin. In which direction do you walk from the origin to reach it?
Stretch: two lines that miss
\(L_1\): \(\mathbf{r} = (1, 2, 0) + t(1, 0, 1)\). \(L_2\): \(\mathbf{r} = (0, 0, 3) + s(0, 1, 1)\). Show they neither meet nor run parallel. The shortest segment joining them is perpendicular to both, so it runs along \(\mathbf{d}_1\times\mathbf{d}_2\). Find the distance between the lines.
Answers
- \(\mathbf{r} = (1, 2, 3) + t(2, -1, 1)\), that is \(x = 1 + 2t\), \(y = 2 - t\), \(z = 3 + t\). \((5, 0, 5)\) is on it: \(t = 2\) fits all three. \((3, 1, 5)\) is not: \(x\) and \(y\) need \(t = 1\), which gives \(z = 4\).
- (a) \(\mathbf{u}\times\mathbf{v} = (12 - 15,\; 12 - 6,\; 5 - 8) = (-3, 6, -3)\). Dots: \(-3 + 12 - 9 = 0\) and \(-12 + 30 - 18 = 0\). (b) \(u_1(u_2 v_3 - u_3 v_2) + u_2(u_3 v_1 - u_1 v_3) + u_3(u_1 v_2 - u_2 v_1)\): the six terms cancel in pairs, for example \(u_1 u_2 v_3\) and \(-u_2 u_1 v_3\).
- Directions \((2, 1, 0)\) and \((0, 1, 2)\); cross product \((2, -4, 2)\), or \((1, -2, 1)\) after dividing by 2. Through \((1, 1, 1)\): \(d = 1 - 2 + 1 = 0\), so \(x - 2y + z = 0\). Distance from \((4, 0, 1)\): \(\lvert 4 - 0 + 1\rvert/\sqrt6 = 5/\sqrt6 \approx 2.04\).
- (a) Same normal \((2, -1, 2)\), so same tilt. Pick a point on the first, say \((0, -3, 0)\), and use the formula with the second: \(\lvert 0 + 3 + 0 - 9\rvert/\sqrt9 = 2\). (b) Walk along the normal: the point is \(t(1, 2, 2)\) with \(t + 4t + 4t = 9\), so \(t = 1\) and the point is \((1, 2, 2)\), at distance \(3\). The formula agrees: \(9/\sqrt9 = 3\).
Stretch. Meeting would need \(1 + t = 0\), \(2 = s\), \(t = 3 + s\): the first two force \(t = -1\), \(s = 2\), and the third fails. The directions are not multiples of each other, so the lines are skew. \(\mathbf{n} = (1, 0, 1)\times(0, 1, 1) = (-1, -1, 1)\). The vector from one line to the other, \((0, 0, 3) - (1, 2, 0) = (-1, -2, 3)\), has shadow on \(\mathbf{n}\) of length \(\lvert 1 + 2 + 3\rvert/\sqrt3 = 6/\sqrt3 = 2\sqrt3 \approx 3.46\). That is the distance: it is the plane formula, applied to a point of \(L_2\) and the plane through \(L_1\) parallel to \(L_2\).