Lines & Planes in Space

Why: if $P_0$ lies on the plane and $P$ is any other plane point, the displacement $P - P_0$ lies in the plane, and the equation says exactly $\mathbf n\cdot(P - P_0) = 0$ — the normal is perpendicular to every in-plane direction. So building planes is mechanical: given a point and a normal, dot them; given three points, form two edge vectors and take their CROSS PRODUCT for the normal. Given the equation, read the normal straight off the coefficients. Parallel planes share a normal and differ only in $d$; perpendicular planes have perpendicular normals — every geometric relationship between planes translates into a dot- or cross-product statement about their normals.

The numerator is the (signed) un-normalized projection; dividing by $|\mathbf n|$ finishes it. The same idea handles parallel planes ($\tfrac{|d_1 - d_2|}{|\mathbf n|}$) and, with a cross product replacing the dot, point-to-line distances. One projection, every distance formula in the chapter.

Why a parameter: a line has one degree of freedom, and $t$ is that freedom. Each value of $t$ picks one point, and equal steps in $t$ are equal steps along the line. Skew lines are the case with no counterpart in the plane: there, two lines that are not parallel must cross, but in space one line can pass over another like two roads at an overpass. To decide which case holds, set the two parametrizations equal. That gives three equations in only two unknowns, $t$ and $s$, and the question is whether all three can hold at once.

As a check, $t = 0$ gives $P$ and $t = 1$ gives $Q$. The $xy$-plane is where $z = 0$, so solve $3 - t = 0$: $t = 3$. The crossing point is $(1 + 9,\ -2 + 6,\ 0) = (10, 4, 0)$. Any line question reduces to choosing $t$: first write the condition on the point, then solve for $t$.

The first equation gives $t = 1$, and the second then gives $s = 0$. The third equation has not been used yet, so it is the real test: $2(1) = 2 - 0$ holds. All three equations are satisfied, so the lines meet, at $t = 1$ on $L_1$: the point $(2, 0, 2)$, which is also $s = 0$ on $L_2$.

Every segment joining the two lines has the same component along $\mathbf n$: moving either endpoint along its line adds a multiple of $(1, 2, 0)$ or $(1, 0, 1)$, and both are perpendicular to $\mathbf n$. So take the easiest segment, from $(1, 0, 2)$ to $(0, 1, -1)$: it is $(-1, 1, -3)$, and $(-1, 1, -3)\cdot(2, -1, -2) = -2 - 1 + 6 = 3$. The distance is $\tfrac{|3|}{3} = 1$. It is not zero, and the directions are not proportional, so the lines are skew. The closest points are $t = \tfrac23$ on $L_1$ and $s = \tfrac73$ on $L_2$, namely $\left(\tfrac53, \tfrac43, 2\right)$ and $\left(\tfrac73, 1, \tfrac43\right)$.

Why normals: a plane's tilt is completely described by its normal, so a question about how two planes sit is a question about two vectors. Substituting a line into a plane works because the plane's equation is linear and each coordinate of $\mathbf r(t)$ is linear in $t$, so $\mathbf n\cdot\mathbf r(t) = d$ is a linear equation in $t$. It has exactly one solution unless the coefficient of $t$, which is $\mathbf n\cdot\mathbf v$, is zero; that happens exactly when the line runs perpendicular to the normal, that is, parallel to the plane.

The point is $\mathbf r(2) = (3, 0, 6)$; check $3 + 0 + 6 = 9$. The angle $\alpha$ between the line and the plane uses a sine, $\sin\alpha = \dfrac{|\mathbf n\cdot\mathbf v|}{|\mathbf n||\mathbf v|} = \dfrac{|1 - 1 + 3|}{\sqrt3\,\sqrt{11}} = \dfrac{3}{\sqrt{33}} \approx 0.522$, so $\alpha \approx 31.5^\circ$. It is a sine because the angle between $\mathbf v$ and the normal is the complement of the angle between $\mathbf v$ and the plane.

So $\theta = 60^\circ$. The two planes meet in a line, and its direction must be perpendicular to both normals, so it is their cross product: $(1, 1, 0)\times(0, 1, 1) = (1\cdot1 - 0\cdot1,\ 0\cdot0 - 1\cdot1,\ 1\cdot1 - 1\cdot0) = (1, -1, 1)$. Normals answer both questions: the dot product gives the angle, the cross product gives the hinge.

The distance is $\tfrac{|-4|}{3} = \tfrac43$; the negative sign only says that $P$ lies on the side the normal points away from. To find the nearest point of the plane, walk from $P$ along the normal: $F = P - \dfrac{\mathbf n\cdot P - d}{|\mathbf n|^2}\,\mathbf n = (3, 1, -2) + \tfrac49(2, -1, 2) = \left(\tfrac{35}{9}, \tfrac59, -\tfrac{10}{9}\right)$. It lies on the plane, since $\tfrac{70 - 5 - 20}{9} = 5$, and $|F - P| = \tfrac49\cdot3 = \tfrac43$, as the formula said.

Step 1 — the direction of the line is perpendicular to BOTH normals, so take the cross product $(1,1,1)\times(1,-1,2) = (1\cdot2 - 1\cdot(-1),\; 1\cdot1 - 1\cdot2,\; 1\cdot(-1) - 1\cdot1) = (3, -1, -2)$.

Step 2 — find one point on both planes: set $z = 0$ and solve $x + y = 6$, $x - y = 3$: $x = 4.5$, $y = 1.5$.

Step 3 — assemble: $\mathbf r(t) = (4.5, 1.5, 0) + t(3, -1, -2)$.

Step 4 — verify at $t = 1$: the point $(7.5, 0.5, -2)$ gives $7.5 + 0.5 - 2 = 6$ ✓ and $7.5 - 0.5 - 4 = 3$ ✓. Two planes, one cross product, one small system.

Step 1 — displacement $\mathbf w = P - A = (1, 2, 2)$.

Step 2 — the perpendicular distance is $\dfrac{|\mathbf w\times\mathbf v|}{|\mathbf v|}$ (the parallelogram area divided by its base). Compute $\mathbf w\times\mathbf v = (2\cdot0 - 2\cdot1,\; 2\cdot1 - 1\cdot0,\; 1\cdot1 - 2\cdot1) = (-2, 2, -1)$, magnitude $3$.

Step 3 — divide by $|\mathbf v| = \sqrt2$: distance $\dfrac{3}{\sqrt2} = \dfrac{3\sqrt2}{2}$.

Step 4 — cross-check by projection: the component of $\mathbf w$ along $\mathbf v$ is $\tfrac{3}{\sqrt2}$, and $|\mathbf w|^2 - \left(\tfrac{3}{\sqrt2}\right)^2 = 9 - 4.5 = 4.5 = \left(\tfrac{3\sqrt2}{2}\right)^2$ ✓ — Pythagoras agrees with the cross product.