2.5 Equations of Lines and Planes in Space

← Chapter 2 §2.5 Lines and Planes in Space 空间直线与平面 · Chapter 2 · Vectors in Space

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Objective	Takeaways; say them first, then click
2.5.1 Write the equations of a line.	$\mathbf r=\mathbf r_0+t\mathbf v$ ; parametric, symmetric; $\mathbf v=Q-P$ ; segment $0\le t\le 1$ ; skew lines
2.5.2 Find the distance from a point to a line.	area ÷ base: $\\|\overrightarrow{PM}\times\mathbf v\\|\,/\,\\|\mathbf v\\|$
2.5.3 Write the equations of a plane.	$\mathbf n\cdot\overrightarrow{PQ}=0$ ; $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
2.5.4 Find the distance from a point to a plane.	shadow on $\mathbf n$ : $\|ax_0+by_0+cz_0+k\|\,/\,\\|\mathbf n\\|$
2.5.5 Find the angle between two planes.	angle of the normals: $\cos\theta=\|\mathbf n_1\cdot\mathbf n_2\|/(\\|\mathbf n_1\\|\\|\mathbf n_2\\|)$

Mistake	Result	Fix
A point used as the direction	$\langle 1,4,-2\rangle$ in Example 2.45: a different line	direction = $Q-P$
Dividing by a zero component	$\frac{y-1}{0}$ in Exercise 249	write $y=1$ separately
One letter $t$ for both lines	forces them to be at the same place at the same "time"	$t$ on one line, $s$ on the other
"Not intersecting, so parallel"	Example 2.48a called parallel	compare directions first; different and apart is skew
Sign of the constant in Theorem 2.14	$x-2y+z=5$ read as $k=+5$ : $8/\sqrt6$	move it over: $k=-5$ , $2/\sqrt6$
Dropping the absolute value	$-2/\sqrt6$ , a negative distance	$\|\,\cdot\,\|$ in the numerator
$\cos\theta$ without $\|\mathbf n_1\cdot\mathbf n_2\|$	1.70 rad for Checkpoint 2.50	take the acute angle, 1.44 rad

#	task	answer
2.43	line through $(1,-3,2)$ , $(5,-2,8)$	$x=1+4t,\ y=-3+t,\ z=2+6t$ ; $\ \frac{x-1}{4}=y+3=\frac{z-2}{6}$
2.44	segment $(-1,3,6)$ to $(-8,2,4)$	$x=-1-7t,\ y=3-t,\ z=6-2t,\ 0\le t\le1$
2.45	$(0,3,6)$ to $x=1-t,\ y=1+2t,\ z=5+3t$	$\sqrt{10/7}\approx 1.20$
2.46	$x=1-4t,\ y=3+t,\ z=8-6t$ ; $x=2+3s,\ y=2s,\ z=-1-3s$	skew
2.47	plane containing $x=-y=z$ and $\frac{x-3}{2}=y=z-2$	$-2x+y+3z=0$
2.48	$(5,-1,0)$ to $4x+2y-z=3$	$15/\sqrt{21}\approx 3.27$
2.49	$x+y-z=3$ meets $3x-y+3z=5$	$x=t,\ y=7-3t,\ z=4-2t$
2.50	angle between those planes	1.44 rad
2.51	$5x-2y+z=6$ and $5x-2y+z=-3$	$9/\sqrt{30}\approx 1.64$

#	task	answer	key step
243	line through $(-3,5,9)$ , $(4,-7,2)$	$\frac{x+3}{7}=\frac{y-5}{-12}=\frac{z-9}{-7}$	$\mathbf v=\langle 7,-12,-7\rangle$
247	$(1,-2,3)$ , $\mathbf v=\langle 1,2,3\rangle$ : meets $z=0$ ?	$(0,-4,0)$	$3+3t=0$
249	$(3,1,5)$ , $\mathbf v=\langle 1,0,0\rangle$	$y=1,\ z=5$ ; never meets $z=0$	no $z$ -part
251	origin to $x=1+t,\ y=3+t,\ z=5+4t$	$\sqrt3$	$\\|\langle 7,1,-2\rangle\\|/\sqrt{18}$
253	$A(-3,1,1)$ to $x=-y=-z$	$2\sqrt2/\sqrt3$	Theorem 2.12
255	$L_1$ : $x=1+t,\ y=t,\ z=2+t$ ; $L_2$ : $x-3=y-1=z-3$	parallel, $\sqrt2/\sqrt3$ apart	both along $\langle 1,1,1\rangle$
259	$x=-2y=3z$ meets $x=-5-t,\ y=-1+t,\ z=t-11$	$(-12,6,-4)$	$r=-2,\ t=7$
261	$x=y-1=-z$ , $x-2=-y=\frac z2$	skew	$\mathbf w\cdot(\mathbf v_1\times\mathbf v_2)=5\neq0$
263	(p187)	equal	same direction, shared point $(1,4,7)$
269	$P(1,2,3)$ , $\mathbf n=\langle 1,2,3\rangle$	$x+2y+3z-14=0$	scalar form, expand
271	$4x+5y+10z-20=0$	$\mathbf n=4\mathbf i+5\mathbf j+10\mathbf k$ ; $(5,0,0),(0,4,0),(0,0,2)$	two coordinates zero
277	through $(-2,1,3)$ , $\perp\ 2x-3y+z=7$	$x=-2+2t,\ y=1-3t,\ z=3+t$	direction = normal
281	$(1,1,1),(2,4,3),(-1,-2,-1)$	$-2y+3z-1=0$	$\overrightarrow{PQ}\times\overrightarrow{PR}=\langle 0,-2,3\rangle$
287	$x=1+2t,\ y=-2t,\ z=2+t$ and $3x+4y+6z-7=0$ (p189)	$(-3,4,0)$	$8+4t=0$
289	$(1,5,-4)$ to $3x-y+2z-6=0$	$16/\sqrt{14}$	Theorem 2.14
291	$x+y+z=0$ , $2x-y+z-7=0$	neither; 62°	$\cos\theta=2/\sqrt{18}$
295	skew lines (p190)	$1/\sqrt6$	$\mathbf N=\langle -2,1,1\rangle$
297	sphere at $(-3,2,4)$ tangent to $2x+4y-3z=8$	$r=18/\sqrt{29}$ ; $\big(-\tfrac{51}{29},\tfrac{130}{29},\tfrac{62}{29}\big)$	point-plane lab, preset Exercise 297

Equations of Lines and Planes in Space

A line is one point plus every multiple of a direction: ${\mathbf r=\mathbf r_0+t\mathbf v}$

Parallel vectors are scalar multiples of each other

Read off components for parametric equations; solve each for $t$ for symmetric ones

Two points give the direction: ${\mathbf v=\overrightarrow{PQ}}$

A segment is the line with the clock stopped at $t=0$ and $t=1$

Walk a line: every $t$ is one point, and $0\le t\le 1$ is the segment

In space, lines with different directions need not meet: skew lines

Directions first; if not parallel, solve for a common point with two different parameters

A skew pair is an intersecting pair pushed apart along $\mathbf v_1\times\mathbf v_2$

Distance to a line = parallelogram area ÷ base

Read $P$ and $\mathbf v$ off the symmetric form, cross, divide

The distance does not depend on which point of the line you pick

A plane is every point $Q$ with ${\overrightarrow{PQ}\perp\mathbf n}$

Expand $\mathbf n\cdot\overrightarrow{PQ}=0$ : the coefficients of $x,y,z$ are the normal

Three points: cross two arrows in the plane to get the normal

A point and a line: the line gives one arrow in the plane, the point the other

Distance to a plane = the shadow of $\overrightarrow{QP}$ on $\mathbf n$

Pick any point in the plane, project onto $\mathbf n$ : ${d=2/\sqrt6}$

Plug the point into the plane's equation, then divide by $\|\mathbf n\|$

The gap between parallel planes is a point-to-plane distance

Only the part of $\overrightarrow{QP}$ along $\mathbf n$ is the distance

Theorem 2.14 is the projection formula with the arithmetic done

The gap between skew lines is the shadow of a connecting arrow on $\mathbf v_1\times\mathbf v_2$

Two distinct planes are either parallel or meet in a line

Line of intersection: solve the two equations together, one variable free

The angle between two planes is the angle between their normals

Classify two planes by their normals: multiples, zero dot product, or compute the angle

Turn plane 2: the angle follows the normals, the line follows $\mathbf n_1\times\mathbf n_2$

Normals measure the angle because each is square to its own plane

A line needs a point and a direction; a plane needs a point and a normal

Common mistakes

Practice and answers