Shortest Distance Between Lines, with Mixed Practice — Worked Solutions

Ques­tion 1: Direc­tion cosines of a com­mon per­pen­dic­u­lar#

The prob­lem#

Find the direc­tion cosines of a line per­pen­dic­u­lar to the lines with direc­tion ratios $1, 2, 3$ and $-2, 1, 4$.

Under­stand­ing the prob­lem#

You need a direc­tion that is at right angles to both given direc­tions, and then its direc­tion cosines $l, m, n$ (which sat­isfy $l^2 + m^2 + n^2 = 1$).

The idea#

The cross prod­uct of two vec­tors is per­pen­dic­u­lar to both. So take $(1, 2, 3) \times (-2, 1, 4)$, sim­plify the ratios, and divide by the length.

Step-by-step solu­tion#

Step 1. Com­pute the cross prod­uct using the deter­mi­nant pat­tern.

$$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 3 \\ -2 & 1 & 4 \end{vmatrix} = (8 - 3)\hat i - (4 + 6)\hat j + (1 + 4)\hat k = 5\hat i - 10\hat j + 5\hat k.$$

Step 2. Divide by the com­mon fac­tor $5$ to get sim­pler direc­tion ratios.

$$1,\ -2,\ 1.$$

Step 3. Find the length of $(1, -2, 1)$.

$$\sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}.$$

Step 4. Divide each ratio by $\sqrt{6}$.

$$\displaystyle l = \frac{1}{\sqrt{6}}, \quad m = -\frac{2}{\sqrt{6}}, \quad n = \frac{1}{\sqrt{6}}.$$

Check­ing the answer#

Per­pen­dic­u­lar­ity: $(1)(1) + (2)(-2) + (3)(1) = 0$ ✓ and $(-2)(1) + (1)(-2) + (4)(1) = 0$ ✓. Also $\displaystyle \tfrac{1 + 4 + 1}{6} = 1$ ✓.

Answer#

$\displaystyle \frac{1}{\sqrt{6}},\ -\frac{2}{\sqrt{6}},\ \frac{1}{\sqrt{6}}$ (or all three with the oppo­site sign).

Ques­tion 2: A line per­pen­dic­u­lar to two given lines#

The prob­lem#

Find the equa­tion of the line through $(1, 2, -4)$ per­pen­dic­u­lar to both $\displaystyle \tfrac{x - 8}{3} = \tfrac{y + 19}{-16} = \tfrac{z - 10}{7}$ and $\displaystyle \tfrac{x - 15}{3} = \tfrac{y - 29}{8} = \tfrac{z - 5}{-5}$.

Under­stand­ing the prob­lem#

You know a point on the required line. You need its direc­tion, which must be per­pen­dic­u­lar to the direc­tions of both given lines.

The idea#

Read the direc­tion ratios from the denom­i­na­tors: $(3, -16, 7)$ and $(3, 8, -5)$. Their cross prod­uct gives the required direc­tion.

Step-by-step solu­tion#

Step 1. Cross prod­uct.

$$\begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = (80 - 56)\hat i - (-15 - 21)\hat j + (24 + 48)\hat k = 24\hat i + 36\hat j + 72\hat k.$$

Step 2. Divide by $12$.

$$\text{direction ratios } 2,\ 3,\ 6.$$

Step 3. Write the line through $(1, 2, -4)$.

$$\displaystyle \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}.$$

Check­ing the answer#

$3(2) - 16(3) + 7(6) = 6 - 48 + 42 = 0$ ✓; $3(2) + 8(3) - 5(6) = 6 + 24 - 30 = 0$ ✓.

Answer#

$\displaystyle \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$

Com­mon mis­take to avoid#

The point $(1, 2, -4)$ gives $z - (-4) = z + 4$ in the third frac­tion, not $z - 4$.

Ques­tion 3: Angle between two lines#

The prob­lem#

Find the angle between the lines $\displaystyle \tfrac{x}{2} = \tfrac{y}{2} = \tfrac{z}{1}$ and $\displaystyle \tfrac{x - 5}{4} = \tfrac{y - 2}{1} = \tfrac{z - 3}{8}$.

Under­stand­ing the prob­lem#

The angle between two lines is the angle between their direc­tions. The points on the lines do not mat­ter.

The idea#

Use $\displaystyle \cos\theta = \frac{\lvert a_1a_2 + b_1b_2 + c_1c_2 \rvert}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$ with direc­tions $(2, 2, 1)$ and $(4, 1, 8)$.

Step-by-step solu­tion#

Step 1. Dot prod­uct of the direc­tions.

$$2(4) + 2(1) + 1(8) = 8 + 2 + 8 = 18.$$

Step 2. Lengths.

$$\sqrt{4 + 4 + 1} = 3, \qquad \sqrt{16 + 1 + 64} = 9.$$

Step 3. Cosine of the angle.

$$\displaystyle \cos\theta = \frac{18}{3 \times 9} = \frac{2}{3}.$$

Step 4. So

$$\displaystyle \theta = \cos^{-1}\frac{2}{3}.$$

Check­ing the answer#

$\displaystyle \tfrac{2}{3}$ lies between $0$ and $1$, so the angle is acute (about $48^\circ$), which is rea­son­able since the dot prod­uct is pos­i­tive.

Answer#

$\displaystyle \theta = \cos^{-1}\left(\frac{2}{3}\right)$

Ques­tion 4: Short­est dis­tance between skew lines (vec­tor form)#

The prob­lem#

Find the short­est dis­tance between $\vec r = \hat i + 2\hat j + 3\hat k + \lambda(\hat i - 3\hat j + 2\hat k)$ and $\vec r = 4\hat i + 5\hat j + 6\hat k + \mu(2\hat i + 3\hat j + \hat k)$.

Under­stand­ing the prob­lem#

The direc­tions $(1, -3, 2)$ and $(2, 3, 1)$ are not pro­por­tional, so the lines are not par­al­lel. Use the skew-lines for­mula.

The idea#

Read off $\vec a_1 = (1, 2, 3)$, $\vec b_1 = (1, -3, 2)$, $\vec a_2 = (4, 5, 6)$, $\vec b_2 = (2, 3, 1)$. Com­pute $\vec b_1 \times \vec b_2$, its length, and its dot prod­uct with $\vec a_2 - \vec a_1$.

Step-by-step solu­tion#

Step 1. Cross prod­uct of the direc­tions.

$$\vec b_1 \times \vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = (-3 - 6)\hat i - (1 - 4)\hat j + (3 + 6)\hat k = -9\hat i + 3\hat j + 9\hat k.$$

Step 2. Its length.

$$\lvert\vec b_1 \times \vec b_2\rvert = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}.$$

Step 3. Vec­tor join­ing the given points.

$$\vec a_2 - \vec a_1 = 3\hat i + 3\hat j + 3\hat k.$$

Step 4. Dot prod­uct.

$$(-9)(3) + (3)(3) + (9)(3) = -27 + 9 + 27 = 9.$$

Step 5. Dis­tance.

$$\displaystyle d = \frac{\lvert 9 \rvert}{3\sqrt{19}} = \frac{3}{\sqrt{19}}.$$

Check­ing the answer#

$d \ne 0$, so the lines are gen­uinely skew (they do not meet). $\displaystyle \tfrac{3}{\sqrt{19}} \approx 0.69$ units.

Answer#

$\displaystyle d = \frac{3}{\sqrt{19}}$ units

Ques­tion 5: Short­est dis­tance between skew lines (Carte­sian form)#

The prob­lem#

Find the short­est dis­tance between $\displaystyle \tfrac{x + 1}{7} = \tfrac{y + 1}{-6} = \tfrac{z + 1}{1}$ and $\displaystyle \tfrac{x - 3}{1} = \tfrac{y - 5}{-2} = \tfrac{z - 7}{1}$.

Under­stand­ing the prob­lem#

First con­vert each line to a point and a direc­tion. Care­ful with signs: $x + 1 = x - (-1)$, so the first point is $(-1, -1, -1)$.

The idea#

$\vec a_1 = (-1, -1, -1)$, $\vec b_1 = (7, -6, 1)$; $\vec a_2 = (3, 5, 7)$, $\vec b_2 = (1, -2, 1)$. Apply the skew-lines for­mula.

Step-by-step solu­tion#

Step 1. Cross prod­uct.

$$\vec b_1 \times \vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = (-6 + 2)\hat i - (7 - 1)\hat j + (-14 + 6)\hat k = -4\hat i - 6\hat j - 8\hat k.$$

Step 2. Length.

$$\sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}.$$

Step 3. Join­ing vec­tor.

$$\vec a_2 - \vec a_1 = (3 + 1)\hat i + (5 + 1)\hat j + (7 + 1)\hat k = 4\hat i + 6\hat j + 8\hat k.$$

Step 4. Dot prod­uct.

$$(-4)(4) + (-6)(6) + (-8)(8) = -16 - 36 - 64 = -116.$$

Step 5. Dis­tance (take the absolute value).

$$\displaystyle d = \frac{\lvert -116 \rvert}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}.$$

(Since $58 = 2 \times 29$, $\displaystyle \tfrac{58}{\sqrt{29}} = 2\sqrt{29}$.)

Check­ing the answer#

Here $\vec a_2 - \vec a_1$ hap­pens to be par­al­lel to $\vec b_1 \times \vec b_2$ (it is exactly its neg­a­tive), so the whole join­ing seg­ment lies along the com­mon per­pen­dic­u­lar, and $d$ equals its full length $\sqrt{116} = 2\sqrt{29}$ ✓.

Answer#

$d = 2\sqrt{29}$ units

Ques­tion 6: Show­ing two lines inter­sect#

The prob­lem#

Show that the lines $\displaystyle \tfrac{x - 1}{2} = \tfrac{y - 2}{3} = \tfrac{z - 3}{4}$ and $\displaystyle \tfrac{x - 4}{5} = \tfrac{y - 1}{2} = z$ inter­sect.

Under­stand­ing the prob­lem#

Two non-par­al­lel lines in space inter­sect exactly when the short­est dis­tance between them is $0$. You can also find the actual meet­ing point.

The idea#

Write a gen­eral point on each line using para­me­ters $\lambda$ and $\mu$, set the coor­di­nates equal, solve two equa­tions, and check the third. (This is equiv­a­lent to $d = 0$.)

Step-by-step solu­tion#

Step 1. Gen­eral point on the first line (each frac­tion $= \lambda$).

$$(2\lambda + 1,\ 3\lambda + 2,\ 4\lambda + 3).$$

Step 2. Gen­eral point on the sec­ond line. Note $\displaystyle z = \tfrac{z - 0}{1}$, so each frac­tion $= \mu$ gives

$$(5\mu + 4,\ 2\mu + 1,\ \mu).$$

Step 3. Equate the $x$- and $y$-coor­di­nates.

$$2\lambda + 1 = 5\mu + 4 \;\Rightarrow\; 2\lambda - 5\mu = 3, \qquad 3\lambda + 2 = 2\mu + 1 \;\Rightarrow\; 3\lambda - 2\mu = -1.$$

Step 4. Solve. Mul­ti­ply the first by $3$ and the sec­ond by $2$, then sub­tract.

$$\begin{aligned} 6\lambda - 15\mu &= 9\\ 6\lambda - 4\mu &= -2\\ \text{subtract: } -11\mu &= 11 \;\Rightarrow\; \mu = -1. \end{aligned}$$

Then $2\lambda - 5(-1) = 3$ gives $2\lambda = -2$, $\lambda = -1$.

Step 5. Check the $z$-coor­di­nates with these val­ues.

$$4\lambda + 3 = -4 + 3 = -1, \qquad \mu = -1.$$

They agree, so the lines share a point.

Step 6. The com­mon point is $(2(-1) + 1,\ 3(-1) + 2,\ -1) = (-1, -1, -1)$.

Check­ing the answer#

$(-1, -1, -1)$ in the first line: $\displaystyle \tfrac{-2}{2} = \tfrac{-3}{3} = \tfrac{-4}{4} = -1$ ✓. In the sec­ond: $\displaystyle \tfrac{-5}{5} = \tfrac{-2}{2} = -1$ ✓. Equiv­a­lently, the short­est-dis­tance for­mula gives $0$.

Answer#

The lines meet at $(-1, -1, -1)$ (their short­est dis­tance is $0$), so they inter­sect.

Ques­tion 7: Dis­tance between par­al­lel lines#

The prob­lem#

Find the dis­tance between the par­al­lel lines $\vec r = \lambda(\hat i + 2\hat j + 2\hat k)$ and $\vec r = 3\hat i + \mu(\hat i + 2\hat j + 2\hat k)$.

Under­stand­ing the prob­lem#

Both lines have direc­tion $\vec b = \hat i + 2\hat j + 2\hat k$, so they are par­al­lel. The first passes through the ori­gin, $\vec a_1 = \vec 0$; the sec­ond through $\vec a_2 = 3\hat i$.

The idea#

For par­al­lel lines, $\vec b_1 \times \vec b_2 = \vec 0$, so the skew for­mula fails. Use $\displaystyle d = \frac{\lvert\vec b \times (\vec a_2 - \vec a_1)\rvert}{\lvert\vec b\rvert}$.

Step-by-step solu­tion#

Step 1. Join­ing vec­tor.

$$\vec a_2 - \vec a_1 = 3\hat i.$$

Step 2. Cross prod­uct.

$$\vec b \times 3\hat i = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 2 \\ 3 & 0 & 0 \end{vmatrix} = (0 - 0)\hat i - (0 - 6)\hat j + (0 - 6)\hat k = 6\hat j - 6\hat k.$$

Step 3. Lengths.

$$\lvert 6\hat j - 6\hat k \rvert = \sqrt{36 + 36} = 6\sqrt{2}, \qquad \lvert\vec b\rvert = \sqrt{1 + 4 + 4} = 3.$$

Step 4. Dis­tance.

$$\displaystyle d = \frac{6\sqrt{2}}{3} = 2\sqrt{2}.$$

Check­ing the answer#

$d$ must be less than the dis­tance $3$ between the two cho­sen points, since those points are not directly oppo­site each other: $2\sqrt{2} \approx 2.83 < 3$ ✓.

Answer#

$d = 2\sqrt{2}$ units

Ques­tion 8: Foot and length of a per­pen­dic­u­lar to a line#

The prob­lem#

Find the foot of the per­pen­dic­u­lar from $(1, 6, 3)$ to the line $\displaystyle \tfrac{x}{1} = \tfrac{y - 1}{2} = \tfrac{z - 2}{3}$ and the length of the per­pen­dic­u­lar.

Under­stand­ing the prob­lem#

The foot $F$ is the point on the line clos­est to $P(1, 6, 3)$. At $F$, the seg­ment $PF$ is per­pen­dic­u­lar to the line's direc­tion $(1, 2, 3)$.

The idea#

Write a gen­eral point $F$ of the line in terms of $\lambda$. Make $\overrightarrow{PF}$ per­pen­dic­u­lar to $(1, 2, 3)$ (dot prod­uct $0$), solve for $\lambda$, then find $F$ and $\lvert PF \rvert$.

Step-by-step solu­tion#

Step 1. Gen­eral point on the line (each frac­tion $= \lambda$).

$$F = (\lambda,\ 2\lambda + 1,\ 3\lambda + 2).$$

Step 2. Vec­tor from $P$ to $F$.

$$\overrightarrow{PF} = (\lambda - 1,\ 2\lambda + 1 - 6,\ 3\lambda + 2 - 3) = (\lambda - 1,\ 2\lambda - 5,\ 3\lambda - 1).$$

Step 3. Per­pen­dic­u­lar to $(1, 2, 3)$.

$$\begin{aligned} 1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) &= 0\\ \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 &= 0\\ 14\lambda - 14 &= 0\\ \lambda &= 1. \end{aligned}$$

Step 4. The foot.

$$F = (1,\ 3,\ 5).$$

Step 5. Length of the per­pen­dic­u­lar.

$$PF = \sqrt{(1 - 1)^2 + (3 - 6)^2 + (5 - 3)^2} = \sqrt{0 + 9 + 4} = \sqrt{13}.$$

Check­ing the answer#

$F$ is on the line: $\displaystyle \tfrac{1}{1} = \tfrac{3 - 1}{2} = \tfrac{5 - 2}{3} = 1$ ✓. $\overrightarrow{PF} = (0, -3, 2)$ and $(0)(1) + (-3)(2) + (2)(3) = 0$ ✓.

Answer#

Foot $(1, 3, 5)$; length of the per­pen­dic­u­lar $\sqrt{13}$ units.