Lines in Space and the Angle Between Them — Worked Solutions

Ques­tion 1: Direc­tion cosines from direc­tion ratios#

The prob­lem#

Find the direc­tion cosines of the line whose direc­tion ratios are $6, -2, 3$.

Under­stand­ing the prob­lem#

Direc­tion ratios $a, b, c$ only tell us the direc­tion up to a scale. Direc­tion cosines $l, m, n$ are the par­tic­u­lar mul­ti­ple for which $l^2 + m^2 + n^2 = 1$.

The idea#

Divide each ratio by $\sqrt{a^2 + b^2 + c^2}$:

$$\displaystyle l = \frac{a}{\sqrt{a^2 + b^2 + c^2}},\quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}},\quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}.$$

Step-by-step solu­tion#

Step 1. Com­pute the length.

$$\sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7.$$

Step 2. Divide each ratio by $7$.

$$\displaystyle l = \frac67,\quad m = -\frac27,\quad n = \frac37.$$

Check­ing the answer#

$\displaystyle \tfrac{36}{49} + \tfrac{4}{49} + \tfrac{9}{49} = \tfrac{49}{49} = 1$ ✓.

Answer#

$\displaystyle \frac67, -\frac27, \frac37$ (or all signs reversed, $\displaystyle -\tfrac67, \tfrac27, -\tfrac37$, for the oppo­site sense of the line).

Ques­tion 2: Direc­tion cosines of a line through two points#

The prob­lem#

Find the direc­tion cosines of the line through $(-1, 2, 5)$ and $(3, 0, 1)$.

Under­stand­ing the prob­lem#

We are given two points, not ratios. First get direc­tion ratios from the points, then con­vert them to cosines.

The idea#

The line through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ has direc­tion ratios $x_2 - x_1$, $y_2 - y_1$, $z_2 - z_1$.

Step-by-step solu­tion#

Step 1. Direc­tion ratios from $P(-1, 2, 5)$ to $Q(3, 0, 1)$:

$$3 - (-1) = 4,\qquad 0 - 2 = -2,\qquad 1 - 5 = -4.$$

Step 2. Length: $\sqrt{16 + 4 + 16} = \sqrt{36} = 6$.

Step 3. Divide by $6$.

$$\displaystyle l = \frac46 = \frac23,\quad m = -\frac26 = -\frac13,\quad n = -\frac46 = -\frac23.$$

Check­ing the answer#

$\displaystyle \tfrac49 + \tfrac19 + \tfrac49 = 1$ ✓. The dis­tance $PQ$ is $6$, which is exactly the length we divided by.

Answer#

$\displaystyle \frac23, -\frac13, -\frac23$ (or the neg­a­tives of all three).

Ques­tion 3: Prov­ing three points collinear#

The prob­lem#

Show that the points $(1, -1, 2)$, $(3, 3, 0)$ and $(-2, -7, 5)$ are collinear.

Under­stand­ing the prob­lem#

Collinear means the three points lie on one straight line. We must prove it.

The idea#

As in Exam­ple 2: find the direc­tion ratios of $AB$ and $BC$. If they are pro­por­tional, the two seg­ments are par­al­lel; since they share $B$, they lie on the same line.

Step-by-step solu­tion#

Step 1. Name the points $A(1, -1, 2)$, $B(3, 3, 0)$, $C(-2, -7, 5)$.

Step 2. Direc­tion ratios of $AB$:

$$3 - 1,\ 3 - (-1),\ 0 - 2 = 2,\ 4,\ -2.$$

Step 3. Direc­tion ratios of $BC$:

$$-2 - 3,\ -7 - 3,\ 5 - 0 = -5,\ -10,\ 5.$$

Step 4. Com­pare the ratios term by term.

$$\displaystyle \frac{-5}{2} = \frac{-10}{4} = \frac{5}{-2} = -\frac52.$$

All three ratios are equal, so the ratios are pro­por­tional and $AB \parallel BC$.

Step 5. $AB$ and $BC$ are par­al­lel and share the point $B$, so $A$, $B$, $C$ lie on one line. $\blacksquare$

Check­ing the answer#

The line through $A$ with ratios $1, 2, -1$ is $(1 + t, -1 + 2t, 2 - t)$. $t = 2$ gives $B(3, 3, 0)$ and $t = -3$ gives $C(-2, -7, 5)$ ✓.

Answer#

The direc­tion ratios of $AB$ ($2, 4, -2$) and $BC$ ($-5, -10, 5$) are pro­por­tional, so the points are collinear.

Ques­tion 4: A line equally inclined to the axes#

The prob­lem#

A line makes equal angles with all three coor­di­nate axes. Find its direc­tion cosines.

Under­stand­ing the prob­lem#

If the angles $\alpha, \beta, \gamma$ with the axes are equal, their cosines are equal too: $l = m = n$.

The idea#

Use $l^2 + m^2 + n^2 = 1$ with $l = m = n$.

Step-by-step solu­tion#

Step 1. Put $l = m = n$.

$$\displaystyle l^2 + l^2 + l^2 = 1 \;\Longrightarrow\; 3l^2 = 1 \;\Longrightarrow\; l^2 = \frac13.$$

Step 2. Take the square root.

$$\displaystyle l = \pm\frac{1}{\sqrt3}.$$

Step 3. Since all three are equal, they share one sign.

$$\displaystyle (l, m, n) = \left(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right)\ \text{or}\ \left(-\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}\right).$$

Check­ing the answer#

$\displaystyle 3 \times \tfrac13 = 1$ ✓. The line with ratios $1, 1, 1$ (such as the line through the ori­gin and $(1, 1, 1)$) is the obvi­ous exam­ple.

Answer#

$\displaystyle l = m = n = \pm\frac{1}{\sqrt3}$ (all with the same sign).

Com­mon mis­take to avoid#

Mix­ing signs such as $\displaystyle \tfrac{1}{\sqrt3}, -\tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt3}$. Then the cosines dif­fer in sign, so the angles with the pos­i­tive axes are not all equal (one is acute, another obtuse).

Ques­tion 5: Vec­tor and Carte­sian equa­tions of a line#

The prob­lem#

Find the vec­tor and Carte­sian equa­tions of the line through $(2, 0, -3)$ par­al­lel to $\hat i - 4\hat j + 5\hat k$.

Under­stand­ing the prob­lem#

A point and a direc­tion are given — exactly what a line equa­tion needs.

The idea#

Vec­tor form: $\vec r = \vec a + \lambda\vec b$. Carte­sian form: $\displaystyle \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, with $a, b, c$ the com­po­nents of $\vec b$.

Step-by-step solu­tion#

Step 1. Posi­tion vec­tor of the point: $\vec a = 2\hat i + 0\hat j - 3\hat k = 2\hat i - 3\hat k$. Direc­tion: $\vec b = \hat i - 4\hat j + 5\hat k$.

Step 2. Vec­tor equa­tion.

$$\vec r = 2\hat i - 3\hat k + \lambda(\hat i - 4\hat j + 5\hat k).$$

Step 3. Carte­sian equa­tion, with $(x_1, y_1, z_1) = (2, 0, -3)$ and ratios $1, -4, 5$.

$$\displaystyle \frac{x - 2}{1} = \frac{y - 0}{-4} = \frac{z + 3}{5}.$$

Check­ing the answer#

$\lambda = 0$ gives the point $(2, 0, -3)$ ✓. In the Carte­sian form, $x = 2, y = 0, z = -3$ makes every frac­tion $0$ ✓.

Answer#

$\vec r = 2\hat i - 3\hat k + \lambda(\hat i - 4\hat j + 5\hat k)$; $\displaystyle \frac{x - 2}{1} = \frac{y}{-4} = \frac{z + 3}{5}$.

Ques­tion 6: Carte­sian equa­tion through two points#

The prob­lem#

Find the Carte­sian equa­tion of the line through $(1, 2, 3)$ and $(4, 0, -1)$.

Under­stand­ing the prob­lem#

We need a point (either one will do) and direc­tion ratios (from the two points).

The idea#

Direc­tion ratios $x_2 - x_1, y_2 - y_1, z_2 - z_1$; then the stan­dard Carte­sian form through the first point.

Step-by-step solu­tion#

Step 1. Direc­tion ratios: $4 - 1 = 3$, $0 - 2 = -2$, $-1 - 3 = -4$.

Step 2. Through $(1, 2, 3)$:

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

Check­ing the answer#

The sec­ond point $(4, 0, -1)$ gives $\displaystyle \tfrac{3}{3} = \tfrac{-2}{-2} = \tfrac{-4}{-4} = 1$ — all equal, so it lies on the line ✓.

Answer#

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

Ques­tion 7: Putting a line in stan­dard form#

The prob­lem#

Write $\displaystyle \frac{x + 3}{2} = \frac{4 - y}{5} = \frac{3z + 1}{3}$ in stan­dard form, and give a point on the line and its direc­tion ratios.

Under­stand­ing the prob­lem#

Stan­dard form needs each numer­a­tor to be exactly $x - x_1$, $y - y_1$, $z - z_1$ — the vari­able with coef­fi­cient $+1$. Here the $y$-term is $4 - y$ (coef­fi­cient $-1$) and the $z$-term is $3z + 1$ (coef­fi­cient $3$), so two frac­tions need fix­ing, just as in Exam­ple 4.

The idea#

Rewrite each numer­a­tor so the vari­able has coef­fi­cient $1$, adjust­ing the denom­i­na­tor to keep each frac­tion unchanged.

Step-by-step solu­tion#

Step 1. The $x$-frac­tion is already stan­dard: $\displaystyle \frac{x - (-3)}{2}$.

Step 2. The $y$-frac­tion: $4 - y = -(y - 4)$, so

$$\displaystyle \frac{4 - y}{5} = \frac{-(y - 4)}{5} = \frac{y - 4}{-5}.$$

Step 3. The $z$-frac­tion: $\displaystyle 3z + 1 = 3\left(z + \tfrac13\right)$, so

$$\displaystyle \frac{3z + 1}{3} = \frac{3\left(z + \frac13\right)}{3} = \frac{z + \frac13}{1}.$$

Step 4. Stan­dard form:

$$\displaystyle \frac{x + 3}{2} = \frac{y - 4}{-5} = \frac{z + \frac13}{1}.$$

Step 5. Read off: point $\displaystyle \left(-3, 4, -\tfrac13\right)$; direc­tion ratios $2, -5, 1$.

Check­ing the answer#

Put $\displaystyle \left(-3, 4, -\tfrac13\right)$ into the orig­i­nal: $\displaystyle \tfrac{0}{2} = \tfrac{0}{5} = \tfrac{-1 + 1}{3} = 0$ ✓.

Answer#

$\displaystyle \frac{x + 3}{2} = \frac{y - 4}{-5} = \frac{z + \frac13}{1}$; point $\displaystyle \left(-3, 4, -\tfrac13\right)$, direc­tion ratios $2, -5, 1$.

Com­mon mis­take to avoid#

Read­ing the direc­tion as $2, 5, 3$ straight from the given form. Only the stan­dard form shows the true point and direc­tion.

Ques­tion 8: Angle between two lines in vec­tor form#

The prob­lem#

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

Under­stand­ing the prob­lem#

The angle between two lines depends only on their direc­tions — the vec­tors mul­ti­plied by $\lambda$ and $\mu$. The points $\hat i$ and $3\hat j$ are irrel­e­vant here.

The idea#

$$\displaystyle \cos\theta = \frac{\lvert a_1a_2 + b_1b_2 + c_1c_2 \rvert}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}.$$

Step-by-step solu­tion#

Step 1. Direc­tions: $\vec b_1 = 2\hat i + \hat j + 2\hat k$ (ratios $2, 1, 2$), $\vec b_2 = \hat i + 2\hat j - 2\hat k$ (ratios $1, 2, -2$).

Step 2. Dot prod­uct: $2(1) + 1(2) + 2(-2) = 2 + 2 - 4 = 0$.

Step 3. Lengths: $\sqrt{4 + 1 + 4} = 3$ and $\sqrt{1 + 4 + 4} = 3$.

Step 4.

$$\displaystyle \cos\theta = \frac{\lvert 0 \rvert}{3 \times 3} = 0 \;\Longrightarrow\; \theta = 90^\circ.$$

Check­ing the answer#

A zero dot prod­uct is exactly the per­pen­dic­u­lar­ity con­di­tion $a_1a_2 + b_1b_2 + c_1c_2 = 0$ ✓.

Answer#

$\theta = 90^\circ$ — the lines are per­pen­dic­u­lar.

Ques­tion 9: Find­ing p for per­pen­dic­u­lar lines#

The prob­lem#

Find $p$ so that the lines $\displaystyle \frac{x - 1}{3} = \frac{y - 2}{2p} = \frac{z - 3}{2}$ and $\displaystyle \frac{x - 1}{3p} = \frac{y - 1}{1} = \frac{z - 6}{-5}$ are per­pen­dic­u­lar.

Under­stand­ing the prob­lem#

Both lines are already in stan­dard form, so their direc­tion ratios can be read from the denom­i­na­tors. They con­tain the unknown $p$.

The idea#

Per­pen­dic­u­lar lines sat­isfy $a_1a_2 + b_1b_2 + c_1c_2 = 0$. This gives a lin­ear equa­tion in $p$.

Step-by-step solu­tion#

Step 1. Direc­tion ratios: first line $3, 2p, 2$; sec­ond line $3p, 1, -5$.

Step 2. Per­pen­dic­u­lar­ity con­di­tion.

$$3(3p) + (2p)(1) + 2(-5) = 0.$$

Step 3. Sim­plify and solve.

$$\displaystyle 9p + 2p - 10 = 0 \;\Longrightarrow\; 11p = 10 \;\Longrightarrow\; p = \frac{10}{11}.$$

Check­ing the answer#

With $\displaystyle p = \tfrac{10}{11}$: ratios $\displaystyle 3, \tfrac{20}{11}, 2$ and $\displaystyle \tfrac{30}{11}, 1, -5$. Dot prod­uct $\displaystyle \tfrac{90}{11} + \tfrac{20}{11} - 10 = \tfrac{110}{11} - 10 = 0$ ✓.

Answer#

$\displaystyle p = \frac{10}{11}$.

Ques­tion 10: Two par­al­lel lines#

The prob­lem#

Show that the line through $(2, 3, 4)$ and $(4, 7, 8)$ is par­al­lel to the line through $(0, 1, 2)$ and $(1, 3, 4)$.

Under­stand­ing the prob­lem#

Each line is given by two points. We must prove the two lines have the same direc­tion.

The idea#

Lines are par­al­lel when their direc­tion ratios are pro­por­tional.

Step-by-step solu­tion#

Step 1. Direc­tion ratios of the first line: $4 - 2,\ 7 - 3,\ 8 - 4 = 2,\ 4,\ 4$.

Step 2. Direc­tion ratios of the sec­ond line: $1 - 0,\ 3 - 1,\ 4 - 2 = 1,\ 2,\ 2$.

Step 3. Com­pare.

$$\displaystyle \frac{2}{1} = \frac{4}{2} = \frac{4}{2} = 2.$$

The ratios are pro­por­tional (the first set is twice the sec­ond), so the lines are par­al­lel. $\blacksquare$

Check­ing the answer#

Make sure the lines are gen­uinely two lines, not one: the vec­tor from $(2, 3, 4)$ to $(0, 1, 2)$ is $-2, -2, -2$, which is not pro­por­tional to $1, 2, 2$. So $(0, 1, 2)$ is not on the first line — the lines are par­al­lel and dis­tinct.

Answer#

The direc­tion ratios $2, 4, 4$ and $1, 2, 2$ are pro­por­tional, so the lines are par­al­lel.