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Chapter 11 · Three Dimensional Geometry

Example 6: Vector and Cartesian equations of the line through (5,2,−4)(5, 2, -4) parallel to 3i^+2j^−8k^3\hat i + 2\hat j - 8\hat k

ExamplesEquation of a line in space · Section 11.3.1
The question

Find the vector and the Cartesian equations of the line through the point (5,2,−4)(5, 2, -4) and which is parallel to the vector 3i^+2j^−8k^3\hat i + 2\hat j - 8\hat k⁠.

Exam answer

Here a⃗=5i^+2j^−4k^\vec a = 5\hat i + 2\hat j - 4\hat k and b⃗=3i^+2j^−8k^\vec b = 3\hat i + 2\hat j - 8\hat k⁠.

Vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠:

r⃗=5i^+2j^−4k^+λ(3i^+2j^−8k^)\begin{aligned} \vec r &= 5\hat i + 2\hat j - 4\hat k \\ &\quad + \lambda(3\hat i + 2\hat j - 8\hat k) \end{aligned}

where λ\lambda is a real number.

Cartesian equation. Put r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠:

xi^+yj^+zk^=(5+3λ)i^+(2+2λ)j^+(−4−8λ)k^\begin{aligned} &x\hat i + y\hat j + z\hat k \\ &= (5 + 3\lambda)\hat i + (2 + 2\lambda)\hat j \\ &\quad + (-4 - 8\lambda)\hat k \end{aligned}

Equating the coefficients of i^,j^,k^\hat i, \hat j, \hat k⁠:

x=5+3λx = 5 + 3\lambda⁠, y=2+2λy = 2 + 2\lambda⁠, z=−4−8λz = -4 - 8\lambda⁠.

Eliminating λ\lambda⁠, the Cartesian equation is

x−53=y−22=z+4−8\dfrac{x - 5}{3} = \dfrac{y - 2}{2} = \dfrac{z + 4}{-8}⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A line is a point plus a direction. Start at the given point AA (position vector a⃗\vec a⁠) and walk any multiple λ\lambda of the direction b⃗\vec b⁠: you reach every point of the line, r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. Writing this vector equation coordinate by coordinate and removing λ\lambda gives the Cartesian form.

Given, to find, and the plan
Given
A point A(5,2,−4)A(5, 2, -4) on the line, and the direction b⃗=3i^+2j^−8k^\vec b = 3\hat i + 2\hat j - 8\hat k⁠.
To find
The vector equation and the Cartesian equation of the line.
Plan
(1) Write a⃗\vec a and b⃗\vec b⁠. (2) Put them into r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. (3) Put r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k and equate components. (4) Eliminate λ\lambda⁠. (5) Check with points of the line.

Basics you need

1Vector equation of a line

The line through the point with position vector a⃗\vec a⁠, parallel to b⃗\vec b⁠, is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, λ∈R\lambda \in \mathbf R (Section 11.3.1, equation (1)). Basics: go to A, then walk λ copies of b

2Position vector of a point

The point (x,y,z)(x, y, z) has position vector xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠. Basics: coordinates in 3D

3Equal vectors have equal components

If pi^+qj^+rk^=p′i^+q′j^+r′k^p\hat i + q\hat j + r\hat k = p'\hat i + q'\hat j + r'\hat k⁠, then p=p′p = p'⁠, q=q′q = q'⁠, r=r′r = r'⁠. This turns one vector equation into three ordinary ones.

4Cartesian form

The line through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c is x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} (Section 11.3.1, equation (3)). Basics: a line is a point plus a direction

Solution

  1. Write a⃗\vec a and b⃗\vec b

    a⃗=5i^+2j^−4k^,b⃗=3i^+2j^−8k^.\begin{aligned} \vec a &= 5\hat i + 2\hat j - 4\hat k, \\ \vec b &= 3\hat i + 2\hat j - 8\hat k. \end{aligned}
    a⃗\vec a is the position vector of the given point (5,2,−4)(5, 2, -4)⁠: its coordinates become the coefficients of i^,j^,k^\hat i, \hat j, \hat k⁠. b⃗\vec b is the given vector the line is parallel to. By the Remark in Section 11.3.1, its components 3,2,−83, 2, -8 are direction ratios of the line.
    Position vector of a point. Basics: three numbers fix a point

  2. The vector equation

    r⃗=a⃗+λb⃗=5i^+2j^−4k^+λ(3i^+2j^−8k^),\begin{aligned} \vec r &= \vec a + \lambda\vec b \\ &= 5\hat i + 2\hat j - 4\hat k + \lambda(3\hat i + 2\hat j - 8\hat k), \end{aligned}

    where λ\lambda is any real number.

    X Y Z A(5, 2, −4) P(2, 0, 4) λ = −1 along b a r O
    Oblique projection, not to scale. a⃗\vec a goes from OO to A(5,2,−4)A(5, 2, -4)⁠. The point P(2,0,4)P(2, 0, 4) of the line has position vector r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b with λ=−1\lambda = -1⁠; the arrow head on the line shows the direction of b⃗\vec b⁠.
    For any point PP of the line, AP→\overrightarrow{AP} is parallel to b⃗\vec b⁠, so AP→=λb⃗\overrightarrow{AP} = \lambda\vec b for some real λ\lambda⁠. Since AP→=r⃗−a⃗\overrightarrow{AP} = \vec r - \vec a⁠, this is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. Conversely, every real λ\lambda gives a point of the line: λ=0\lambda = 0 gives AA itself, positive λ\lambda points along b⃗\vec b⁠, negative λ\lambda the other way.
    Section 11.3.1, equation (1), and Fig 11.3. Lesson: start somewhere, then walk

  3. Write it with coordinates

    The point P(x,y,z)P(x, y, z) has r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠. Collecting the terms on the right:

    xi^+yj^+zk^=(5+3λ)i^+(2+2λ)j^+(−4−8λ)k^.\begin{aligned} x\hat i + y\hat j + z\hat k &= (5 + 3\lambda)\hat i + (2 + 2\lambda)\hat j \\ &\quad + (-4 - 8\lambda)\hat k. \end{aligned}

    Equating the coefficients of i^\hat i⁠, j^\hat j and k^\hat k⁠:

    x=5+3λ,y=2+2λ,z=−4−8λ.\begin{aligned} &x = 5 + 3\lambda, \quad y = 2 + 2\lambda, \\ &z = -4 - 8\lambda. \end{aligned}
    λ(3i^+2j^−8k^)=3λi^+2λj^−8λk^\lambda(3\hat i + 2\hat j - 8\hat k) = 3\lambda\hat i + 2\lambda\hat j - 8\lambda\hat k⁠, and each of these joins the matching term of a⃗\vec a⁠. Two vectors are equal only when their components are equal, so one vector equation becomes three equations. These are the parametric equations of the line (Section 11.3.1, equation (2)).
    Equal vectors have equal components. Basics: a⃗ + λb⃗

  4. Eliminate λ\lambda

    Solving each equation for λ\lambda⁠:

    λ=x−53,λ=y−22,λ=z+4−8.\lambda = \frac{x - 5}{3}, \quad \lambda = \frac{y - 2}{2}, \quad \lambda = \frac{z + 4}{-8}.

    All three equal the same λ\lambda⁠, so

    x−53=y−22=z+4−8.\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}.
    From z=−4−8λz = -4 - 8\lambda we get z+4=−8λz + 4 = -8\lambda⁠, so λ=z+4−8\lambda = \frac{z + 4}{-8}⁠; the −8-8 stays in the denominator. This is the Cartesian equation (3) of Section 11.3.1 with (x1,y1,z1)=(5,2,−4)(x_1, y_1, z_1) = (5, 2, -4) and direction ratios 3,2,−83, 2, -8⁠: the numerators are "xx minus the point's coordinate", the denominators are the direction ratios.
    Section 11.3.1, equation (3). Lesson: Example 6

  5. Check with points of the line

    At λ=0\lambda = 0⁠: (5,2,−4)(5, 2, -4)⁠, the given point. At λ=−1\lambda = -1⁠: (2,0,4)(2, 0, 4)⁠, and

    2−53=0−22=4+4−8=−1.\frac{2 - 5}{3} = \frac{0 - 2}{2} = \frac{4 + 4}{-8} = -1.
    A correct equation must contain the given point, and every point it produces must satisfy the Cartesian form with all three fractions equal (to the same λ\lambda⁠). Moving from λ=0\lambda = 0 to λ=1\lambda = 1 changes the coordinates by 3,2,−83, 2, -8⁠, the components of b⃗\vec b⁠, so the direction is right too.
    Each value of λ\lambda gives one point of the line. Basics: a line is a point plus a direction
Result

Vector equation: r⃗=5i^+2j^−4k^\vec r = 5\hat i + 2\hat j - 4\hat k + λ(3i^+2j^−8k^)+\ \lambda(3\hat i + 2\hat j - 8\hat k)⁠, λ∈R\lambda \in \mathbf R⁠.

Cartesian equation: x−53=y−22=z+4−8\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}⁠.

(Textbook answer: the same vector equation, and x−53=y−22=z+4−8\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}⁠.)

Check it with numbers

Take λ=0.5\lambda = 0.5⁠: the point is (6.5,3,−8)(6.5, 3, -8)⁠. Then 6.5−53=0.5\frac{6.5 - 5}{3} = 0.5⁠, 3−22=0.5\frac{3 - 2}{2} = 0.5 and −8+4−8=0.5\frac{-8 + 4}{-8} = 0.5⁠, all equal to λ\lambda⁠.

The point (8,4,−12)(8, 4, -12) (λ=1\lambda = 1⁠) is at distance ∣b⃗∣=9+4+64=77≈8.774964|\vec b| = \sqrt{9 + 4 + 64} = \sqrt{77} \approx 8.774964 from AA⁠, as it should be one full step b⃗\vec b away.

Common mistakes
  1. Swapping a⃗\vec a and b⃗\vec b⁠. a⃗\vec a is the point, b⃗\vec b is the direction; r⃗=b⃗+λa⃗\vec r = \vec b + \lambda\vec a is a different line.
  2. Sign errors in the numerators. The point has z1=−4z_1 = -4⁠, so the numerator is z−(−4)=z+4z - (-4) = z + 4⁠.
  3. Dropping the minus sign of the direction ratio, writing z+48\frac{z + 4}{8}⁠. That is a different line; the denominator is −8-8⁠.
  4. Leaving out λ\lambda⁠, or not saying what it is. The vector equation needs the parameter λ\lambda⁠, with λ\lambda any real number.
What to take away
  1. Vector form: r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, point first, direction second.
  2. Cartesian form: x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠, the point in the numerators, the direction ratios in the denominators.
  3. To go from vector to Cartesian: equate components, then eliminate λ\lambda⁠.

Every fact used, and where it comes from

FactUsed inSource
Components of b⃗\vec b are direction ratios of the lineStep 1Remark in Section 11.3.1
r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, from AP→=λb⃗\overrightarrow{AP} = \lambda\vec bStep 2Section 11.3.1, equation (1), Fig 11.3 (Lesson)
Parametric equations x=x1+λax = x_1 + \lambda a⁠, y=y1+λby = y_1 + \lambda b⁠, z=z1+λcz = z_1 + \lambda cStep 3Section 11.3.1, equation (2) (Basics)
Cartesian equation x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}Step 4Section 11.3.1, equation (3) (Lesson)
Magnitude ∣b⃗∣=b12+b22+b32|\vec b| = \sqrt{b_1^2 + b_2^2 + b_3^2}CheckSection 10.5.1
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