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Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q5: Vector and Cartesian equations of the line through 2i^−j^+4k^2\hat i - \hat j + 4\hat k along i^+2j^−k^\hat i + 2\hat j - \hat k

Exercise 11.2Equation of a line in space · Section 11.3.1
The question

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i^−j^+4k^2\hat i - \hat j + 4\hat k and is in the direction i^+2j^−k^\hat i + 2\hat j - \hat k⁠.

Exam answer

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

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

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

where λ\lambda is a real number.

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

xi^+yj^+zk^=(2+λ)i^+(−1+2λ)j^+(4−λ)k^\begin{aligned} &x\hat i + y\hat j + z\hat k \\ &= (2 + \lambda)\hat i + (-1 + 2\lambda)\hat j \\ &\quad + (4 - \lambda)\hat k \end{aligned}

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

x=2+λx = 2 + \lambda⁠, y=−1+2λy = -1 + 2\lambda⁠, z=4−λz = 4 - \lambda⁠.

Eliminating λ\lambda⁠, the Cartesian form is

x−21=y+12=z−4−1\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z - 4}{-1}⁠.

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. From the point a⃗\vec a⁠, walking any multiple λ\lambda of the direction b⃗\vec b reaches every point of the line: r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. Writing this coordinate by coordinate and removing λ\lambda gives the Cartesian form.

Given, to find, and the plan
Given
The point with position vector a⃗=2i^−j^+4k^\vec a = 2\hat i - \hat j + 4\hat k⁠, that is A(2,−1,4)A(2, -1, 4)⁠, and the direction b⃗=i^+2j^−k^\vec b = \hat i + 2\hat j - \hat k⁠.
To find
The equation of the line in vector form and in Cartesian form.
Plan
(1) Write a⃗\vec a and b⃗\vec b⁠. (2) The vector form 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

2Equal 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.

3Cartesian 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=2i^−j^+4k^,b⃗=i^+2j^−k^.\vec a = 2\hat i - \hat j + 4\hat k, \qquad \vec b = \hat i + 2\hat j - \hat k.
    a⃗\vec a is the position vector of the point the line passes through, so the point is A(2,−1,4)A(2, -1, 4)⁠. b⃗\vec b gives the direction, so its components 1,2,−11, 2, -1 are direction ratios of the line (Remark in Section 11.3.1). The book prints the middle term of a⃗\vec a as "jj" without a hat; it is the unit vector j^\hat j⁠, as the key's answer shows.
    Position vector of a point. Basics: three numbers fix a point

  2. The vector form

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

    where λ\lambda is any real number.

    X Y Z A(2, −1, 4) P(3, 1, 3) λ = 1 b a O
    Oblique projection, not to scale. a⃗\vec a goes from OO to A(2,−1,4)A(2, -1, 4)⁠. One step b⃗\vec b from AA (λ=1\lambda = 1⁠) reaches P(3,1,3)P(3, 1, 3)⁠; the arrow head on the line shows the direction of b⃗\vec b⁠.
    For any point PP of the line, AP→=r⃗−a⃗\overrightarrow{AP} = \vec r - \vec a is parallel to b⃗\vec b⁠, so r⃗−a⃗=λb⃗\vec r - \vec a = \lambda\vec b for some real λ\lambda⁠. Conversely, every real λ\lambda gives a point of the line.
    Section 11.3.1, equation (1). Lesson: start somewhere, then walk

  3. Write it with coordinates

    Put r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k and collect the terms on the right:

    xi^+yj^+zk^=(2+λ)i^+(−1+2λ)j^+(4−λ)k^.\begin{aligned} x\hat i + y\hat j + z\hat k &= (2 + \lambda)\hat i + (-1 + 2\lambda)\hat j \\ &\quad + (4 - \lambda)\hat k. \end{aligned}

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

    x=2+λ,y=−1+2λ,z=4−λ.\begin{aligned} x &= 2 + \lambda, \\ y &= -1 + 2\lambda, \\ z &= 4 - \lambda. \end{aligned}
    λ(i^+2j^−k^)=λi^+2λj^−λk^\lambda(\hat i + 2\hat j - \hat k) = \lambda\hat i + 2\lambda\hat j - \lambda\hat k⁠, and each term joins the matching term of a⃗\vec a⁠. Two vectors are equal only when their components are equal, so one vector equation becomes three: 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−21,λ=y+12,λ=z−4−1.\lambda = \frac{x - 2}{1}, \quad \lambda = \frac{y + 1}{2}, \quad \lambda = \frac{z - 4}{-1}.

    All three equal the same λ\lambda⁠, so

    x−21=y+12=z−4−1.\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}.
    From y=−1+2λy = -1 + 2\lambda we get y+1=2λy + 1 = 2\lambda⁠; from z=4−λz = 4 - \lambda we get z−4=−λz - 4 = -\lambda⁠, so λ=z−4−1\lambda = \frac{z - 4}{-1}⁠. The denominators 1 and −1-1 are kept so that the denominators show the direction ratios 1,2,−11, 2, -1⁠, as in the standard form; the numerators show the point (2,−1,4)(2, -1, 4)⁠.
    Section 11.3.1, equation (3). Exercise 11.2 player

  5. Check with points of the line

    At λ=0\lambda = 0⁠: (2,−1,4)(2, -1, 4)⁠, the given point. At λ=1\lambda = 1⁠: (3,1,3)(3, 1, 3)⁠, and

    3−21=1+12=3−4−1=1.\frac{3 - 2}{1} = \frac{1 + 1}{2} = \frac{3 - 4}{-1} = 1.
    The equation must contain the given point, and one step of λ\lambda must change the coordinates by 1,2,−11, 2, -1⁠, the components of b⃗\vec b⁠. Both hold.
    Each value of λ\lambda gives one point of the line. Basics: a line is a point plus a direction
Result

Vector form: r⃗=2i^−j^+4k^\vec r = 2\hat i - \hat j + 4\hat k + λ(i^+2j^−k^)+\ \lambda(\hat i + 2\hat j - \hat k)⁠, where λ\lambda is a real number.

Cartesian form: x−21=y+12=z−4−1\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}⁠.

(Textbook answer: r⃗=2i^−j^+4k^\vec r = 2\hat i - \hat j + 4\hat k + λ(i^+2j^−k^)+\ \lambda(\hat i + 2\hat j - \hat k) and x−21=y+12=z−4−1\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}⁠.)

Check it with numbers

Take λ=2.5\lambda = 2.5⁠: the point is (4.5,4,1.5)(4.5, 4, 1.5)⁠. Then 4.5−21=2.5\frac{4.5 - 2}{1} = 2.5⁠, 4+12=2.5\frac{4 + 1}{2} = 2.5 and 1.5−4−1=2.5\frac{1.5 - 4}{-1} = 2.5⁠, all equal to λ\lambda⁠.

One full step has length ∣b⃗∣=1+4+1=6≈2.449490|\vec b| = \sqrt{1 + 4 + 1} = \sqrt6 \approx 2.449490⁠.

Common mistakes
  1. Sign errors in the numerators. The point has y1=−1y_1 = -1⁠, so the numerator is y−(−1)=y+1y - (-1) = y + 1⁠.
  2. Dropping the denominator −1-1 and writing z−4z - 4 alone, or z−41\frac{z - 4}{1}⁠. The direction ratio is −1-1⁠.
  3. Swapping a⃗\vec a and b⃗\vec b: the point goes first, the direction is multiplied by λ\lambda⁠.
What to take away
  1. Vector form: r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.
  2. Cartesian form: numerators "xx minus the point's coordinate", denominators the direction ratios, including 1 and −1-1⁠.
  3. 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 bStep 2Section 11.3.1, equation (1) (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) (Example 6)
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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