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Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q4: Line through (1,2,3)(1, 2, 3) parallel to 3i^+2j^−2k^3\hat i + 2\hat j - 2\hat k

Exercise 11.2Vector equation of a line · Section 11.3.1
The question

Find the equation of the line which passes through the point (1,2,3)(1, 2, 3) and is parallel to the vector 3i^+2j^−2k^3\hat i + 2\hat j - 2\hat k⁠.

Exam answer

Here a⃗=i^+2j^+3k^\vec a = \hat i + 2\hat j + 3\hat k (the position vector of the point) and b⃗=3i^+2j^−2k^\vec b = 3\hat i + 2\hat j - 2\hat k⁠.

The line through the point with position vector a⃗\vec a and parallel to b⃗\vec b is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.

∴\therefore r⃗=i^+2j^+3k^\vec r = \hat i + 2\hat j + 3\hat k + λ(3i^+2j^−2k^)+\ \lambda(3\hat i + 2\hat j - 2\hat k)⁠, where λ\lambda is a real number.

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 (position vector a⃗\vec a⁠) and walk any multiple λ\lambda of the given vector b⃗\vec b⁠: this reaches every point of the line, so the line is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.

Given, to find, and the plan
Given
A point A(1,2,3)A(1, 2, 3) of the line and a vector b⃗=3i^+2j^−2k^\vec b = 3\hat i + 2\hat j - 2\hat k parallel to it.
To find
The 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) The same line in Cartesian form. (4) 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

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=i^+2j^+3k^,b⃗=3i^+2j^−2k^.\vec a = \hat i + 2\hat j + 3\hat k, \qquad \vec b = 3\hat i + 2\hat j - 2\hat k.
    a⃗\vec a is the position vector of the given point (1,2,3)(1, 2, 3)⁠: its coordinates become the coefficients of i^,j^,k^\hat i, \hat j, \hat k⁠. b⃗\vec b is the given vector that the line is parallel to; by the Remark in Section 11.3.1, its components 3,2,−23, 2, -2 are direction ratios of the line.
    Position vector of a point. Basics: three numbers fix a point

  2. The vector equation

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

    where λ\lambda is any real number.

    X Y Z A(1, 2, 3) P(4, 4, 1) λ = 1 b a O
    Oblique projection, not to scale. a⃗\vec a goes from OO to A(1,2,3)A(1, 2, 3)⁠. One step b⃗\vec b from AA (that is, λ=1\lambda = 1⁠) reaches P(4,4,1)P(4, 4, 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 gives r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. Conversely, each real λ\lambda gives a point of the line: λ=0\lambda = 0 gives AA⁠, positive λ\lambda moves along b⃗\vec b⁠, negative λ\lambda the other way. This vector equation is the answer the question asks for.
    Section 11.3.1, equation (1), and Fig 11.3. Lesson: start somewhere, then walk

  3. The same line in Cartesian form

    Writing r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k and equating coefficients: x=1+3λx = 1 + 3\lambda⁠, y=2+2λy = 2 + 2\lambda⁠, z=3−2λz = 3 - 2\lambda⁠. Eliminating λ\lambda⁠:

    x−13=y−22=z−3−2.\frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{-2}.
    The question says "the equation" without naming a form, and the answer key gives the vector form. The Cartesian form is the same line written with coordinates: each of the three equations solved for λ\lambda gives one fraction, and all three equal the same λ\lambda⁠. The −2-2 stays in the denominator.
    Section 11.3.1, equations (2) and (3). Exercise 11.2 player

  4. Check with points of the line

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

    4−13=4−22=1−3−2=1.\frac{4 - 1}{3} = \frac{4 - 2}{2} = \frac{1 - 3}{-2} = 1.
    A correct equation must contain the given point, and moving from λ=0\lambda = 0 to λ=1\lambda = 1 must change the coordinates by 3,2,−23, 2, -2⁠, the components of b⃗\vec b⁠. Both hold, and the Cartesian form agrees with the vector form at λ=1\lambda = 1⁠.
    Each value of λ\lambda gives one point of the line. Basics: a line is a point plus a direction
Result

r⃗=i^+2j^+3k^\vec r = \hat i + 2\hat j + 3\hat k + λ(3i^+2j^−2k^)+\ \lambda(3\hat i + 2\hat j - 2\hat k)⁠, λ∈R\lambda \in \mathbf R⁠; in Cartesian form, x−13=y−22=z−3−2\frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{-2}⁠.

(Textbook answer: r⃗=i^+2j^+3k^\vec r = \hat i + 2\hat j + 3\hat k + λ(3i^+2j^−2k^)+\ \lambda(3\hat i + 2\hat j - 2\hat k)⁠, where λ\lambda is a real number.)

Check it with numbers

Take λ=0.5\lambda = 0.5⁠: the point is (2.5,3,2)(2.5, 3, 2)⁠. Then 2.5−13=0.5\frac{2.5 - 1}{3} = 0.5⁠, 3−22=0.5\frac{3 - 2}{2} = 0.5 and 2−3−2=0.5\frac{2 - 3}{-2} = 0.5⁠, all equal to λ\lambda⁠.

The point P(4,4,1)P(4, 4, 1) is at distance ∣b⃗∣=9+4+4=17≈4.123106|\vec b| = \sqrt{9 + 4 + 4} = \sqrt{17} \approx 4.123106 from AA⁠, one full step b⃗\vec b⁠.

Common mistakes
  1. Swapping a⃗\vec a and b⃗\vec b⁠. a⃗\vec a is the point, b⃗\vec b the direction; r⃗=b⃗+λa⃗\vec r = \vec b + \lambda\vec a is a different line.
  2. Leaving out λ\lambda, or writing r⃗=a⃗+b⃗\vec r = \vec a + \vec b⁠, which is a single point, not a line.
  3. Dropping the minus sign of −2k^-2\hat k (or of the denominator −2-2 in the Cartesian form).
What to take away
  1. Line through a point parallel to a vector: r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠, point first, direction second.
  2. The components of b⃗\vec b are direction ratios of the line.
  3. Equate components and eliminate λ\lambda to get the Cartesian form.

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 and Cartesian equationsStep 3Section 11.3.1, equations (2) and (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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