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Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q7: Vector form of x−53=y+47=z−62\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}

Exercise 11.2Vector and Cartesian equations of a line · Section 11.3.1
The question

The Cartesian equation of a line is x−53=y+47=z−62\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}⁠. Write its vector form.

Exam answer

Let x−53=y+47=z−62=λ\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} = \lambda⁠.

Then x=5+3λx = 5 + 3\lambda⁠, y=−4+7λy = -4 + 7\lambda⁠, z=6+2λz = 6 + 2\lambda⁠.

The position vector of any point (x,y,z)(x, y, z) of the line is

r⃗=xi^+yj^+zk^=(5+3λ)i^+(−4+7λ)j^+(6+2λ)k^=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\begin{aligned} \vec r &= x\hat i + y\hat j + z\hat k \\ &= (5 + 3\lambda)\hat i + (-4 + 7\lambda)\hat j \\ &\quad + (6 + 2\lambda)\hat k \\ &= (5\hat i - 4\hat j + 6\hat k) \\ &\quad + \lambda(3\hat i + 7\hat j + 2\hat k) \end{aligned}

So the line passes through the point with position vector 5i^−4j^+6k^5\hat i - 4\hat j + 6\hat k and is parallel to 3i^+7j^+2k^3\hat i + 7\hat j + 2\hat k⁠.

∴ r⃗=(5i^−4j^+6k^)\therefore\ \vec r = (5\hat i - 4\hat j + 6\hat k) + λ(3i^+7j^+2k^)+\ \lambda(3\hat i + 7\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

The Cartesian form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} and the vector form r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b carry the same two pieces of information: a point (x1,y1,z1)(x_1, y_1, z_1) of the line (the numerators) and a direction a,b,ca, b, c (the denominators). To change form, read off the point and the direction and write them as vectors.

Given, to find, and the plan
Given
The line x−53=y+47=z−62\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}⁠.
To find
Its vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.
Plan
(1) Write the numerators as x−x1x - x_1⁠, y−y1y - y_1⁠, z−z1z - z_1 to read the point. (2) Read the direction ratios. (3) Write a⃗\vec a and b⃗\vec b and the vector form. (4) Check by setting each fraction equal to λ\lambda⁠.

Basics you need

1Cartesian form of a line

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

2Vector form 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\lambda real (Section 11.3.1, equation (1)). Basics: go to A, then walk λ copies of b

3Position 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: three numbers fix a point

Solution

  1. Read the point from the numerators

    x−53=y−(−4)7=z−62.\frac{x - 5}{3} = \frac{y - (-4)}{7} = \frac{z - 6}{2}.

    So the line passes through A(5,−4,6)A(5, -4, 6)⁠.

    Matching with x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠, every numerator must have the form "coordinate minus a number". The middle one is y+4=y−(−4)y + 4 = y - (-4)⁠, so y1=−4y_1 = -4⁠, not +4+4⁠.
    Cartesian equation (3), Section 11.3.1. Basics: point plus direction

  2. Read the direction from the denominators

    The direction ratios are 3,7,23, 7, 2⁠, so the line is parallel to b⃗=3i^+7j^+2k^\vec b = 3\hat i + 7\hat j + 2\hat k⁠.

    The denominators are the direction ratios a,b,ca, b, c⁠. By the Remark in Section 11.3.1, if a,b,ca, b, c are direction ratios of a line, then ai^+bj^+ck^a\hat i + b\hat j + c\hat k is parallel to it. The coefficients of xx⁠, yy and zz in the numerators are all 1, so the denominators can be read off as they stand.
    Direction ratios and a direction vector (Remark, Section 11.3.1). Basics: direction cosines and ratios

  3. Write the vector form

    The position vector of AA is a⃗=5i^−4j^+6k^\vec a = 5\hat i - 4\hat j + 6\hat k⁠. So

    r⃗=a⃗+λb⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^),\begin{aligned} \vec r &= \vec a + \lambda\vec b \\ &= (5\hat i - 4\hat j + 6\hat k) + \lambda(3\hat i + 7\hat j + 2\hat k), \end{aligned}

    where λ\lambda is a real number.

    Each real value of λ\lambda gives one point of the line: start at AA and walk λ\lambda copies of b⃗\vec b⁠. The brackets around a⃗\vec a are not needed, but they keep the point and the direction apart, as in the answer key.
    Vector equation (1), Section 11.3.1. Basics: r = a + λb

  4. Check: back to the Cartesian form

    Put each fraction equal to λ\lambda⁠: x=5+3λx = 5 + 3\lambda⁠, y=−4+7λy = -4 + 7\lambda⁠, z=6+2λz = 6 + 2\lambda⁠. Then

    xi^+yj^+zk^=(5i^−4j^+6k^)+λ(3i^+7j^+2k^),\begin{aligned} x\hat i + y\hat j + z\hat k &= (5\hat i - 4\hat j + 6\hat k) \\ &\quad + \lambda(3\hat i + 7\hat j + 2\hat k), \end{aligned}

    the same equation. For λ=1\lambda = 1 the point is P(8,3,8)P(8, 3, 8)⁠, and 8−53=3+47=8−62=1\frac{8 - 5}{3} = \frac{3 + 4}{7} = \frac{8 - 6}{2} = 1⁠.

    X Y Z A(5, −4, 6) P(8, 3, 8) λ = 1 b a O
    Oblique projection, not to scale. The line through A(5,−4,6)A(5, -4, 6) parallel to b⃗=3i^+7j^+2k^\vec b = 3\hat i + 7\hat j + 2\hat k⁠: the position vector a⃗=OA→\vec a = \overrightarrow{OA}⁠, then one full step b⃗\vec b to P(8,3,8)P(8, 3, 8) at λ=1\lambda = 1⁠.
    This is the derivation of the Cartesian form from the vector form (Section 11.3.1) run backwards: the parametric equations x=x1+λax = x_1 + \lambda a⁠, y=y1+λby = y_1 + \lambda b⁠, z=z1+λcz = z_1 + \lambda c link the two forms. Both forms describe the same set of points.
    Parametric equations (2), Section 11.3.1. Exercise 11.2 player
Result

r⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\vec r = (5\hat i - 4\hat j + 6\hat k) + \lambda(3\hat i + 7\hat j + 2\hat k)⁠, λ∈R\lambda \in \mathbf R⁠.

(Textbook answer: r⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\vec r = (5\hat i - 4\hat j + 6\hat k) + \lambda(3\hat i + 7\hat j + 2\hat k)⁠.)

Check it with numbers

Take λ=0.5\lambda = 0.5⁠: the point is (5+1.5,−4+3.5,6+1)(5 + 1.5, -4 + 3.5, 6 + 1) =(6.5,−0.5,7)= (6.5, -0.5, 7)⁠. Then 6.5−53=0.5\frac{6.5 - 5}{3} = 0.5⁠, −0.5+47=0.5\frac{-0.5 + 4}{7} = 0.5 and 7−62=0.5\frac{7 - 6}{2} = 0.5⁠, all equal to λ\lambda⁠.

The direction has length ∣b⃗∣=9+49+4=62≈7.874008|\vec b| = \sqrt{9 + 49 + 4} = \sqrt{62} \approx 7.874008⁠.

Common mistakes
  1. Taking the point as (5,4,6)(5, 4, 6)⁠. y+4=y−(−4)y + 4 = y - (-4)⁠, so the yy-coordinate is −4-4⁠.
  2. Swapping point and direction, writing r⃗=(3i^+7j^+2k^)+λ(5i^−4j^+6k^)\vec r = (3\hat i + 7\hat j + 2\hat k) + \lambda(5\hat i - 4\hat j + 6\hat k)⁠. The numerators give the point; the denominators give the direction.
  3. Leaving out λ\lambda or not saying it is a real number: without the parameter the equation describes one point, not a line.
What to take away
  1. Cartesian to vector: numerators give a⃗\vec a⁠, denominators give b⃗\vec b⁠, then r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.
  2. Make each numerator "coordinate minus number" before reading the point; a plus sign means a negative coordinate.
  3. Setting each fraction equal to λ\lambda turns one form into the other (compare Exercise 11.2 Q5, which goes the other way).

Every fact used, and where it comes from

FactUsed inSource
Cartesian equation x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}Steps 1, 2Section 11.3.1, equation (3) (Basics)
Direction ratios a,b,ca, b, c give the parallel vector ai^+bj^+ck^a\hat i + b\hat j + c\hat kStep 2Remark, Section 11.3.1 (Basics)
Vector equation r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec bStep 3Section 11.3.1, equation (1) (Basics)
Parametric equations x=x1+λax = x_1 + \lambda a⁠, y=y1+λby = y_1 + \lambda b⁠, z=z1+λcz = z_1 + \lambda cStep 4Section 11.3.1, equation (2)
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