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Chapter 11 · Three Dimensional Geometry

Miscellaneous · Q3: Value of kk that makes two lines perpendicular

Miscellaneous ExercisePerpendicular lines · Section 11.4
The question

If the lines x−1−3=y−22k=z−32\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} and x−13k=y−11=z−6−5\frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} are perpendicular, find the value of kk⁠.

Exam answer

The direction ratios of the lines are −3,2k,2-3, 2k, 2 and 3k,1,−53k, 1, -5⁠.

The lines are perpendicular, so a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0⁠:

(−3)(3k)+(2k)(1)+2(−5)=0⇒ −9k+2k−10=0⇒ −7k=10\begin{aligned} &(-3)(3k) + (2k)(1) \\ &\quad + 2(-5) = 0 \\ \Rightarrow\ &-9k + 2k - 10 = 0 \\ \Rightarrow\ &-7k = 10 \end{aligned}

∴ k=−107\therefore\ k = -\dfrac{10}{7}⁠.

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

The one idea this problem is built on

Two lines are perpendicular exactly when a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0 for their direction ratios. Both lines are already in the standard form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠, so the direction ratios are the denominators, and the condition becomes a linear equation in kk⁠.

Given, to find, and the plan
Given
x−1−3=y−22k=z−32\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} and x−13k=y−11=z−6−5\frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} are perpendicular.
To find
The value of kk⁠.
Plan
(1) Check the standard form and read the direction ratios. (2) Apply the perpendicularity condition and solve for kk⁠. (3) Check by substituting back.

Basics you need

1Standard Cartesian form

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠: point (x1,y1,z1)(x_1, y_1, z_1)⁠, direction ratios a,b,ca, b, c (Section 11.3.1). The coefficients of xx⁠, yy⁠, zz must be +1+1⁠. Basics: a line is a point plus a direction

2Perpendicular lines

Lines with direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2 are perpendicular if a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0 (Section 11.4, condition (i)). Lesson: perpendicular and parallel lines

3Direction ratios up to a multiple

Multiplying a set of direction ratios by the same nonzero number gives direction ratios of the same line (Section 11.2). We use this in the check. Basics: direction cosines and ratios

Solution

  1. Read the direction ratios

    In both equations the numerators are x−…x - \ldots⁠, y−…y - \ldots⁠, z−…z - \ldots⁠, so both are in standard form:

    the first line passes through (1,2,3)(1, 2, 3) with direction ratios −3,2k,2-3, 2k, 2⁠;

    the second line passes through (1,1,6)(1, 1, 6) with direction ratios 3k,1,−53k, 1, -5⁠.

    The coefficients of xx⁠, yy and zz are already +1+1⁠, so no rewriting is needed (compare Exercise 11.2 Q10, where the lines had to be rewritten first). The denominators, in order, are the direction ratios; a negative denominator such as −3-3 or −5-5 is simply a negative direction ratio.
    Cartesian equation (3), Section 11.3.1. Lesson: the equation of a line

  2. Apply the condition and solve

    (−3)(3k)+(2k)(1)+(2)(−5)=0⇒ −9k+2k−10=0⇒ −7k=10⇒ k=−107.\begin{aligned} &(-3)(3k) + (2k)(1) + (2)(-5) = 0 \\ \Rightarrow\ &-9k + 2k - 10 = 0 \\ \Rightarrow\ &-7k = 10 \\ \Rightarrow\ &k = -\frac{10}{7}. \end{aligned}
    Multiply the xx-ratios, the yy-ratios and the zz-ratios, add, and set the sum to 0. The equation is linear in kk⁠, so there is exactly one value. Dividing −7k=10-7k = 10 by −7-7 makes kk negative. Since k≠0k \ne 0⁠, the denominators 2k2k and 3k3k are not 0.
    Section 11.4, condition (i) for perpendicular lines. Lesson: perpendicular lines

  3. Check by substituting back

    With k=−107k = -\frac{10}{7}⁠: 2k=−2072k = -\frac{20}{7} and 3k=−3073k = -\frac{30}{7}⁠. The direction ratios become −3,−207,2-3, -\frac{20}{7}, 2 and −307,1,−5-\frac{30}{7}, 1, -5⁠; multiplying each set by 7 gives −21,−20,14-21, -20, 14 and −30,7,−35-30, 7, -35⁠. Then

    (−21)(−30)+(−20)(7)+14(−35)=630−140−490=0.\begin{aligned} &(-21)(-30) + (-20)(7) + 14(-35) \\ &= 630 - 140 - 490 = 0. \end{aligned}
    Scaling each set by 7 keeps the same lines and clears the fractions, so the check uses whole numbers. The sum is 0, so with k=−107k = -\frac{10}{7} the lines are perpendicular.
    Direction ratios are fixed only up to a nonzero multiple (Section 11.2). Basics: direction ratios
Result

k=−107k = -\frac{10}{7}⁠.

(Textbook answer: k=−107k = -\frac{10}{7}⁠.)

Check it with numbers

k=−107≈−1.428571k = -\frac{10}{7} \approx -1.428571⁠. Then 2k≈−2.8571432k \approx -2.857143 and 3k≈−4.2857143k \approx -4.285714⁠, and

(−3)(−4.285714)(-3)(-4.285714) + (−2.857143)(1)+2(−5)+\ (-2.857143)(1) + 2(-5) =12.857142−2.857143−10= 12.857142 - 2.857143 - 10 =−0.000001= -0.000001⁠,

which is 0 up to the rounding of the decimals.

Common mistakes
  1. Pairing the wrong ratios, for example −3-3 with 11⁠. Pair xx with xx⁠: −3-3 with 3k3k⁠, 2k2k with 11⁠, 22 with −5-5⁠.
  2. Sign slips: (−3)(3k)=−9k(-3)(3k) = -9k and 2(−5)=−102(-5) = -10⁠.
  3. Losing the minus sign at the end: −7k=10-7k = 10 gives k=−107k = -\frac{10}{7}⁠, not 107\frac{10}{7}⁠.
  4. Using the points (1,2,3)(1, 2, 3) and (1,1,6)(1, 1, 6) in the condition. The angle between lines depends only on the directions.
What to take away
  1. First make sure each line is in standard form; then the denominators are the direction ratios.
  2. Perpendicular: a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0 gives an equation for the unknown.
  3. Check by substituting back, scaling the direction ratios to whole numbers.

Every fact used, and where it comes from

FactUsed inSource
Standard Cartesian form; its denominators are direction ratiosStep 1Section 11.3.1, equation (3) (Lesson)
Perpendicular lines: a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0Steps 2, 3Section 11.4, condition (i) (Lesson)
Any nonzero multiple of direction ratios servesStep 3Section 11.2 (Basics)
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