Chapter 11 · Three Dimensional Geometry

Miscellaneous · Q5: Vector equation of the line through $(1, 2, -4)$ perpendicular to two given lines

Miscellaneous ExerciseA line perpendicular to two lines · Sections 11.3.1 and 11.4

The question

Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines:

$\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ ⁠.

Exam answer

The direction ratios of the given lines are $3, -16, 7$ and $3, 8, -5$ ⁠.

Let $\vec b_1 = 3\hat i - 16\hat j + 7\hat k$ and $\vec b_2 = 3\hat i + 8\hat j - 5\hat k$ ⁠.

The required line is perpendicular to both lines, so it is parallel to $\vec b_1\times\vec b_2$ ⁠.

\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \\ &= (80 - 56)\hat i \\ &\quad - (-15 - 21)\hat j \\ &\quad + (24 + 48)\hat k \\ &= 24\hat i + 36\hat j + 72\hat k \\ &= 12(2\hat i + 3\hat j + 6\hat k) \end{aligned}

So take $\vec b = 2\hat i + 3\hat j + 6\hat k$ ⁠.

The line passes through $(1, 2, -4)$ ⁠, so $\vec a = \hat i + 2\hat j - 4\hat k$ ⁠.

Its vector equation is $\vec r = \vec a + \lambda\vec b$ ⁠:

$\therefore\ \vec r = \hat i + 2\hat j - 4\hat k$ $+\ \lambda(2\hat i + 3\hat j + 6\hat k)$ ⁠, where $\lambda$ 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 perpendicular to two lines must have a direction perpendicular to both of their directions. The cross product $\vec b_1\times\vec b_2$ is exactly such a vector. So the direction comes from the cross product, the point is given, and $\vec r = \vec a + \lambda\vec b$ finishes the job.

Given, to find, and the plan

Given: The point $A(1, 2, -4)$ ⁠; the lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ ⁠.
To find: The vector equation of the line through $A$ perpendicular to both given lines.
Plan: (1) Read the two directions. (2) Find $\vec b_1\times\vec b_2$ ⁠. (3) Remove the common factor. (4) Write the line through $A$ ⁠. (5) Check with dot products, and by a second method.

Basics you need

1Cartesian form and its direction ratios

In $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ the denominators $a, b, c$ are direction ratios, and $a\hat i + b\hat j + c\hat k$ is parallel to the line (Section 11.3.1). Basics: a line is a point plus a direction

2The cross product

$\vec b_1\times\vec b_2$ is perpendicular to both $\vec b_1$ and $\vec b_2$ ⁠, and it is $\vec 0$ only when they are parallel (Section 10.6.3). Basics: a⃗ × b⃗ stands up from both

3Perpendicular directions

Two lines are perpendicular when the dot product of their directions is 0: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ (Section 11.4, condition (i)). Lesson: perpendicular lines

4Vector form of a line

$\vec r = \vec a + \lambda\vec b$ ⁠, through the point with position vector $\vec a$ ⁠, parallel to $\vec b$ (Section 11.3.1, equation (1)). Basics: go to A, then walk λ copies of b⃗

Solution

Read the two directions
$\begin{aligned} \vec b_1 &= 3\hat i - 16\hat j + 7\hat k, \\ \vec b_2 &= 3\hat i + 8\hat j - 5\hat k. \end{aligned}$
Both equations are in standard form (each numerator is $x - \ldots$ ⁠, $y - \ldots$ ⁠, $z - \ldots$ ⁠), so the denominators are the direction ratios. The points $(8, -19, 10)$ and $(15, 29, 5)$ of the given lines are not needed: "perpendicular" is about directions only. The two directions are not parallel, since $\frac33 = 1$ but $\frac{-16}{8} = -2$ ⁠.

Cartesian equation (3), Section 11.3.1. Lesson: the equation of a line
The cross product $\vec b_1\times\vec b_2$
$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \\ &= \big((-16)(-5) - (7)(8)\big)\hat i \\ &\quad - \big((3)(-5) - (7)(3)\big)\hat j \\ &\quad + \big((3)(8) - (-16)(3)\big)\hat k \\ &= 24\hat i + 36\hat j + 72\hat k. \end{aligned}$
We need a vector $\vec b$ with $\vec b\cdot\vec b_1 = 0$ and $\vec b\cdot\vec b_2 = 0$ ⁠; the cross product is one, and it is not $\vec 0$ because $\vec b_1$ and $\vec b_2$ are not parallel. In the $\hat j$ term, $-(-15 - 21) = +36$ ⁠: the minus sign of the middle term and the negative bracket cancel. In the $\hat k$ term, $24 - (-48) = 72$ ⁠.

Determinant form of the cross product (Section 10.6.3). Basics: the cross product
Remove the common factor
$24\hat i + 36\hat j + 72\hat k = 12(2\hat i + 3\hat j + 6\hat k),$
so we take $\vec b = 2\hat i + 3\hat j + 6\hat k$ ⁠, with direction ratios $2, 3, 6$ ⁠.

24, 36 and 72 have highest common factor 12. A nonzero multiple of a direction is the same direction, so $2, 3, 6$ describes the same line with smaller numbers. Here $|\vec b| = \sqrt{4 + 9 + 36} = 7$ ⁠, so the direction cosines are $\frac27, \frac37, \frac67$ ⁠.

Direction ratios are fixed only up to a nonzero multiple (Section 11.2). Basics: direction ratios
Write the line through $(1, 2, -4)$
$\begin{aligned} \vec a &= \hat i + 2\hat j - 4\hat k, \\ \vec r &= \vec a + \lambda\vec b \\ &= \hat i + 2\hat j - 4\hat k + \lambda(2\hat i + 3\hat j + 6\hat k). \end{aligned}$
The position vector of $(1, 2, -4)$ is $\hat i + 2\hat j - 4\hat k$ ⁠. Here $\lambda$ runs through all real numbers; $\lambda = 0$ gives the point $A$ itself. The question asks only for the vector equation; in Cartesian form the same line is $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ ⁠.

Section 11.3.1, equation (1). Miscellaneous exercise player
Check, and a second method

The new direction is perpendicular to both given directions:
$\begin{aligned} \vec b\cdot\vec b_1 &= 2(3) + 3(-16) + 6(7) \\ &= 6 - 48 + 42 = 0, \\ \vec b\cdot\vec b_2 &= 2(3) + 3(8) + 6(-5) \\ &= 6 + 24 - 30 = 0. \end{aligned}$
Second method: let the direction ratios be $p, q, r$ ⁠. Perpendicular to both lines means
$\begin{aligned} 3p - 16q + 7r &= 0, \\ 3p + 8q - 5r &= 0. \end{aligned}$
Subtracting, $-24q + 12r = 0$ ⁠, so $r = 2q$ ⁠. Then $3p = 16q - 14q = 2q$ ⁠, so $p = \frac{2q}{3}$ ⁠, and $p : q : r = \frac23 : 1 : 2 = 2 : 3 : 6$ ⁠.

Oblique projection, not to scale (right angles are not shown true). From $A(1, 2, -4)$ ⁠: unit arrows along the two given directions $\vec b_1$ and $\vec b_2$ ⁠, and along the new direction $2\hat i + 3\hat j + 6\hat k$ ⁠, which is perpendicular to both.

Both dot products are 0, so the line is perpendicular to both given lines (Section 11.4, condition (i)). The second method solves the two perpendicularity conditions directly and gets the same direction $2, 3, 6$ ⁠, which confirms the cross product. Two equations in three unknowns fix only the ratio $p : q : r$ ⁠, which is all a direction needs.

Section 11.4, condition (i). Lesson: perpendicular lines

Result

$\vec r = \hat i + 2\hat j - 4\hat k + \lambda(2\hat i + 3\hat j + 6\hat k)$ ⁠, $\lambda \in \mathbf R$ ⁠.

(Textbook answer: $\vec r = \hat i + 2\hat j - 4\hat k + \lambda(2\hat i + 3\hat j + 6\hat k)$ ⁠.)

Check it with numbers

$\lambda = 0.5$ gives the point $(1 + 1, 2 + 1.5, -4 + 3) = (2, 3.5, -1)$ ⁠, and $\frac{2 - 1}{2} = \frac{3.5 - 2}{3} = \frac{-1 + 4}{6} = 0.5$ ⁠.

The unit vector along $\vec b$ is $\left(\frac27, \frac37, \frac67\right)$ $\approx (0.285714,\ 0.428571,\ 0.857143)$ ⁠; its dot product with $\vec b_1$ is $\frac{6 - 48 + 42}{7} = 0$ ⁠.

Common mistakes

Sign of the $\hat j$ term: $-\big((3)(-5) - (7)(3)\big) = -(-36) = +36$ ⁠, not $-36$ ⁠.
Using a point of a given line, such as $(8, -19, 10)$ ⁠, instead of $(1, 2, -4)$ ⁠.
Using $\vec b_1 + \vec b_2$ or $\vec b_1 - \vec b_2$ as the direction. Those are not perpendicular to both lines; only the cross product (or a multiple of it) is.
Taking the dot product instead of the cross product. $\vec b_1\cdot\vec b_2 = 9 - 128 - 35 = -154$ is a number, not a direction.

What to take away

"Perpendicular to two lines" means "parallel to $\vec b_1\times\vec b_2$ ".
Take out the common factor of the cross product to get small direction ratios.
Check by dotting the answer's direction with $\vec b_1$ and $\vec b_2$ ⁠: both must give 0.

Every fact used, and where it comes from

Fact	Used in	Source
Direction ratios from the Cartesian form; $a\hat i + b\hat j + c\hat k$ is parallel to the line	Step 1	Section 11.3.1 (Basics)
$\vec b_1\times\vec b_2$ as a determinant; it is perpendicular to both vectors	Step 2	Section 10.6.3 (Basics)
Any nonzero multiple of direction ratios serves	Step 3	Section 11.2 (Basics)
Vector equation $\vec r = \vec a + \lambda\vec b$	Step 4	Section 11.3.1, equation (1) (Basics)
Perpendicular lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$	Step 5	Section 11.4, condition (i) (Lesson)

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Miscellaneous · Q5: Vector equation of the line through (1,2,−4)(1, 2, -4)(1,2,−4) perpendicular to two given lines

Basics you need

Solution

Read the two directions

The cross product b⃗1×b⃗2\vec b_1\times\vec b_2b1​×b2​

Remove the common factor

Write the line through (1,2,−4)(1, 2, -4)(1,2,−4)

Check, and a second method

Every fact used, and where it comes from

Miscellaneous · Q5: Vector equation of the line through $(1, 2, -4)$ perpendicular to two given lines

The cross product $\vec b_1\times\vec b_2$

Write the line through $(1, 2, -4)$