Intersecting and Perpendicular Lines in 3D

This lesson works through two common Class 12 questions on straight lines in three dimensions: first showing that two given lines intersect and locating their common point, then finding the value of a parameter that makes two lines perpendicular.

Question 1: showing two lines intersect

We are given the two lines

$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \qquad (\text{line 1})$

$\frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} \qquad (\text{line 2})$

and we must show they intersect and find the point of intersection.

A general point on line 1

Set each ratio of line 1 equal to a parameter $k$:

$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = k.$

Then $x - 1 = 2k$, $y - 2 = 3k$, $z - 3 = 4k$, so any point on line 1 is

$P = (2k+1,\; 3k+2,\; 4k+3).$

Forcing the point onto line 2

The point $P$ lies on line 2 exactly when its coordinates satisfy line 2:

$\frac{(2k+1)-4}{5} = \frac{(3k+2)-1}{2} = \frac{4k+3}{1},$

which simplifies to

$\frac{2k-3}{5} = \frac{3k+1}{2} = \frac{4k+3}{1}.$

For $P$ to lie on line 2, all three ratios must agree, so we take them in pairs and check that both give the same $k$.

First pair.

$\frac{2k-3}{5} = \frac{3k+1}{2} \;\Rightarrow\; 2(2k-3) = 5(3k+1)$

$4k - 6 = 15k + 5 \;\Rightarrow\; -11k = 11 \;\Rightarrow\; k = -1.$

Second pair.

$\frac{3k+1}{2} = \frac{4k+3}{1} \;\Rightarrow\; 3k+1 = 2(4k+3)$

$3k + 1 = 8k + 6 \;\Rightarrow\; -5k = 5 \;\Rightarrow\; k = -1.$

Both pairs give $k = -1$, so the point on line 1 does lie on line 2. The two lines therefore intersect.

The point of intersection

Put $k = -1$ back into $P = (2k+1,\; 3k+2,\; 4k+3)$:

$P = (2(-1)+1,\; 3(-1)+2,\; 4(-1)+3) = (-1,\; -1,\; -1).$

The lines meet at $(-1, -1, -1)$.

Question 2: making two lines perpendicular

Find the value of $p$ so that the lines

$\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2} \qquad (\text{line 1})$

$\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \qquad (\text{line 2})$

are at right angles.

Putting each line in standard form

Direction ratios can only be read off once each line is in the form $\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$.

Line 1. Pull the minus sign out of $1-x$, and factor $7$ out of $7y-14$:

$\frac{x-1}{-3} = \frac{y-2}{\tfrac{2p}{7}} = \frac{z-3}{2},$

so its direction ratios are $\left(-3,\; \tfrac{2p}{7},\; 2\right)$.

Line 2. Pull $-7$ out of $7-7x$ and the minus out of $6-z$:

$\frac{x-1}{-\tfrac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5},$

so its direction ratios are $\left(-\tfrac{3p}{7},\; 1,\; -5\right)$.

Applying the perpendicularity condition

Two lines are perpendicular when $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$:

$(-3)\!\left(-\tfrac{3p}{7}\right) + \left(\tfrac{2p}{7}\right)(1) + (2)(-5) = 0$

$\frac{9p}{7} + \frac{2p}{7} - 10 = 0 \;\Rightarrow\; \frac{11p}{7} = 10.$

$11p = 70 \;\Rightarrow\; p = \frac{70}{11}.$

Key takeaways

• Write a general point on a line as $(x_1 + a k,\; y_1 + b k,\; z_1 + c k)$, then substitute it into the other line; if a single $k$ satisfies all the ratios, the lines intersect there.
• The common value of $k$ gives the exact point of intersection when put back into the general point.
• Always rewrite a line in standard symmetric form before reading off its direction ratios, and use $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ to find unknowns that make two lines perpendicular.