Sure Questions of 3D: Lines in Space (Part 2)

This lesson covers three frequently asked Class 12 questions on lines in three dimensions: proving two lines intersect, making two lines perpendicular, and finding direction cosines.

Showing two lines intersect

We are given the two lines

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

$\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=t$

From the first line, set each ratio equal to $k$ to write the coordinates in terms of one parameter:

$x=2k+1,\qquad y=3k+2,\qquad z=4k+3$

So the general point $(2k+1,\;3k+2,\;4k+3)$ lies on line 1. It also lies on line 2 if and only if these coordinates satisfy the second line.

Substituting into line 2

Replacing $x,y,z$ in line 2 gives

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

which simplifies to

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

Take any two of these expressions. Using the first two:

$2(2k-3)=5(3k+1)$

$4k-6=15k+5$

$-11k=11\quad\Rightarrow\quad k=-1$

Finding the point of intersection

Put $k=-1$ back into the coordinates from line 1:

$x=2(-1)+1=-1,\qquad y=3(-1)+2=-1,\qquad z=4(-1)+3=-1$

A quick check confirms $k=-1$ also gives $\tfrac{2k-3}{5}=-1$ and $\tfrac{3k+1}{2}=-1$, so all three ratios agree. The lines therefore intersect at

$P=(-1,\,-1,\,-1)$

Finding p so the lines are perpendicular

We must find $p$ so that the lines

$\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$

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

are at right angles. Neither is in standard form, so rewrite each first.

Line 1 in standard form

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

Direction ratios: $a_1=-3,\;b_1=\tfrac{2p}{7},\;c_1=2$.

Line 2 in standard form

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

Direction ratios: $a_2=-\tfrac{3p}{7},\;b_2=1,\;c_2=-5$.

Applying the perpendicularity condition

Two lines are perpendicular when $a_1a_2+b_1b_2+c_1c_2=0$:

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

$\frac{9p}{7}+\frac{2p}{7}-10=0$

$\frac{11p}{7}=10\quad\Rightarrow\quad p=\frac{70}{11}$

Direction cosines of a parallel line

Given the line

$\frac{2x-5}{4}=\frac{y+4}{3}=\frac{6-z}{6}$

find the direction cosines of a line parallel to it. First put it in standard form by factoring:

$\frac{2\left(x-\tfrac{5}{2}\right)}{4}=\frac{y-(-4)}{3}=\frac{z-6}{-6}$

Cancelling the $2$ in the first fraction gives

$\frac{x-\tfrac{5}{2}}{2}=\frac{y+4}{3}=\frac{z-6}{-6}$

So the direction ratios are $2,\;3,\;-6$, and a parallel line has the same direction ratios. The magnitude is

$\sqrt{2^2+3^2+(-6)^2}=\sqrt{4+9+36}=\sqrt{49}=7$

Dividing each ratio by $7$ gives the direction cosines

$l=\frac{2}{7},\qquad m=\frac{3}{7},\qquad n=-\frac{6}{7}$

Key takeaways

• Two lines intersect when their parametric coordinates satisfy both equations for the same parameter value; solve, then substitute back to get the point.
• Two lines are perpendicular when the dot product of their direction ratios, $a_1a_2+b_1b_2+c_1c_2$, equals zero.
• Always rewrite a line in standard form first, then divide the direction ratios by their magnitude to get the direction cosines.