Q.Prove that the lines $x = py + q$, $z = ry + s$ and $x = p'y + q'$, $z = r'y + s'$ are perpendicular if $pp' + rr' + 1 = 0$.

Concept: Two lines are perpendicular if the dot product of their direction vectors is zero.

Steps:

1. Rewrite each line in symmetric form. From $x = py + q$, $z = ry + s$, treat $y$ as the parameter:

   $\frac{x - q}{p} = \frac{y - 0}{1} = \frac{z - s}{r}$.

   Direction vector of first line: $\vec{d}_1 = (p, 1, r)$.

2. Similarly, second line: $x = p'y + q'$, $z = r'y + s'$ gives

   $\frac{x - q'}{p'} = \frac{y - 0}{1} = \frac{z - s'}{r'}$. …

The key idea is that any line in 3D can be described by two linear equations, and the form given here — $x = py + q$, $z = ry + s$ — is a clever way of writing the line using $y$ as a parameter. This is called the symmetric form where one variable acts as the free parameter.

Let’s understand why this works. If we set $y = t$ (a parameter), then $x = pt + q$ and $z = rt + s$. So the line passes through the point $(q, 0, s)$ when $t = 0$, and its direction vector is given by the coefficients of $t$: $(p, 1, r)$. The 1 in the $y$-component comes from the fact that $y$ itself changes at rate 1 as the parameter increases.

Similarly, the second line $x = p'y + q'$, $z = r'y + s'$ has direction vector $(p', 1, r')$.

Now, two lines in space are perpendicular if and only if their direction vectors are perpendicular — that is, their dot product is zero.

1. Write the direction vector of the first line.

   From $x = py + q$, $z = ry + s$, treat $y$ as parameter.

   When $y$ increases by 1, $x$ increases by $p$, $y$ increases by 1, $z$ increases by $r$.

   So direction vector $\vec{d}_1 = (p, 1, r)$.

2. Write the direction vector of the second line.

   Similarly, $\vec{d}_2 = (p', 1, r')$.

3. Condition for perpendicularity:

   $\vec{d}_1 \cdot \vec{d}_2 = 0$ …

Method: Reading direction ratios from the form $x=py+q,\ z=ry+s$

Use this when a line is given as two equations expressing $x$ and $z$ in terms of $y$, and you must test perpendicularity or parallelism.

Steps

Step 1: Treat $y$ as the parameter.

Put $y=t$. Then $x=pt+q$, $y=t$, $z=rt+s$. As $t$ increases by $1$: $x$ changes by $p$, $y$ by $1$, $z$ by $r$. So the direction ratios are

The middle entry is $1$, NOT $0$ — $y$ is the parameter itself. …

Common Mistakes

Mistake 1: Taking the direction ratios as $(p,0,r)$.

Why it's wrong: $y$ is the parameter, so its coefficient is $1$, not $0$; the direction is $(p,1,r)$. Correct approach: from $y=t$, the $y$-rate is $1$.

Mistake 2: Forgetting the $+1$ from the two unit $y$-coefficients in the dot product.

Why it's wrong: $(p,1,r)\cdot(p',1,r')=pp'+1+rr'$, and dropping the $(1)(1)$ term loses the crucial $+1$ in the condition. Correct approach: the perpendicularity condition is exactly $pp'+rr'+1=0$. …