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

Given: x = py + q, z = ry + s, and x = p’y + q’, z = r’y + s’ are perpendicular.

To Prove: pp’ + rr’ + 1 = 0.

Proof:

Let us take x = py + q and z = ry + s.

From x = py + q;

py = x - q

$\Rightarrow y=\frac{x-q}{p}$

From z = ry + s;

ry = z - s

$\Rightarrow y=\frac{z-s}{r}$

So, $\frac{x-q}{p}=y=\frac{z-s}{r}$

$\frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r}$            Or, … (i)

Now, if we take x = p’y + q’ and z = r’y + s’

From x = p’y + q’;

p’y = x - q’

$\Rightarrow y=\frac{x-{q}'}{{p}'}$

From z = r’y + s’;

r’y = z - s’

$\Rightarrow y=\frac{z-{s}'}{{r}'}$

So,

$\frac{x-{q}'}{{p}'}=y=\frac{z-{s}'}{{r}'}$

Or,

$L_{2}:\frac{x-{q}'}{{p}'}=\frac{y}{1}=\frac{z-{s}'}{{r}'}.......(ii)$

From (i),

Line L1 is parallel to $p\hat{i}+\hat{j}+r\hat{k}$  (from the denominators of the equation (i))

From (ii),

Line L2 is parallel to ${p}'\hat{i}+\hat{j}+{r}'\hat{k}$ (from the denominators of the  equation (ii))

According to the question, L1 and L2 are perpendicular.

Therefore, the dot product of the vectors should equate to 0.

Or,

$\left (p\hat{i}+\hat{j}+r\hat{k} \right ).\left ({p}'\hat{i}+\hat{j}+{r}'\hat{k} \right )\\ \Rightarrow p{p}'+1+r{r}'=0$

(since, in vector dot product, $\left (x\hat{i}+y\hat{j}+z\hat{k} \right )\left ({x}'\hat{i}+{y}'\hat{j}+{z}'\hat{k} \right )= x{x}'+y{y}'+z{z}'=0$

Or,

$p{p}'+r{r}'+1=0$

Therefore, the lines are perpendicular if pp’ + rr’ + 1 = 0.