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  • 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.

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.

Answers (1)

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.

Posted by

infoexpert24

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