First slide

Introduction to 3-D Geometry

Question

$The equation of the line of intersection of planes 4 x + 4 y - 5 z = 12 and 8 x + 12 y - 13 z = 32$ can be written as

Moderate

A

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

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

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

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

Solution

$G i v e n e q u a t i o n o f p l a n e s a r e$
$\begin{array}{l} 4 x + 4 y - 5 z = 12 - - - - - (i) \\ And 8 x + 12 y - 13 z = 32 - - - - - (ii) \end{array}$
$Let DR's of required line be (l, m, n)$
From Eq’s (i) and (ii) we get
$4 l + 4 m - 5 n = 0$
$And 8 l + 12 m - 13 n = 0$
$\begin{aligned} \Rightarrow \frac{l}{8} = \frac{m}{12} = \frac{n}{16} \\ \Rightarrow \frac{l}{2} = \frac{m}{3} = \frac{n}{4} \end{aligned}$
$Now we take intersection point with z = 0 is given by$
$\begin{array}{l} 4 x + 4 y = 12 - - - - (iii) \\ And 8 x + 12 y = 32 - - - - (iv) \end{array}$
$\begin{array}{l} On solving Equations (i) and (ii) we get the point (1, 2, 0) \\ Required line is \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z}{4} \end{array}$

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