Question

Find the equation of the plane passing through the line of intersection of the planes x+2y+3z-5=0 and 3x-2y-z+1=0 and cutting off equal intercepts on x-axis and z-axis.

Answer 1

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Answer 2

Given :

Equations of the two intersecting planes :

x + 2y + 3z -5 =0 and 3x - 2y - z + 1 = 0

The required plane passes through the line of  intersection of these two planes and cuts off equal intercepts on X and Z axes .

To Find :

Equation of the plane passing through the intersection the given two planes and cutting off equal intercepts on X and Z axis = ?

Solution :

∴We know that for any two planes L and M equation of the plane passing through these planes is given as :

L +$\lambda$M = 0

So as per above formula the equation of required plane will be given as :$(x+2y+3x-5) +\lambda (3x-2y-z+1)=0$

$Or, (1+3\lambda) x+(2-2\lambda)y + (3-\lambda)z-(5-\lambda)=0$

$Or, \frac{(1+3\lambda)}{(5-\lambda)} x + \frac{(2-2\lambda)}{(5-\lambda)} y +\frac{(3-\lambda)}{(5-\lambda)} z=0$

Since x and z intercepts are equal :

$\frac{1+3\lambda}{5-\lambda} = \frac{3-\lambda}{5-\lambda}$

$So, 4\lambda = 2 \\\lambda = \frac{1}{2}$

So on putting the value of $\lambda$ we get the required equation as :

$(1+\frac{3}{2} )x+(2-1)y +(3-\frac{1}{2} )z = 0$

Or,$\frac{5}{2} x+ y +\frac{5}{2}z = 0$

Or, 5x + 2y = 5z = 0

So the required equation of plane is 5x + 2y + 5z = 0