Question

Find the equations of the two planes passing through the points (0, 4, -3) and (6,-4, 3),
if the sum of their intercepts on the three axes is zero.​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Given : two points (0, 4, -3) and (6, -4, 3).

To find :  two equation of plane passing through (0,4,-3) and (6,-4,3).

Step-by-step explanation:

Let the equation of the plane is,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\hfill (1)$

where a, b, c are the intercepts with three axis of which sum in this problem is zero, that is,

$a+b+c=0\hfill (2)$

Now since (0,4,-3) and (6,-4,3) are lying on (1) so it satisfied that is,

$\frac{4}{b}-\frac{3}{c}=1\hfill (3)$

$\frac{6}{a}-\frac{4}{b}-\frac{3}{c}=1\hfill (4)$

Adding (3) and (4) we get,

$\frac{6}{a}=2\implies a=3$

From (2) we get,

$a+b+c=0\implies b+c=-a\implies -c=b+3$

From (2),

$\frac{4}{b}+\frac{3}{-b}=1\implies \frac{4}{b}+\frac{3}{b+3}=1$

$\implies b^2-4b-12=0$

$\implies(b-6)(b+2)=0$

$\therefore b=6, -2$ so that c=-3-b=-9,-1 respectively.

Hence two equation of planes are,

$\frac{x}{3}+\frac{y}{6}-\frac{z}{9}=1$

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