Find the equation of the plane passing through the point (-1,3,2) and perpendicular to the planes x+2y+3z= 5\:\; and\:\; 3x+3y+z= 0\cdot

 

 

 

 
 
 
 
 

Answers (1)

Let A,B,C be the d,r's of normal to the required plane. so A\left ( x+1 \right )+B\left ( y-3 \right )+C\left ( z-2 \right )= 0---(i)
As (i) is \perp ^{r}  to the planes x+2y+3z= 5\; \; and\; \; 3x+3y+z= 0   so, 
A+2B+3C= 0---(ii)
3A+3B+C= 0---(iii)
solving (ii) and (iii) we get \frac{A}{-7}= \frac{B}{8}= \frac{C}{-3}
ie the d.r's are 7,-8,3
by (i) \left ( 7x+1 \right )-8\left ( y-3 \right )+3\left ( z-2 \right )= 0
ie  7x-8y+3z+25= 0

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