Find the vector and Cartesian equations of the plane passing through the points(2, 2 –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.

 

 

 

 
 
 
 
 

Answers (1)

Given

Let A(2,2,-1), B(3,4,2) & C(7,0,6)

The equation of plane passsing through A(2,2,-1)

a(x -2) +b(y -2) + c(z +1) = 0 \quad - (i)

\because (3,7,2) & (7,0,6) lie of the plane

        a + 2b + 3c = 0 \quad -(ii)

        5a - 2b + 7c = 0 \quad -(iii)

Solving equation (ii) and (iii), we get

    \frac{a}{14 + 6} = \frac{-b}{7-15} = \frac{c}{-2-10} = k \ (\text{assumption})

    \frac{a}{20} = \frac{-b}{-8} = \frac{c}{-12} = k

    putting the value of a, b and c in (i) we get

20k(x-2) + 8k(y-2)-12k(z + 1)=0 \\\Rightarrow 4k[5k-10 + 2y -4 -3z -3]= 0 \\\Rightarrow 5x +2y -3z - 17 = 0

This is the required equation of the plane.

Now, the second plane passes through the point (4,3,1)

\because this plane is parallel to the above plane,

\thereffore\therefore DR of the second plane be <5,2,-3>

So, the equation of the second plane

5(x-4) +2(y-3) -3(z-1) = 0 \\\Rightarrow 5x - 20 +2y -6 -3x + 3 =0 \\\Rightarrow 5x + 2y -3z -23 = 0

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