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)

R Ravindra Pindel

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

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