9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

Answers (1)

The equation of any plane through the intersection of the planes,

3x-y+2z-4=0\ and\ x+y+z-2=0

Can be written in the form of; (3x-y+2z-4)\ +\alpha( x+y+z-2)= 0, where \alpha \epsilon R

So, the plane passes through the point (2,2,1), will satisfy the above equation.

(3\times2-2+2\times1-4)+\alpha(2+2+1-2) = 0

That implies 2+3\alpha= 0

\alpha = \frac{-2}{3}

Now, substituting the value of \alpha in the equation above we get the final equation of the plane;

(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0

(3x-y+2z-4)\ +\frac{-2}{3}( x+y+z-2)= 0

\Rightarrow 9x-3y+6z-12\ -2 x-2y-2z+4= 0

\Rightarrow 7x-5y+4z-8= 0 is the required equation of the plane.

 

 

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