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  • Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 , which contains the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y + 5 = 0.

Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 , which contains the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0.

Answers (1)

Given, a plane is perpendicular to another plane 5x + 3y + 6z + 8 = 0,and also contains line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0.

We must find the equation of this plane.
We know, the equation of a plane passing through the line of intersection of the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given as,

( a_{1}x + b_{1}y + c_{1}z + d_{1}) +\lambda(a_{2}x + b_{2}y + c_{2}z + d_{2}) = 0

Similarly, the equation of a plane through the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0. is given by,

(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0 \\ \Rightarrow x + 2y + 3z - 4 + 2\lambda x + \lambda - \lambda z + 5\lambda = 0\\ \Rightarrow x + 2 \lambda x + 2y + \lambda y + 3z - \lambda z - 4 + 5 \lambda = 0\\ \Rightarrow (1 + 2 \lambda)x + (2 + \lambda)y + (3 - \lambda)z - 4 + 5 \lambda = 0 ....(i)

Thus, the direction ratio of plane in (i) is,

(1 + 2\lambda, 2 + \lambda, 3 - \lambda)

Since the plane in equation (i) is perpendicular to the plane 5x + 3y + 6z   + 8 = 0;

 we can replace x, y, z with (1 + 2λ), (2 + λ) and (3 - λ) respectively in the plane 5x + 3y + 6z + 8 = 0 (neglecting 8) and equating to 0.

This gives us,
5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0\\ \Rightarrow 5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0\\ \Rightarrow 10\lambda + 3\lambda- 6\lambda + 5 + 6 + 18 = 0\\ \Rightarrow 7\lambda + 29 = 0\\ \Rightarrow 7\lambda = -29\\ \Rightarrow \lambda=- \frac{29}{7}

Substituting this value of \lambda in equation (i) we get

\left ( 1+2\left (-\frac{29}{7} \right ) \right )x+\left ( 2-\frac{29}{7} \right )y+\left ( 3+\frac{29}{7} \right )z-4+5\left (-\frac{29}{7} \right )=0\\ \\ \\ \Rightarrow \left ( 1+\frac{58}{7} \right )x+\left ( 2-\frac{29}{7} \right )y+\left ( 3+\frac{29}{7} \right )z-4-\frac{145}{7}=0\\ \\ \\ \Rightarrow \left ( \frac{7-58}{7} \right )x+\left ( \frac{14-29}{7} \right )y+\left ( \frac{21+29}{7} \right )z+\left ( \frac{-28-145}{7} \right )=0 \\ \\ \\ \Rightarrow -\frac{51}{7}x-\frac{15}{7}y+\frac{50}{7}z-\frac{173}{7}=0\\ \\ \\ \Rightarrow -51x - 15y + 50z - 173 = 0 \\ \\ \Rightarrow 51x + 15y - 50z + 173 = 0

Thus, the required equation of the plane is 51x + 15y - 50z + 173 = 0.

Posted by

infoexpert24

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