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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 3

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 3

Answers (1)

Answer:

 Option (d)

Hint:

 \lambda is a scalar quantity.

Given:

 x+2y+3z=4 \text { and }2x+y-z=-5

are perpendicular to the plane

5x+3y+6z+8=0

Solution:

The equation of the plane through the intersection of the planes

x+2y+3z=4 \text { or } x+2y+3x-4=0 \text { and }2x+y-z=-5 \text { or } 2x+y-z+5=0

is given as,

\begin{aligned} &(x+2y+3z-4)-\lambda (2x+y-z+5)=0\\ &x(1+2\lambda )+y(2+\lambda )+z(3-\lambda )-4+5\lambda =0, \end{aligned}

where \begin{aligned} &\lambda \end{aligned} is a scalar.

Given, that the required plane is perpendicular to the plane 

\begin{aligned} &5x+3y+6z+8=0 \end{aligned}

so we have,

\begin{aligned} &5(1+2\lambda )+3(2+\lambda )+6(3-\lambda )=0\\ &5+10\lambda +6+3\lambda +18-6\lambda =0\\ &29+7\lambda =0\\ &\lambda =-\frac{29}{7} \end{aligned}

Therefore, the equation of the required plane is,

\begin{aligned} &(x+2y+3z-4)-\frac{29}{7}(2x+y-z+5)=0\\ &7(x+2y+3z-4)-29(2x+y-z+5)=0\\ &7x+14y+21z-28-58x-29y+29z-145=0\\ &-51x-15y+50z-173=0\\ &51x+15y-50z+173=0 \end{aligned}

Posted by

Gurleen Kaur

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