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Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 6

Answers (1)

The answer of the given question is 33x+45y+50z-41=0.

Hint:-  We know that, equation of a plane passing through the line of intersection of two planes

a_{1}x+b_{1}y+c_{1}z+d_{1}=0 and a_{2}x+b_{2}y+c_{2}z+d_{2}=0  is given by

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

Given:- x+2y+3z-4=0 and 2x+y-z+5=0

Solution:-  So equation of plane passing through the line of intersection of given two planes

x+2y+3z-4=0 and 2x+y-z+5=0 is given by
\left (x+2y+3z-4 \right )+k\left (2x+y-z+5 \right )=0\\ x+2y+3z-4+2kx+ky-kz+5k=0\\ x(1+2k)+y(2+k)+z(3-k)-4+5k=0                                    … (i)

We know that, two planes are perpendicular if

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0                                                                       … (ii)

Given, plane (i) is perpendicular to plane,

5x+3y-6z+8=0                                                                             … (iii)

Using (i) and (iii) in equation (ii)

5(1+2k)+3(2+k)+-6(3-k)=0\\ 5+10k+6+3k-18+6k=0\\ k=\frac{7}{19}

Putting the value of k in equation (iii)
\begin{aligned} &x\left(1+\frac{14}{19}\right)+y\left(2+\frac{7}{19}\right)+z\left(3-\frac{7}{19}\right)-4+\frac{35}{19}=0 \\ \end{aligned}

\begin{aligned} &x \frac{(19+14)}{19}+y\left(\frac{38+7}{19}\right)+z\left(\frac{57-7}{19}\right)+\frac{-76+35}{19}=0 \\ \end{aligned}

\begin{aligned} &\qquad x\left(\frac{33}{19}\right)+y\left(\frac{45}{19}\right)+z\left(\frac{50}{19}\right)-\frac{41}{19}=0 \end{aligned}

Multiplying with 19 we get

33x+45y+50z-41=0

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