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  • Please solve RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 9 maths textbook solution

Please solve RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 9 maths textbook solution

Answers (1)

Answer:

 Option (a)

Hint:

 Solve the given equations simultaneously.

Given:

 x+y+z+3=0,\: 2x-y+3z+1=0 \text { and } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}

Solution:

Equation of line passing through the line

x+y+z+3=0,\: 2x-y+3z+1=0

is given by,

\begin{aligned} &x+y+z+3+k(2x-y+3z+1)=0 \qquad \qquad \dots (1)\\ &x(1+2k)+y(1-k)+z(1+3k)+3+k=0 \end{aligned}

where k is constant

Again, the required plane is parallel to the line

\begin{aligned} &\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \end{aligned}

So, we have

\begin{aligned} &\left [ 1\times (1+2k) \right ]+\left [ 2\times (1-k) \right ]+\left [ 3\times (1+3k) \right ]=0\\ &1+2k+2-2k+3+9k=0\\ &9k=-6\\ &k=-\frac{2}{3} \end{aligned}

Put k in eq. (1)

\begin{aligned} &(x+y+z+3)-\frac{-2}{3}(2x-y+3z+1)=0\\ &3(x+y+z+3)-2(2x-y+3z+1)=0\\ &3x+3y+3z+9-4x+2y-6z-2=0\\ &x-5y+3z=7 \end{aligned}

Therefore, the equation of the plane through the line x + y + z + 3 = 0, 2x - y + 3z + 1 = 0 and the parallel to the line 

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}

x - 5y + 3z = 7

Posted by

Gurleen Kaur

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