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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.5 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.5 question 5 maths textbook solution

Answers (1)

Answer:

 The required vector equation of plane is 9x + 2y - 7z - 15 = 0

Hint:

 Using

\frac{a}{c_2b_1-b_1c_2}=\frac{b}{a_2c_1-a_1c_2}=\frac{c}{b_2a_1-b_1a_2}

Given:

 Evaluation of the plane passing through the points

3\hat{i}+4\hat{j}+2\hat{k}, \; 2\hat{i}-2\hat{j}-\hat{k},\; 7\hat{i}+6\hat{k}

Solution:

Co-ordinates of the given plane are A (3, 4, 2), B (2, -2, -1) and C (7, 0, 6)

General equation of plane is given by-

\begin{aligned} &a(x-3)+b(y-4)+c(z-2)=0 \qquad \qquad \dots (i)\\ &a(2-3)+b(-2-4)+c(-1-2)=0\\ &a(-1)+b(-6)+c(-3)=0\\ &a+6b+3c=0 \qquad \qquad \dots (ii) \end{aligned}

Again putting point C in general equation

\begin{aligned} &a(7-3)+b(0-4)+c(6-2)=0\\ &4a-4b+4c=0\\ &a-b+c=0 \qquad \qquad \dots (ii) \end{aligned}

Let’s cross multiply;

\begin{aligned} &\frac{a}{6+3}=\frac{b}{3-1}=\frac{c}{-1-6}\\ &\frac{a}{9}=\frac{b}{2}=\frac{c}{-7}=\lambda \\ &\therefore a=9\lambda ,\; b=2\lambda ,\: c=-7\lambda \end{aligned}

Substituting a, b and c by their values in equation (i)

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

Divide by λ

\begin{aligned} &\therefore 9x+2y-7z-21=0 \end{aligned}

Posted by

Gurleen Kaur

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