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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.3 question 21 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.3  question 21 maths textbook solution

Answers (1)

Answer:

The answer of given question is \vec{r} \cdot(4 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})=12

Hint:

Let the equation be Ax+By+Cz+D=0

Given:

Plane with intercepts 3,-4 and 2 on x, y and z axes

Solution:

Let the equation be Ax+By+Cz+D=0….(i)

Let the plane meets the  x,y,z axes (3,0,0), (0,-4,0) and (0,0,2) respectively.

Putting (0,0,2), we get

\begin{aligned} &A(0)+B(0)+C(2)+D=0 \\ &\Rightarrow C=-\frac{D}{2} \end{aligned}

Putting (0,-4,0), we get

\begin{aligned} &A(0)+B(-4)+C(0)+D=0 \\ &\Rightarrow B=\frac{D}{4} \end{aligned}

And putting (3,0,0) we get

\begin{aligned} &A(3)+B(0)+C(0)+D=0 \\ &\Rightarrow A=-\frac{D}{3} \end{aligned}

Substituting the values of A, B, C in equation (i) we get by putting B(0,-4,0), we get

\begin{aligned} &\left(-\frac{D}{3}\right) x+\frac{D}{4} y+\left(-\frac{D}{2}\right) z+D=0 \\ &\Rightarrow \frac{-4 D x+3 D y-6 D z+12 D}{2}=0 \\ &\Rightarrow(-D)(4 x-3 y+6 z-12)=0 \\ &\Rightarrow 4 x-3 y+6 z=12 \end{aligned}

This is the cartesian form of equation of the required plane.

Now vector equation form of the plane4x-3y+6z=12

\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(4 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})=12 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})=12 \end{aligned}

 

Posted by

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