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Please solve RD Sharma class 12 chapter The Plane exercise 28.3  question 5 maths textbook solution

Answers (1)

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Answer:

\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+4 \hat{k})=7, \quad 2 x-y+4 z=7

Hint:

Equation of plane, a(x-1)+b(y+1)+c(z-1)=0

Given:

Plane passing through the point (1,-1,1) and normal to the line joining the points (1,2,5) and (-1,3,1).

Solution:

Equation of plane, a(x-1)+b(y+1)+c(z-1)=0

Where a, b, c are direction ratio

A = (1,2,5) and B = (-1,3,1)

A B=\frac{x}{-2}=\frac{y}{1}=\frac{z}{-4}

(-2, 1, -4) are direction ratio.

Direction ratio of the plane will be equal to direction ratio of the line joining two point

\therefore a=-2, b=1, c=-4

Put the value of a, b, c in equ (1)

Equ of plane

\begin{aligned} &-2(x-1)+1(y+1)+(-4)(z-1)=0 \\\\ &\Rightarrow-2 x+2+y+1-4 z+4=0 \\\\ &\Rightarrow-2 x+y-4 z=-7 \end{aligned}

Vector equation

\vec{r} \cdot(2 \hat{\imath}-\hat{\jmath}+4 \hat{k})=7, \quad[\therefore \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}]

For cartesian equation, substitute \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}

\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(2 \hat{\imath}-\hat{\jmath}+4 \hat{k})=7 \\\\ &2 \mathrm{x}-\mathrm{y}+4 \mathrm{z}=7 \end{aligned}

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