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Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 12 Maths Textbook Solution.

Answers (1)

Answer: $12 x+15 y-14 z-68=0$

Hint: Use formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$

Given: Points $(2,2,-1) \&(3,4,2)$ ratios $7, 0, 6$

Solution: We know that the equation of plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ is

$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ ……………….. (1)

So, equation of plane passing through $(2,2,-1)$ is

$a(x-2)+b(y-2)+c(z+1)=0$ ………………… (2)

It also passes through $(3,4,2)$

So, equation (2) passes through $(3,4,2)$

$\begin{aligned} &a(3-2)+b(4-2)+c(2+1)=0 \\ & \end{aligned}$

$\Rightarrow a+2 b+3 c=0$ …………………. (3)

We know that line

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$

if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …………… (4)

Here, the plane (2) is parallel to line having direction ratios 7, 0 ,6

So, $\begin{aligned} &a \times 7+b \times 0+c \times 6=0 \\ & \end{aligned}$

$\Rightarrow 7 a+6 c=0 \\$

$\Rightarrow a=\frac{-6 c}{7}$ ………………… (5)

Putting the value of a in equation (3) we get

$\begin{aligned} &a+2 b+3 c=0 \\ & \end{aligned}$

$\Rightarrow \frac{-6 c}{7}+2 b+3 c=0 \\$

$\Rightarrow 14 b+15 c=0 \\$

$\Rightarrow b=\frac{-15 c}{14}$

Putting the value of a and b in equation (2) we get

$\begin{aligned} &a(x-2)+b(y-2)+c(z+1)=0 \\ & \end{aligned}$

$\Rightarrow \frac{-6 c}{7}(x-2)+\frac{-15 c}{14}(y-2)+c(z+1)=0 \\$

$\Rightarrow \frac{-6 x c}{7}+\frac{12 c}{7}-\frac{15 c y}{14}+\frac{30 c}{14}+c z+c=0$

Multiplying by $\frac{14}{c}$ we have

$\begin{aligned} &\Rightarrow-12 x+24-15 y+30+14 z+14=0 \\ \end{aligned}$

$\Rightarrow-12 x-15 y+14 z+68=0 \\$

$\Rightarrow 12 x+15 y-14 z-68=0$

Equation of required plane is $12 x+15 y-14 z-68=0$

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