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Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 7

Answers (1)

Answer: $7 y+4 z-5=0$

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

Given: $(2,3,-4) \&(1,-1,3)$

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

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

So, equation of plane passing through (2,3,-4) is

$a(x-2)+b(y-3)+c(z+4)=0$ …………….…… (2)

It also passes through (1,-1,3)

$\Rightarrow-a-4 b+7 c=0 \\$

$a(1-2)+b(-1-3)+c(3+4)=0 \\$

$\begin{aligned} & &\Rightarrow a+4 b-7 c=0 \end{aligned}$ ……………….... (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 equation (2) is parallel to x- axis

$\frac{x}{1}=\frac{y}{0}=\frac{x}{0}$ …………………… (5)

Using (2) and (5) in equation (4) we get

$\begin{aligned} &a \times 1+b \times 0+c \times 0=0 \\ &\Rightarrow a=0 \end{aligned}$

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

$\begin{aligned} &\Rightarrow a-4 b+7 c=0 \\ & \end{aligned}$

$\Rightarrow 0-4 b+7 c=0 \\$

$\Rightarrow-4 b=-7 c \\$

$\Rightarrow b=\frac{7 c}{4}$

Now, putting the value of a and b in (2) we get

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

$\Rightarrow 0(x-2)+\frac{7 c}{4}(y-3)+c(z+4)=0$

$\begin{aligned} &\Rightarrow 0+\frac{7 c y}{4}-\frac{21 c}{4}+c z+4 c=0 \\ & \end{aligned}$

$\Rightarrow 7 c y-21 c+4 c z+16 c=0$

Dividing by c we have

$\begin{aligned} &7 y-21+4 z+16=0 \\ & \end{aligned}$

$\Rightarrow 7 y+4 z-5=0$

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