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  • Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 8

Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 8

Answers (1)

Answer:  x-19 y-11 z=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  (0,0,0) \&(3,-1,2)  and line  \frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}

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 (0,0,0) is

            a(x-0)+b(y-0)+c(z-0)=0                                

            \begin{aligned} &\\ &a x+b y+c z=0 \end{aligned}                                                                       …………………. (2)

It also passes through  (3,-1,2)

So, equation (2) must satisfy the point  (3,-1,2)

            3 a-b+2 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 is parallel to line

       \frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7} \\

So, a \times 1+b \times(-4)+c \times 7=0 \\

      \begin{aligned} & &\Rightarrow a-4 b+7 c=0 \end{aligned}                                                                           …………….… (5)

Solving equation (3) and (5) by cross multiplication we have,

            \begin{aligned} &\frac{\mathrm{a}}{-1 \times 7-(-4) \times 2}=\frac{\mathrm{b}}{1 \times 2-3 \times 7}=\frac{\mathrm{c}}{3 \times(-4)-1 \times(-1)} \\ & \end{aligned}

           \Rightarrow \frac{\mathrm{a}}{-7+8}=\frac{\mathrm{b}}{2-21}=\frac{\mathrm{c}}{-12+1}

           \begin{aligned} &\Rightarrow \frac{a}{1}=\frac{b}{-19}=\frac{c}{-11}=k \\ & \end{aligned}

           a=k, b=-19 k, c=-11 k

Putting the value in equation (2) we get

            \begin{aligned} &a x+b y+c z=0 \\\\ & \end{aligned}

            k x-19 k y-11 k z=0

Dividing by k we have

            x-19 y-11 z=0

The required equation is  x-19 y-11 z=0

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