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  • Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 19

Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 19

Answers (1)

Answer: Therefore, required equation of plane is  8 x-13 y+15 z+13=0

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

Given:  \frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}

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  \left ( 3,4,1 \right ) is

            a(x-3)+b(y-4)+c(z-1)=0                                 ……………… (2)

It also passes through \left ( 0,1,0 \right )

So, equation (2) must satisfy the point  \left ( 0,1,0 \right )

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

            \Rightarrow-3 a-3 b-c=0 \\

            \Rightarrow 3 a+3 b+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

            \begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\ & \end{aligned}                                                   ……………… (4)

So,       a \times 2+b \times 7+c \times 5=0 \\

            \Rightarrow 2 a+7 b+5 c=0                                                      ………………. (5)

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

            \begin{aligned} &\frac{\mathrm{a}}{3 \times 5-(7) \times 1}=\frac{\mathrm{b}}{2 \times 1-3 \times 5}=\frac{\mathrm{c}}{3 \times 7-2 \times 3} \\ & \end{aligned}

            \Rightarrow \frac{\mathrm{a}}{15-7}=\frac{\mathrm{b}}{2-15}=\frac{\mathrm{c}}{21-6}

            \begin{aligned} &\Rightarrow \frac{a}{8}=\frac{b}{-13}=\frac{c}{15}=k(s a y) \\ \end{aligned}

                 a=8 k, b=-13 k, c=15 k

Putting the value in equation (2) we get

            \begin{aligned} &8 k(x-3)-13 k(y-4)+15 k(z-1)=0 \\ & \end{aligned}

            8 k x-24 k-13 k y+52 k+15 k z-15 k=0

Dividing by k we have

            8 x-13 y+15 z+13=0

Equation of required plane is  8 x-13 y+15 z+13=0

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