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  • Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 9

Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 9

Answers (1)

Answer:  \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+k(-3 \hat{i}+5 \hat{j}+4 \hat{k})

                        or

               \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}

Hint: Use formula  \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}

Given: Point \left ( 1,2,3 \right )   and planes

            \begin{aligned} &\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5 \\ & \end{aligned}

            \vec{r} \cdot(3 \hat{i}+\hat{j}+2 \hat{k})=6

Solution: We know that, the equation of line passing through \left ( 1,2,3 \right )  is given by

            \frac{x-1}{a_{1}}=\frac{y-2}{b_{1}}=\frac{z-3}{c_{1}}                            ……………….. (1)

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                     ………………… (2)

Here, line (1) is parallel to plane

       x-y+2 z=5 \\

So, a \times 1+b \times(-1)+c \times 2=0 \\

      \begin{gathered} \Rightarrow a-b+2 c=0 \end{gathered}                                        …………………. (3)

Also, line (1) is parallel to plane,

            3 x+y+z=6

So,  a \times 3+b \times(1)+c \times 1=0 \\   

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

Solving equation (3) and (4) by cross multiplication

We have,

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

             \Rightarrow \frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3} \\

             \Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=k \\

             a=-3 k, b=5 k, c=4 k

Putting the value in equation (1) we have

            \frac{x-1}{-3 k}=\frac{y-2}{5 k}=\frac{z-3}{4 k}

Multiplying by k we have

            \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}

The required equation is

            \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\\ 

                        or

            \begin{aligned} & & &\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+k(-3 \hat{i}+5 \hat{j}+4 \hat{k}) \end{aligned}

 

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