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  • Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.13 Question 1 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 28 The Plane  Exercise 28.13 Question 1 Maths textbook Solution.

Answers (1)

Answer: \vec{r}.\left ( -\hat{i}+2\hat{j}-\hat{k} \right )=7

Hint: use vector cross product to prove

Given:\overline{r_{1}}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k}) \text { and } \vec{r}_{2}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})

Solution:

                It is given that

               \begin{aligned} &\overline{r_{1}}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k}) \\ \end{aligned}

               \begin{aligned} &\vec{r}_{2}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ \end{aligned}

    Consider:\begin{aligned} &\overline{r_{1}}=\overrightarrow{a_{1}}+\lambda \vec{b}_{1} \& \overrightarrow{r_{2}}=\overline{a_{2}}+\lambda \overline{b_{2}} \\ \end{aligned}

                     \begin{aligned} &\overrightarrow{a_{1}}=2 \hat{j}-3 \hat{k} \\ &\overrightarrow{b_{1}}=\hat{i}+2 \hat{j}+3 \hat{k} \\ &\overrightarrow{a_{2}}=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ &\overline{b_{2}}=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ \end{aligned}

It can be written as

                     \begin{aligned} &\overrightarrow{b_{1}} \times \overline{b_{2}}=\left(\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{array}\right) \\ \end{aligned}

on further calculation

                    \begin{aligned} &=\hat{i}(8-9)-\hat{j}(4-6)+\hat{k}(3-4) \\ &=-\hat{i}+2 \hat{j}-\hat{k} \\ \end{aligned}

So we get

\begin{aligned} \vec{a}_{1}\left(\vec{b}_{1} x \overline{b_{2}}\right)=(0 x(-1))+(2 x 2)+((-3) x(-1)) \\ =0+4+3=7 \end{aligned}

\begin{aligned} \vec{a}_{1}\left(\vec{b}_{1} \times \overline{b_{2}}\right)=7 \end{aligned}                                                              .....(i)

Similary :

                \begin{aligned} &\overline{a_{2}} \cdot\left(\vec{b}_{1} \times \overline{b_{2}}\right)=(2 \times(-1))+(6 \times 2)+((3) \times(-1)) \\ &=-2+12-3=7 \\ &\left.\overline{a_{2}} \cdot \vec{b}_{1} \times \overline{b_{2}}\right)=7 \\ \end{aligned}....(ii)

Using both the equation we get

                    \begin{aligned} &\left.\overrightarrow{a_{1}}\left(\vec{b}_{1} x \overline{b_{2}}\right)=\overline{a_{2}} \cdot \overrightarrow{b_{1}} x \overline{b_{2}}\right) \\ \end{aligned}

So the lines \vec{r_{1}} & \vec{r_{2}}  are coplanar

Hence the equation of plane containing  \begin{aligned} &\overline{r_{1}} \& \overrightarrow{r_{2}} \text { is } \vec{r} \cdot\left(\vec{b}_{1} \times \overline{b_{2}}\right)=\overrightarrow{a_{1}} \cdot\left(\vec{b}_{1} \times \overline{b_{2}}\right) \\ \end{aligned}

\begin{aligned} &\vec{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})=7 \\ \end{aligned}

Hence, the given lines are coplanar and the equation of the plane determined by these lines is

\begin{aligned} &\vec{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})=7 \end{aligned}

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