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  • Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28 .14 Question 1 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.14 Question 1 maths textbook solution.

Answers (1)

Answer : \frac{65}{\sqrt{122}}

Hint : Distance between shortest lines

D=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b_{2}}\right|}\right|

Given : Find the shortest distance between lines \frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3} and \frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}

Solution : \frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}

\vec{r}=(2 \hat{\imath}+5 \hat{\jmath})+\lambda(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\vec{r}=\vec{a}+\lambda \vec{b}]

\begin{aligned} &\vec{a}_{1}=2 \hat{i}+5 \hat{j} \\ &\vec{b}_{1}=-\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}

\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\right]

\begin{aligned} &\vec{r}=(-\hat{\jmath}+\hat{k})+\mu(2 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &\vec{a}_{2}=-\hat{j}+\hat{k}, \vec{b}_{2}=2 \hat{i}-\hat{j}+2 \hat{k} \\ &D=\left|\frac{\left.\overrightarrow{(a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right| \end{aligned}

\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & -1 & 2 \end{array}\right|

\begin{aligned} &=\hat{i}(4+3)-\hat{j}(-2-6)+\hat{k}(1-4)=7 \hat{i}+8 \hat{j}-3 \hat{k} \\ &\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{7^{2}+8^{2}+(-3)^{2}}=\sqrt{49+64+9}=\sqrt{122} \\ &D=\left|\frac{(-2 \hat{\imath}-6 \hat{\jmath}+\hat{k}) \cdot(7 \hat{\imath}+8 \hat{\jmath}-3 \hat{k})}{\sqrt{122}}\right| \\ &=\left|\frac{-14-48-3}{\sqrt{122}}\right|=\frac{|-65|}{\sqrt{122}} \\ &D=\frac{65}{\sqrt{122}} \end{aligned}

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