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  • Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28 .14 Question 2 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.14 Question 2 maths textbook solution.

Answers (1)

Answer : 2\sqrt{29}
Hint: Formula of 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+1}{7}=\frac{y+1}{6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Solution : \frac{x+1}{7}=\frac{y+1}{6}=\frac{z+1}{1}

\vec{r}=(-\hat{\imath}-\hat{\jmath}+\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\vec{r}=\vec{a}+\lambda \vec{b}]

\vec{a}=-\hat{i}-\hat{j}-\hat{k}, \vec{b}=7 \hat{i}-6 \hat{j}+\hat{k}

Now \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{7} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\right]

\begin{aligned} &\vec{r}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+\hat{k}) \\ &\vec{a}_{2}=3 \hat{i}+5 \hat{j}+7 \hat{k}, \quad \vec{b}_{2}=\hat{i}-2 \hat{j}+\hat{k} \\ &\therefore\left(\vec{a}_{2}-\vec{a}_{1}\right)=(3 \hat{i}+5 \hat{j}+7 \hat{k})-(-\hat{i}-\hat{j}-\hat{k}) \\ &=4 \hat{i}+6 \hat{j}+8 \hat{k} \end{aligned}

\therefore \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{array}\right|

=\hat{\imath}(-6+2)-\hat{\jmath}(7-1)+\hat{k}((-14)+6))

\begin{aligned} &\vec{b}_{1} \times \vec{b}_{2}=-4 \hat{i}-6 \hat{j}-8 \hat{k} \\ &D=\left|\frac{(4 \hat{i}+6 \hat{j}+8 \hat{k}) \cdot(-4 \hat{i}-6 \hat{j}-8 \hat{k})}{\sqrt{16+36+64}}\right| \\ &D=\left|\frac{-16-36-64}{\sqrt{122}}\right| \end{aligned}

\begin{aligned} &=\frac{-116}{\sqrt{116}}=\sqrt{116}=2 \sqrt{29} \\ &d=2 \sqrt{29} \text { (answer) } \end{aligned}

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