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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (v) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (v) maths textbook solution

Answers (1)

Answer: d=\frac{5}{2} \sqrt{2} \text { units }

Hint: Convert the given equation in the normal form \overrightarrow{\mathrm{\gamma }}=\vec{a}+\lambda \vec{b}

\begin{aligned} \text { Let } \overrightarrow{\mathrm{\gamma}} &=a_{1}+\lambda b_{1} \\\\ \overrightarrow{\mathrm{\gamma}} &=a_{2}+\mu b_{2} \end{aligned}

Given: \vec{\gamma}=(\lambda-1) \hat{\imath}+(\lambda+1) \hat{\jmath}-(1+\lambda) \hat{k} \text { and } \overrightarrow{\mathrm{\gamma}}=(1-\mu) \hat{\imath}+(2 \mu-1) \hat{\jmath}+(\mu+2) \hat{k}

Solution:

\begin{aligned} &\vec{\gamma}=(-\hat{\imath}+\hat{\jmath}-\hat{k})+\lambda(\hat{\imath}+\hat{\jmath}-\hat{k}) \text { and } \\\\ &\vec{\gamma}=(\hat{\imath}-\hat{\jmath}+2 \hat{k})+\mu(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \\\\ &D=\frac{\left|\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)\right|}{\left|\left(b_{1} \times b_{2}\right)\right|} \end{aligned}

Here

\begin{array}{ll} \overrightarrow{a_{1}}=-\hat{\imath}+\hat{\jmath}-\hat{k} & \overrightarrow{b_{1}}=\hat{\imath}+\hat{\jmath}-\hat{k} \\\\ \overrightarrow{a_{2}}=\hat{\imath}-\hat{\jmath}+2 \hat{k} & \overrightarrow{b_{2}}=-\hat{\imath} \\\\ \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=2 \hat{i}-2 \hat{\jmath}+3 \hat{k} & \end{array}

\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & -1 \\ -1 & 2 & 1 \end{array}\right|

\begin{aligned} \overrightarrow{b_{1}} \times \overrightarrow{b_{2}} &=\hat{\imath}(1+2)-\hat{\jmath}(-1-1)+\hat{k}(2+1) \\\\ &=3 \hat{\imath}+3 \hat{k} \end{aligned}

\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(3)^{2}+(3)^{2}} \Rightarrow 3 \sqrt{2} \\\\ &\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)=(2 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}) \cdot(3 \hat{\imath}+3 \hat{k}) \Rightarrow 6+9 \Rightarrow 15 \\\\ &D=\frac{15}{3 \sqrt{2}} \Rightarrow \frac{15}{\sqrt{2}} \Rightarrow \frac{5}{2 \sqrt{2}} \end{aligned}

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