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  • Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (iii)

Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (iii)

Answers (1)

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Answer: Given Lines are not interesting

Hint: Consider the given equation as (1) and (2)

Given: : \frac{x-1}{2}=\frac{y+1}{3}=z \rightarrow(1) \text { and } \frac{x+1}{5}=\frac{y-2}{1} ; z=2 \rightarrow(2)

Solution: Since line (1) passes through the point (1,-1,0)  and has direction ratios proportional to (2,3,1)  its vector equation is \overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}

Here

\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}-1 \hat{\jmath}+0 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{i}+3 \hat{j}+\hat{k} \end{aligned}

Also line (2) passes through the point (-1,2,2)  and has direction ratios proportional to(5,1,0) its vector equation is\overrightarrow{\mathrm{\gamma }}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}

Here,

\begin{aligned} &\overrightarrow{a_{2}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ &\overrightarrow{b_{2}}=5 \hat{i}+1 \hat{j}+0 \hat{k} \end{aligned}

Now,

\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=-2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{array}\right| \Rightarrow-\hat{\imath}+5 \hat{\jmath}-13 \hat{k} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k}) \cdot(-\hat{\imath}+5 \hat{\jmath}-13 \hat{k}) \Rightarrow 9+15-26 \Rightarrow 9 \end{aligned}

We observe,   \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \neq 0

Thus, the given lines are not intersecting.

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