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  • Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 6 sub question (ii)

Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 6 sub question (ii)

Answers (1)

Answer: The given line do not intersect.

 

Hint: If  \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=0   then the lines intersect each other.

 

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

 

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

The first line passes through the point (1,-1,0) and has direction ratio proportional to 2,3,1 and its vector equation is

\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}               …(i)

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

Also, the second line passes through the point (-1,2,2) and has direction ratios proportional to 5,1,0.

Its vector equation  is

\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}                ...........(ii)

Here,

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

Now,

\overrightarrow{a_{2}}-\overrightarrow{a_{1}}=-2 \hat{i}+3 \hat{j}+2 \widehat{k}

And

\begin{aligned} &\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{array}\right| \\\\ &=-\hat{\imath}+\overline{5}-13 \hat{k} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2 \hat{i}+3 \hat{j}+2 \widehat{k}) \cdot(-\hat{i}+5 \hat{j}-13 \widehat{k}) \\ &=2+15-26 \\ &=-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 don't intersect.

Posted by

infoexpert26

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