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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 2

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 2

Answers (1)

Answer: Correct answer is A.

Hint: Use vector cross product.

 

Given:: \frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}

 

Solution: The equation of the given lines are

\begin{aligned} &\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \rightarrow(1) \\\\ &\text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \\\\ &\Rightarrow \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \rightarrow(2) \end{aligned}
Thus the two lines are parallel to the vector \vec{b}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} and pass through the points (0,0,0)and (1,2,3).

Now,

\begin{aligned} &\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right) \times \vec{b} \\\\ &=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \times(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\\\ &=\overrightarrow{0} \\\\ &\Rightarrow(\vec{a} \times \vec{a}=\overrightarrow{0}) \end{aligned}

Since the distance between the two parallel lines is 0, the given lines are coincident.

So, the correct option is (a) coincident.

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