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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 18

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 18

Answers (1)

Answer: Correct answer is d.

Hint: Use vector cross product.

Given:   L_{1}: x=5, \quad \frac{y}{3-2}=\frac{z}{-2}, \quad L_{2}: x=2, \quad \frac{y}{-1}=\frac{z}{2-d}

Solution:

\begin{aligned} &x=5, \frac{y}{3-2}=\frac{z}{-2} \ldots \ldots \ldots . . . \text { (i) }\\ \\ &x=2, \frac{y}{-1}=\frac{z}{2-2}\ldots \ldots \ldots . . . \text { (ii) } \end{aligned}

∴  Parallel vectors b1  and  b2 of line (i) and (ii) are:

\overrightarrow{b_{1}}=0 \hat{\imath}+(3-2) \hat{\jmath}-2 \hat{k}, \overrightarrow{b_{2}}=0 \hat{\imath}-\hat{\jmath}+(2-2) \hat{k}

\begin{aligned} &\text { Lines (i) } \perp(i i) \Rightarrow \overrightarrow{b_{1}} \perp \overrightarrow{b_{2}} \\\\ &\Rightarrow \overrightarrow{b_{1}} \overrightarrow{b_{2}}=0 \\\\ &\Rightarrow(0 \times 0) \hat{\imath}+\{(3-2)(-1)\} \hat{j}+\{(-2)(2-2)\} \hat{k}=0 \end{aligned}

\begin{aligned} &\Rightarrow d 3+2 x-4=0 \\\\ &\Rightarrow 3 x-7=0 \\\\ &\Rightarrow d=7 / 3 \end{aligned}

And, hence the correct option is (D).

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