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  • Need Solution for R.D. Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 18 Maths Textbook Solution.

Need Solution for R.D. Sharma Maths Class 12 Chapter 27 Straight Line in Space  Exercise 27.2 Question 18  Maths Textbook Solution.

Answers (1)

Answer: The equation of the given question will be \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})

Given: Determine the equation of the line passing through the point (1,2,-4) and perpendicular to the two lines;

\frac{x-8}{3}=\frac{y+9}{-16}=\frac{3-0}{7} &\frac{x-15}{3}=\frac{y-29}{8}=\frac{3-5}{5}   

Hint:  \vec{r}=\vec{a}+\lambda \vec{b}(for showing parallel lines)

Solution:

Direction\: ratio _{1}=3 i-16 \hat{j}+7 \hat{k}

Direction \: ratio _{2}=3 i-8 \hat{j}-5 \hat{k}

\begin{aligned} &\therefore D R_{1} \times D R_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & -8 & -5 \end{array}\right| \\ &=24 \hat{i}+36 \hat{j}+72 \hat{k} \\ &=12(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ &\therefore D R=2,3,6 \\ &\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \end{aligned}

The equation will be \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})

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