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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 16 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space  Exercise 27.2 Question 16  Maths Textbook Solution.

Answers (1)

Answer: The equation will be \begin{aligned} &\vec{r}=(\hat{i}+\hat{j}-3 \hat{k})+\lambda(4 \hat{i}-5 \hat{j}+\hat{k}) \\ \end{aligned}

Given: Find the equation of the line passing through the pointi+j-3k and perpendicular to the lines \begin{aligned} &\vec{r}=\hat{i}+\lambda(2 \hat{i}+\hat{j}-3 \hat{k}) \& \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(\hat{i}+\hat{j}+\hat{k}) \\ \end{aligned}

Hint:   \begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \end{aligned}

Solution:

\vec{b}=\vec{b}_{1} \times \overrightarrow{b_{2}} \rightarrow b_{1}=2 \hat{i}+\hat{j}-3 \hat{k} \& b_{2}=\hat{i}+\hat{j}+\hat{k}

\vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right| \\

=\hat{i}(1+3)+\hat{j}(-3-2)+\hat{k}(2-1)

= 4 \hat{i}-5 \hat{j}+\hat{k}

\\ \therefore \vec{r}=(\hat{i}+\hat{j}-3 \hat{k})+\lambda(4 \hat{i}-5 \hat{j}+\hat{k)}

 

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