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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 15 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 15  Maths Textbook Solution.

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Answer: the equation of the line will be \vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k})

Given: find the equation of the line passing through the point (2,1,3) and perpendicular to the

\begin{aligned} &\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-3} \text { and } \frac{x}{-3}=\frac{y}{2}=\frac{z}{5} \\ \end{aligned}

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

Solution:

\begin{aligned} &\vec{b}=\vec{b}_{1} \times \overline{b_{2}} \\ &\vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{array}\right| \\ &=\hat{i}(10-6)-\hat{j}(5+9)+\hat{k}(2+6) \\ &=4 \hat{i}-14 \hat{j}+8 \hat{k} \\ &\vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k}) \end{aligned}

So, the equation of the line will be \vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k})

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