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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 26 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 26 Maths Textbook Solution.

Answers (1)

Answer: the required vector equation is \vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(-4 \hat{i}+4 \hat{j}-\hat{k})

 

Given: find the vector and Cartesian equation of the line which is perpendicular to the lines with the equation

\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{2}and \: \frac{x-1}{2}=\frac{y-2}{3}

                =         \frac{z-3}{4}     and passes through the points (1,1,1). Also find the angle between the given lines

Hint: for showing perpendicular dot product must be equal to ‘0’

Solution: let the cartesian equation of the line passing through the point (1,1,1) will be

\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}

Now parallel vectors are  \bar{b_{1}},\bar{b_{2}},\bar{b_{3}}                                                                                                                                                       

From the question,

 \begin{aligned} &\vec{b}_{1}=\hat{a} \hat{i}+b \hat{j}+c \hat{k} \\ &\overrightarrow{b_{2}}=\hat{i}+2 \hat{j}+4 \hat{k} \text { Here } \\ &\overrightarrow{b_{3}}=2 \hat{i}+3 \hat{j}+4 \hat{k} \end{aligned}

                 \begin{aligned} &a+2 b+4 c=0 \ldots .(\mathrm{lV}) \\ &2 a+3 b+4 c=0 \ldots \ldots(\mathrm{V}) \end{aligned}

                  Solving (iv) and (v) we get

                   \begin{aligned} &\frac{a}{8-12}=\frac{b}{8-4}=\frac{c}{3-4} \\ &\therefore \frac{a}{-4}=\frac{b}{4}=\frac{c}{-1}=\lambda \end{aligned} 

               Therefore a=4\lambda ,b=4\lambda ,c=-\lambda

Now putting the above makes in equation;

  \begin{aligned} &\frac{x-1}{-4 \lambda}=\frac{y-1}{4 \lambda}=\frac{z-1}{-\lambda} \\ &\frac{x-1}{-4}=\frac{y-1}{4}=\frac{z-1}{-1} \end{aligned}           hence the vector equation is \vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(-4 \hat{i}+4 \hat{j}-\hat{k}).

 

 

 

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