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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 12 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 12 Maths Textbook Solution.

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Answer: the equation will be \frac{x-2}{1}=\frac{y+1}{2}=\frac{z+1}{3} which is parallel to \vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})

Given: find the vector equation of the line passing through the point (2,-1,-1) parallel to line 6x-2=3y+1=2z-2

Hint: for vector equation, \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

Solution: 6x-2=3y+1=2z-2

              6(x-1 / 3)=3(y+1 / 3)=2(z-1) \\

               =\frac{x-1 / 3}{1 / 6}=\frac{y+1 / 3}{1 / 3}=\frac{z-1}{1 / 2} \\

     D R=1 / 6,1 / 3,1 / 2=1,2,3

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

            \therefore \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} \\

            \vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k}) \\

So, the vector equation will be

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

 

 

 

 

 

 

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