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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 2 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 2  maths textbook solution

Answers (1)

Answer :-

\vec{r}=(-\hat{\imath}+2 \hat{k})+\lambda(4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})

Hint :-

\vec{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})

Given :-

Points (-1 , 0 , 2) and (3 , 4 , 6)

Solution :-

Position Vector of the point ( -1 , 0 , 2)

\vec{a}=-\hat{i}+2\hat{k}

Position Vector of the point (3 , 4 , 6)

\vec{b}=-3\hat{i}+4\hat{j}+6\hat{k}

Now equation of line passing through the point where position vector are a and b

\begin{aligned} \vec{r} &=\vec{a}+\lambda(\vec{b}-\vec{a}) \\ \vec{r} &=(-\hat{\imath}+2 \hat{k})+\lambda(3 \hat{\imath}+4 \widehat{\jmath}+6 \hat{k}+\hat{\imath}-2 \hat{k}) \\ &=(-\hat{\imath}+2 \hat{k})+\lambda(4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k}) \end{aligned}

 

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