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Explain solution RD Sharma class 12 Chapter 27 Striaght Line in Space Exercise Fill in the blanks question 18

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Answer :   3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda(-2 \hat{\imath}-5 \hat{\jmath}+13 \hat{k})

Hint : Use the vector equation

Given :     (3 , 4 , -7) and ( 1 , -1 , 6)

Solution : position vector of (3 , 4 , -7) is

                  \vec{a}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}

                  And position vector of ( 1 , -1 , 6) is

                  \vec{b}=\hat{\imath}-\hat{\jmath}+6 \hat{k}

                  Also,

                  The vector equation of a line which passes through two

                   Points whose position vectors are  \vec{a} \text { and } \vec{b}   is

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

                    Hence , required equation of line is

           \begin{aligned} &\Rightarrow \vec{r}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda[\hat{\imath}-\hat{\jmath}+6 \hat{k}-(3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k})] \\ & \end{aligned}

           \Rightarrow \vec{r}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda(-2 \hat{\imath}-5 \hat{\jmath}+13 \hat{k})

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