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Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Fill in the blanks Question 19

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Answer :   (5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})

Hint : Compare the equation with the formula

Given :    \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}  

Solution :    We have given line as

                   \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}

                     By comparing with equation

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

                    We get given line passes through the point (x1 , x2, x3)

                    i.e. (5 , -4 , 6) and direction ratios are (a ,b ,c) i.e. ( 3 , 7 , 2)

                    Now , we can write vector equation of the line as

                 \vec{A}=(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})

 

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