**Example, 6**

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example, 6 Find the vector and the Cartesian equations of the line through the point (5, 2, β 4) and which is parallel to the vector 3πΒ Μ + 2πΒ Μ β 8πΒ Μ . Vector equation Equation of a line passing through a point with position vector πΒ β , and parallel to a vector πΒ β is πΒ β = πΒ β + ππΒ β Since line passes through (5, 2, β 4) πΒ β = 5πΒ Μ + 2πΒ Μ β 4πΒ Μ Since line is parallel to 3πΒ Μ + 2πΒ Μ β 8πΒ Μ πΒ β = 3πΒ Μ + 2πΒ Μ β 8πΒ Μ Equation of line πΒ β = πΒ β + ππΒ β πΒ β = (5πΒ Μ + 2πΒ Μ β 4πΒ Μ) + π (3πΒ Μ + 2πΒ Μ β 8πΒ Μ) Therefore, equation of line in vector form is πΒ β = (5πΒ Μ + 2πΒ Μ β 4πΒ Μ) + π (3πΒ Μ + 2πΒ Μ β 8πΒ Μ) Cartesian equation Equation of a line passing through a point (x, y, z) and parallel to a line with direction ratios a, b, c is (π₯ β π₯1)/π = (π¦ β π¦1)/π = (π§ β π§1)/π Since line passes through (5, 2, β4) π₯1 = 5, y1 = 2 , z1 = β4 Also, line is parallel to 3πΒ Μ + 2πΒ Μ β 8πΒ Μ , π = 3, b = 2, c = β 8 Equation of line in Cartesian form is (π₯ β π₯1)/π = (π¦ β π¦1)/π = (π§ β π§1)/π (π₯ β 5)/3 = (π¦ β 2)/2 = (π§ β ( β 4))/( β 8) (π β π)/π = (π β π)/π = (π + π)/(βπ)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.