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# The Cartesian Equation of the Line is 2x - 3 = 3y + 1 = 5 - 6z. Find the Vector Equation of a Line Passing Through (7, –5, 0) and Parallel to the Given Line. - ISC (Commerce) Class 12 - Mathematics

ConceptVector and Cartesian Equation of a Plane

#### Question

The Cartesian equation of the line is 2x - 3 = 3y + 1 = 5 - 6z. Find the vector equation of a line passing through (7, –5, 0) and parallel to the given line.

#### Solution

Cartesian equation of a line is

2x - 3 = 3y + 1= 5 - 6z

i.e 2(x- 3/2) = 3(y + 1/3) = -6(z - 5/6)

Dividing by –6 throughout

i.e  (x - 3/2)/(-3) = (y + 1/3)/(-2) = (2-5/6)/1

∴ D.r.s of the above line is –3, –2, 1

Now, equation of a line passing through point (7, –5, 0) and parallel to the above line whose d.r.s. is –3, –2, 1 is

vecr = (7hati - 5hatj )+ lambda(-3hati -3hatj +  hatk)

:. vecr = (7hati - 5hatj) + lambda (3hati + 2hatj -  hatk)

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Solution The Cartesian Equation of the Line is 2x - 3 = 3y + 1 = 5 - 6z. Find the Vector Equation of a Line Passing Through (7, –5, 0) and Parallel to the Given Line. Concept: Vector and Cartesian Equation of a Plane.
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