HSC Science (General) 12th Board ExamMaharashtra State Board
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# Solution for The Cartesian Equations of Line Are 3x -1 = 6y + 2 = 1 - z. Find the Vector Equation of Line. - HSC Science (General) 12th Board Exam - Mathematics and Statistics

#### Question

The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.

#### Solution

Given equations of the line are:
3x -1 = 6y +2 = 1 - z
Rewriting the above equation, we have,

3(x-1/3)=6(y+2/6)=-(z-1)

((x-1/3))/(1/3)=((y+1/3))/(1/6)=((z-1))/-1 ....(1)

Now consider the general equation of the line:

(x-a)/l=(y-b)/m=(z-c)/n...(2)

where, l,m and n are the direction ratios of the line and the point (a,b,c) lies on the line.
Compare the equation (1), with the general equation (2),

we have l=1/3, m=1/6 and n=-1

Also, a=1/3, b=-1/3 and c=1

This shows that the given line passes through (1/3, -1/3, 1)

Therefore, the given line passes through the point having

position vector bara=1/3hati-i/3hatj+hatk and is parallel to the
vector barb=1/3hati+1/6hatj-hatk

So its vector equation is

barr=(1/3hati-1/3hatj+hatk)+lambda(1/3hati+1/6hatj-hatk)

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#### APPEARS IN

2014-2015 (March) (with solutions)
Question 1.2.4 | 2.00 marks

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Solution for question: The Cartesian Equations of Line Are 3x -1 = 6y + 2 = 1 - z. Find the Vector Equation of Line. concept: Equation of a Line in Space. For the courses HSC Science (General) , HSC Arts, HSC Science (Computer Science), HSC Science (Electronics)
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