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