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The vector equation of the straight line...

The vector equation of the straight line `(1-x)/(3)=(y+1)/(-2)=(3-z)/(-1)` is

A

`r=(hati-hatj+3hatk)+lambda(3hati+2hatj-hatk)`

B

`r=(hati-hatj+3hatk)+lambda(3hati-2hatj-hatk)`

C

`r=(3hati-2hatj-hatk)+lambda(hati-hatj+3hatk)`

D

`r=(3hati+2hatj-hatk)+lambda(hati-hatj+3hatk)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given line is `(x-1)/(-3)=(y+1)/(-2)=(z-3)/(1)`
`Rightarrow` Line is passing through (1,-1,3) and having direction ratios (-3,-2,1) i.e, (3,2,-1)
`therefore` Vector equation of the line is `tau=(hati-hatj+3hatk)+lambda(3hati+2hatj-hatk)`
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