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Find the equation of a straight line in the plane ` vec rdot vec n=d` which is parallel to ` vec r= vec a+lambda vec b` and passes through the foot of the perpendicular drawn from point `P( vec a)to vec rdot vec n=d(w h e r e vec ndot vec b=0)dot` a. ` vec r= vec a+((d- vec adot vec n)/(n^2))n+lambda vec b` b. ` vec r= vec a+((d- vec adot vec n)/n)n+lambda vec b` c. ` vec r= vec a+(( vec adot vec n-d)/(n^2))n+lambda vec b` d. ` vec r= vec a+(( vec adot vec n-d)/n)n+lambda vec b`

A

`vecr=veca+((d-veca.vecn)/(n^(2)))vecn+lamdavecb`

B

`vecr=veca+((d-veca.vecn)/(n))vecn+lamdavecb`

C

`vecr=veca+((veca.vecn-d)/(n^(2)))vecn+lamdavecb`

D

`vecr=veca+((veca.vecn-d)/(n))vecn+lamdavecb`

Text Solution

Verified by Experts

The correct Answer is:
a
Foot of the perpendicular from point `A(veca)` on the plane `vecr*vecn=d` is `veca+ ((d-veca*vecn))/(|vecn|^(2))vecn`
Therefore, equation of the line parallel to `vecr=veca+lamdavecb` in the plane `vecr*vecn =d` is given by
`" "vecr=veca+ ((d-veca*vecn))/(|vecn|^(2))vecn+lamdavecb`
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