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Let P=0 be the equation of a plane passi...

Let `P=0` be the equation of a plane passing through the line of intersection of the planes `2x-y=0a n d3z-y=0` and perpendicular to the plane `4x+5y-3z=8.` Then the points which lie on the plane `P=0` is/are a. `(0,9,17)` b. `(1//7,21//9)` c. `(1,3,-4)` d. `(1//2,1,1//3)`

A

(0,9,17)

B

`(1//7,2,1//9)`

C

(1,3,-4)

D

`(1//2,1,1//3)`

Text Solution

Verified by Experts

The correct Answer is:
a, d
The equation of the plane through the intersection of the planes `2x-y=0 and 3z-y=0` is
`" "2x-y+lamda(3z-y)=0" "`(i)
or `" "2x-y(lamda+1)+3ladmaz=0`
Plane (i) is perpendicular to `4x+5y-3z=8`. Therefore,
`" "4xx2-5(lamda+1)-9lamda=0`
or `" "8-5lamda-5-9lamda=0`
or `" "3-14lamda=0`
or `" "lamda= 3//4`
`therefore" "2x-y+ (3)/(14) (3z-y)=0`
`" "28x-17y+9z=0`
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