AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

The equation of plane joining points (

^

i

+

^

j

) and (3

^

i

+

^

j

−

^

k

) in Cartesian form is,

x−1

2

=

y−1

0

=

z

−1

=r
Then, point on above line is p(2r+1,1−r).
Now the equation of the plane passing through (2

^

i

+4

^

j

) andparallel to the vector (3

^

j

+5

^

k

) and (3

^

i

−

^

k

) in Cartesian form is,
|

x−2	y−4	z
0	3	5
3	0	−1

|=0
−3(x−2)+15(y−4)−9z=0
3x−15y+9z+54=0
Let the point p(2r+1,1−r) on the above plane itself then,
r=14
So, the position vector of point P is,
OP=29

^

i

+

^

j

−14

^

k