TS EAMCET 5 May 2018 Shift 1 Solved Paper

The points in Cartesian form are,
(2,1,1),(-1,-1,1) and (-1,1,-1)
The equation of plane passing through three points is,
|

x−2	y−1	z−1
−1	−2	0
−3	0	−2

| =0
Solve the above determinat then,
2x−y−3z=0
The above equation can be expressed in vector form as,
r.(2

^

i

−

^

j

−3

^

k

)=0
If e is unit vector perpendicular to given plane then,
e=

2 ^ i − ^ j −3 ^ k

√22+(−1)2+(−3)2

=

1

√14

(2

^

i

−

^

j

−3

^

k

)
And, a=2

^

i

−3

^

j

+6

^

k

The projection of a on e is calculated as,
a.e=

4

√14

+

3

√14

−

18

√14

=

−11

√14

The projection vector of a on e is calculated as,
a.e=−

11

√14

(

1

√14

(2

^

i

−

^

j

−3

^

k

))
=

11

14

(−2

^

i

+

^

j

+3

^

k

)