TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, plane 2x−2y+4z+5=0
Direction ratios of the normal =2,−2,4
Equation of line passing through (1,‌

3

2

,2)
and dr's 2, −2,4 is given by ‌

x−1

2

=‌

y− 3 2

−2

=‌

z−2

4

=r
x=2r+1,y=−2r+‌

3

2

,z=4r+2
This point also lies on given plane.
∴2(2r+1)−2(−2r+‌

3

2

)+4(4r+2+5=0
24r+12=0‌‌‌ or ‌‌‌r=‌

−1

2

∴ Required point
(2⋅(‌

−1

2

)+1,−2(‌

−1

2

)+‌

3

2

,4(‌

−1

2

)+2)=(0,‌

5

2

,0)