AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

It is given that the points,
A(−2,1,3),B(1,1,1),C(2,3,4)
The position vectors are,

→

AB

=(

∧

i

+

∧

j

+

∧

k

)−(−2

∧

i

+

∧

j

+

∧

k

)=3

∧

i

−2

∧

j

→

BC

=(2

∧

i

+3

∧

j

+4

∧

k

)−(

∧

i

+

∧

j

+

∧

k

)=

∧

i

+2

∧

j

+3

∧

k

The normal of the plane,

→

n

=

→

AB

×

→

BC

→

n

=|

∧ i	∧ j	∧ k
3	0	−2
1	2	3

|
By solving the above matrix,
n=4

∧

i

−11

∧

j

+6

∧

k

The equationof the plane is,
4(x−2)−11(y−3)+6(z−4)=0
4x−11y−6z+1=0
If ax+bx+cz+d=0 is the equation of plane, then the normal form is,

{(

a

√a2+b2+c2

)x+(‌

b

√a2+b2+c2

)y+(‌

c

√a2+b2+c2

)z+(‌

d

√a2+b2+c2

)}=0

Substitute the value and simplify the equation.

{‌

4x

√173

−‌

11y

√173

+‌

6z

√173

+‌

1

√173

}=0

(‌

−4

√173

)x+(‌

11

√173

)y+(‌

−6

√173

)z=‌

1

√173