AP EAPCET 24-Aug-2021 Shift 1 Paper

Given, A(4,3,5,B(0,6,0),C(−8,1,4) and D are the vertices of a parallelogram
Let D be the point (x,y,z).
∴ Diagonals of parallelogram bisect each other. ⇒ mid-point of AC= mid point of BD ⇒(−2,2,

9

2

)=(

x

2

,

y+6

2

,

z

2

)
⇒

x

2

=−2,

y+6

2

=2,

z

2

=

9

2

x=−4,y=−2,z=9
∴ D is (−4,−2,9).
Now, AC=PV of C−PV of A
=(−8

^

i

+

^

j

+4

^

k

)−(4

^

i

+3

^

j

+5

^

k

)
=−8

^

i

+

^

j

+4

^

k

−4

^

i

−3

^

j

−5

^

k

=−12

^

i

−2

^

j

−

^

k

BD = PV of D - PV of B
=(−4

^

i

−2

^

j

+9

^

k

)−(0

^

i

+6

^

j

+0

^

k

)
=−4

^

i

−8

^

j

+9

^

k

Let θ be the angle between AC and BD.
Then, cos‌θ=|

AC.BD

|AC||BD|

|
⇒cos‌θ=|

48+16−9

√161√149

|
⇒θ=cos−1(

55

√149√161

)