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
(
4+(−8)
2
,
3+1
2
,
5+4
2
)=(
x+0
2
,
y+6
2
,
z+0
2
)
⇒(−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=|