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Mathematics
Mathematics
Problem
In a parallelogram ABCD,
A
B
=
i
^
+
j
^
+
k
^
\mathbf{AB} = \hat{i} + \hat{j} + \hat{k}
AB
=
i
^
+
j
^
+
k
^
and diagonal
A
C
=
i
^
−
j
^
+
k
^
\mathbf{AC} = \hat{i} - \hat{j} + \hat{k}
AC
=
i
^
−
j
^
+
k
^
and area of parallelogram is
8
\sqrt{8}
8
sq units, then
∠
B
A
C
\angle BAC
∠
B
A
C
is equal to a.
π
6
\frac{\pi}{6}
6
π
b.
π
3
\frac{\pi}{3}
3
π
c.
sin
−
1
(
8
3
)
\sin^{-1}\left(\frac{\sqrt{8}}{3}\right)
sin
−
1
(
3
8
)
d.
cos
−
1
(
8
3
)
\cos^{-1}\left(\frac{\sqrt{8}}{3}\right)
cos
−
1
(
3
8
)
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