Find the unit vector perpendicular to the plane containing the vectors ] = (41+ 21 - k) and
Q = (3î – ĵ + k). Also find the area of the parallelogram formed by the two vectors.
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Answer:
We know that a×b is vector perpendicular to the plane of a and b
Therefore unit vector perpendicular to the plane of a and b =
∣a×b∣
a×b
Now a×b=(2i−j+k)×(3i+4j−k)
=
∣
∣
∣
∣
∣
∣
∣
∣
i
2
3
j
−1
4
k
1
−1
∣
∣
∣
∣
∣
∣
∣
∣
=i(1−4)+j(3+2)+k(8+3)
=−3i+5j+11k
⇒ unit vector parallel to a×b
=
∣a×b∣
a×b
=
(3
2
+5
2
+11
2
)
−3i+5j+11k
=
155
−3i+5j+11k
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