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Question

Question: If **a, b, c** are position vector of vertices of a triangle \(ABC\), then unit vector perpendicular...

If a, b, c are position vector of vertices of a triangle ABCABC, then unit vector perpendicular to its plane is

A

a×b+b×c+c×aa \times b + b \times c + c \times a

B

a×b+b×c+c×aa×b+b×c+c×a\frac{a \times b + b \times c + c \times a}{|a \times b + b \times c + c \times a|}

C

a×ba×b\frac{a \times b}{|a \times b|}

D

None of these

Answer

a×b+b×c+c×aa×b+b×c+c×a\frac{a \times b + b \times c + c \times a}{|a \times b + b \times c + c \times a|}

Explanation

Solution

Unit vector perpendicular to plane

=AB×ACAB×AC=(ba)×(ca)(ba)×(ca)= \frac{\overset{\rightarrow}{AB} \times \overset{\rightarrow}{AC}}{|\overset{\rightarrow}{AB} \times \overset{\rightarrow}{AC}|} = \frac{(b - a) \times (c - a)}{\left| (b - a) \times (c - a) \right|}

=a×b+b×c+c×aa×b+b×c+c×a.= \frac{a \times b + b \times c + c \times a}{|a \times b + b \times c + c \times a|}.