Find a unit vector perpendicular to the a-i-2j-k- and b-2i-3j-k-

We know thatunit vector perpendicular to both a and b is diven by x̂=a×b|a×b| a×b=[ î #160; amp; ĵ #160; amp; k̂ #160; 1 ...

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Multiplication with Vectors

Find a unit v...

Question

Find a unit vector perpendicular to the $\to a = \to i + 2 \to j - \to k$ and $\to b = 2 \to i + 3 \to j + \to k$ .

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Solution

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\to a = \to i - 2 \to j - 3 \to k, \to b = 2 \to i + \to j - \to k, \to c = \to i + 3 \to j - 2 \to k

(\to a \times \to b) \times \to c

\to a = 3 \to i + 2 \to j + 6 \to k

\to b = 2 \to i + \to j + \to k

\to c = \to i - \to j + \to k

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k,

\to C = 4 \to i - 3 \to j + 7 \to k

\to R

\to R \times \to B = \to C \times \to B

\to R . \to A = 0

\to r = 3 \to i + 2 \to j - 5 \to k

\to a = 2 \to i - \to j + \to k

\to b = \to i + 3 \to j - 2 \to k

\to c = - 2 \to i + \to j - 3 \to k

\to r = λ \to a + μ \to b + ν \to c

\to a = 2 \to i + \to j + \to k

\to b = \to i + 2 \to j - \to k

\to c

\to c

\to a

\to c =

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