α=2i−3j+kβ=7i−5j+k
Let d=α×β
Consider the computation
d=∣∣i27j−3−5k11∣∣=(−3+5)i−(2−7)j+(−10+21)k=2i+5j+11k
Let c be the unit vector of d.c=(2)2+(5)2+(11)22i+5j+11k=1502i+5j+11k=562i+5j+11k
We claim that c is the desired unit vector perpendicular to α and β . To see this consider
c⋅α=561(2i+5j+11k)⋅(2i−3j+k)=561(4−15+11)=0
Also, consider
c⋅β=561(2i+5j+11k)⋅(7i−5j+k)=561(14−25+11)=0
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