What is the unit vector that is normal to the plane containing 3i+7j-2k and 8i+2j+9k?

Answer 1

The unit vector normal to the plane is
#(1/94.01)(67hati-43hatj+50hatk)#.

Let us consider #vecA=3hati+7hatj-2hatk, vecB= 8hati+2hatj+9hatk# The normal to the plane #vecA, vecB# is nothing but the vector perpendicular i.e., cross product of #vecA, vecB#. #=>vecAxxvecB= hati(63+4)-hatj(27+16)+hatk(6-56) =67hati-43hatj+50hatk#. The unit vector normal to the plane is #+-[vecAxxvecB//(|vecAxxvecB|)]# So#|vecAxxvecB|=sqrt[(67)^2+(-43)^2+(50)^2]=sqrt8838=94.01~~94# Now substitute all in above equation, we get unit vector =#+-{[1/(sqrt8838)][67hati-43hatj+50hatk]}#.

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Answer 2

Nathan Ashcroft

The unit vector normal to the plane containing 3i+7j-2k and 8i+2j+9k is 0.28i - 0.64j + 0.72k.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.