What is the unit vector that is normal to the plane containing #( i +k )# and #(2i+ j - 3k)?

Answer 1

#+-(3hati-3hatj+hatk)/(sqrt19#

If #vecA=hati+hatj and vecB =2hati+hatj-3hatk# then vectors which will be normal to the plane containing #vec A and vecB# are either#vecAxxvecB or vecBxxvecA# .So we are to find out the unit vectors of these two vector . One is opposite to another.

Now #vecAxxvecB=(hati+hatj+0hatk )xx(2hati+hatj-3hatk)# #=(1*(-3)-0*1)hati+(0*2-(-3)*1)hatj+(1*1-1*2)hatk# #=-3hati+3hatj-hatk#

So unit vector of #vecAxxvecB=(vecAxxvecB)/|vecAxxvecB|# #=-(3hati-3hatj+hatk)/(sqrt(3^2+3^2+1^2))=-(3hati-3hatj+hatk)/(sqrt19#

And unit vector of #vecBxxvecA=+(3hati-3hatj+hatk)/sqrt19#

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Answer 2

Charlotte Aaron

To find the unit vector normal to the plane containing the given points, first find the cross product of the vectors formed by subtracting the coordinates of the points. Then, normalize the resulting vector to get the unit vector.

Vector 1 = (1, 0, 1) Vector 2 = (2, 1, -3)

Cross product of Vector 1 and Vector 2 = (-1, 5, 1)

Unit vector normal to the plane = (-1/√27, 5/√27, 1/√27)

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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.