What is the unit vector that is orthogonal to the plane containing # ( i - 2 j + 3 k) # and # (4 i + 4 j + 2 k) #?

Answer 1

There are two steps in solving this question: (1) taking the cross product of the vectors and then (2) normalizing the resultant. In this case, the final unit vector is #(-16/sqrt500i+10/sqrt500j+12/sqrt500k)# or #(-16/22.4i+10/22.4j+12/22.4k)#.

The cross product of the vectors is the first step.

#(i-2j+3k)xx(4i+4j+2k) = (((-2)*2-3*4))i+(3*4-1*2)j+(1*4-(-2)*4)k)=((-4-12)i+(12-2)j+(4-(-8))k)=(-16i+10j+12k)#

Normalizing the resulting vector is the second step.

Each element of a vector is normalized by dividing it by its length, which can be found as follows:

#l=sqrt((-16)^2+10^2+12^2)=sqrt500~~22.4#

When everything is considered, the unit vector orthogonal to the supplied vectors can be expressed as follows:

#(-16/sqrt500i+10/sqrt500j+12/sqrt500k)# or #(-16/22.4i+10/22.4j+12/22.4k)#

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Answer 2

Isabella Alvarez

The unit vector orthogonal to the plane containing (i - 2j + 3k) and (4i + 4j + 2k) is (-6i + 10j - 8k) / sqrt(180).

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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.