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What is the cross product of #<-3,0,1># and #<1,2,-4>#?

Answer 1

The cross product of (-3,0,1) and (1,2,-4) is (-2,11,-6).

We have two vectors #A=(A_x,A_y,A_z)# and #B=(B_x,B_y,B_z)#.

The cross product can be calculated nicely using matrices. It is #det((hat(x),hat(y),hat(z)),(A_x,A_y,A_z),(B_x,B_y,B_z))=#

Unit vectors (length 1) along each of the axes are shown in the first row. The first vector's coordinates are shown next, followed by the second vector's coordinates.

#=hat(x)A_yB_z+hat(y)A_zB_x+hat(z)A_xB_y -hat(x)A_zB_y-hat(y)A_xB_z-hat(z)A_yB_x =hat(x)(A_yB_z-A_zB_y)+hat(y)(A_zB_x-A_xB_z) +hat(z)(A_xB_y-A_yB_x) =(A_yB_z-A_zB_y, A_zB_x-A_xB_z, A_xB_y-A_yB_x)#

This expression is resulting vector. The formula could be memorized by the people, who don't know matrices yet. Notice how Neither #A_x# nor #B_x# has any impact on x coordinate of the result. The same for y and z. That's the sign of perpendicularity of the result to both vectors. Notice how result disappears, when #A~~B#.

Plugging values in: #A=(-3,0,1)# and #B=(1,2,-4)# #A times B# #=(0*(-4)-1*2, 1*1-(-3)(-4), (-3)*2-0*1) =(0-2, 1-12, -6-0)=(-2,11,-6)#

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Answer 2

Nora Arzate

The cross product of <-3,0,1> and <1,2,-4> is <8,11,6>.

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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.

What is the cross product of #&lt;-3,0,1&gt;# and #&lt;1,2,-4&gt;#?

What is the cross product of #<-3,0,1># and #<1,2,-4>#?