find a unit vector orthogonal to -5i+8.9j+7.9k.

2 vectors are orthogonal if dot product is zero unit vector is defined by dividing each component by vector's magnitude Let a = -5i +8.9j +7.9k, b = xi +yj + zk, where b is orthogonal to a u is unit vector of b Result of dot product: \[-5x +8.9y +7.9z = 0\] Let y=z=1 \[-5x +16.8 = 0\] \[\rightarrow x = 3.36\] \[b = 3.36i +j+k\] \[u = \frac{b}{\left| b \right|}\] \[\left| b \right| = \sqrt{3.36^{2}+1^{2}+1^{2}} = \sqrt{13.2896}\] \[u = \frac{3.36}{\sqrt{13.2896}}i + \frac{1}{\sqrt{13.2896}}j + \frac{1}{\sqrt{13.2896}}k\]

Since there are many, u might as well just pick the one in the x.y plane to make ur life a bit simpler.

i wonder if line formats are up and running :) as i see it, the given vector is the normal to a plane. -5(x-Px) +8.9(y-Py) +7.9(z-Pz) = 0 the vectors from (x,y,z) to (Px,Py,Pz) are all the vectors that are orthogonal to the normal

line returns are working again, yay!!!

since a vector can be placed anywhere in the field; we can pick any starting, or ending point to determine the value for our orthogonal vectors. id pick (0,0,0) as one of the points. -5(x-0) +8.9(y-0) +7.9(z-0) = 0 -5x +8.9y +7.9z = 0

this form is the result of a dot product which results in a scalar of 0, its the definition of orthogonal, but we still need to consider the "unit" portion once we determine values for x,y,and z i think

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estudier suggestion is great, we can pick the one the one the resides in the xy plane, which means that z=0

-5x +8.9y +7.9(0) = 0 -5x +8.9y = 0, solve for your pick; y maybe? y = \(\cfrac{5}{8.9}x\)

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