Question

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

Answer 1

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
\[-5x +16.8 = 0\]
\[\rightarrow x = 3.36\]
\[b = 3.36i +j+k\]
\[u = \frac{b}{\left| b \right|}\]
\[\left| b \right| = \sqrt{3.36^{2}+1^{2}+1^{2}} = \sqrt{13.2896}\]
\[u = \frac{3.36}{\sqrt{13.2896}}i + \frac{1}{\sqrt{13.2896}}j + \frac{1}{\sqrt{13.2896}}k\]

Answer 2

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

Answer 3

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

Answer 4

line returns are working again, yay!!!

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

y = \(\cfrac{5}{8.9}x\)