The point (4, -3) is on the graph of which function PLEASE HELP
1 answer:
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Answer:
Step-by-step explanation:
Given:
u = 1, 0, -4
In unit vector notation,
u = i + 0j - 4k
Now, to get all unit vectors that are orthogonal to vector u, remember that two vectors are orthogonal if their dot product is zero.
If v = v₁ i + v₂ j + v₃ k is one of those vectors that are orthogonal to u, then
u. v = 0 [<em>substitute for the values of u and v</em>]
=> (i + 0j - 4k) . (v₁ i + v₂ j + v₃ k) = 0 [<em>simplify</em>]
=> v₁ + 0 - 4v₃ = 0
=> v₁ = 4v₃
Plug in the value of v₁ = 4v₃ into vector v as follows
v = 4v₃ i + v₂ j + v₃ k -------------(i)
Equation (i) is the generalized form of all vectors that will be orthogonal to vector u
Now,
Get the generalized unit vector by dividing the equation (i) by the magnitude of the generalized vector form. i.e
Where;
|v| =
|v| =
=
This is the general form of all unit vectors that are orthogonal to vector u
where v₂ and v₃ are non-zero arbitrary real numbers.