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MAXImum [283]
3 years ago
15

Find all unit vectors that are orthogonal to the vector u = 1, 0, −4 .

Mathematics
1 answer:
KIM [24]3 years ago
3 0

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

\frac{v}{|v|}

Where;

|v| = \sqrt{(4v_3)^2 + (v_2)^2 + (v_3)^2}

|v| = \sqrt{17(v_3)^2 + (v_2)^2}

\frac{v}{|v|} = \frac{4v_3i + v_2j + v_3k}{\sqrt{17(v_3)^2 + (v_2)^2}}

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.

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A box with a square base and open top must have a volume of 32,000 cm3. find the dimensions of the box that mini- mize the amoun
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Alrighty


squaer base so length=width, nice


v=lwh
but in this case, l=w, so replace l with w
V=w²h

and volume is 32000
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the amount of materials is the surface area
note that there is no top
so
SA=LW+2H(L+W)
L=W so
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alrighty

we gots
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we want to minimize the square foottage
get rid of one of the variables
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solve for H
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subsitute

SA=W²+4WH
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take derivitive to find the minimum
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where does it equal 0?

0=2W-1280000/W²
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so sub back
32000/W²=H
32000/(40)²=H
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the box is 20cm height and the width and length are 40cm
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Step-by-step explanation:

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First, disregard the sign for absolute value and solve for x. 

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Answer:

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