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grigory [225]
3 years ago
7

Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

v_1=(\frac{1}{10},-\frac{3}{10})

v_2=(-\frac{1}{10},\frac{3}{10})

Step-by-step explanation:

First we define two generic vectors in our \mathbb{R}^2 space:

  1. v_1 = (x_1,y_1)
  2. v_2 = (x_2,y_2)

By definition we know that Euclidean norm on an 2-dimensional Euclidean space \mathbb{R}^2 is:

\left \| v \right \|= \sqrt{x^2+y^2}

Also we know that the inner product in \mathbb{R}^2 space is defined as:

v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

\left \| v_1 \right \|= \sqrt{x^2+y^2}=1

and

\left \| v_2 \right \|= \sqrt{x^2+y^2}=1

As second condition we have that:

v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0

v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0

Which is the same:

y_1=-3x_1\\y_2=-3x_2

Replacing the second condition on the first condition we have:

\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}

Since x_1^2= \frac{1}{10} we have two posible solutions, x_1=\frac{1}{10} or x_1=-\frac{1}{10}. If we choose x_1=\frac{1}{10}, we can choose next the other solution for x_2.

Remembering,

y_1=-3x_1\\y_2=-3x_2

The two vectors we are looking for are:

v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})

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P(A)* P(B)≠ p(AnB)<br><br> Why does it contain a not equal to sign
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Step-by-step explanation:

Conditional probability is the probability of one event occurring  with some relationship to one or more other events

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The formula for conditional probability of P(D|A) = P(A∩D)/P(A)

The table

              ↓         ↓       ↓

           :  C     :   D    : Total

→ A      :  6     :    2    :   8

→ B      :  1      :    8    :   9

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- P(D) means total of column D

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∴ P(D|A) =

∵ P(A|D) =

∵ P(D|A) =

∵   ≠  

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Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Alguien puede ayudarme en esto
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What adds up to -54 and multiplies to -16
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xy = -16
\frac{xy}{x} = \frac[-16}{x}
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\frac{x^{2}}{x} + \frac{-16}{x} = -54
\frac{x^{2} - 16}{x} = -54
x^{2} - 16 = -54x
x^{2} + 54x - 16 = 0
x = \frac{-(54) \± \sqrt{(54)^{2} - 4(1)(-16)}}{2(1)}
x = \frac{-54 \± \sqrt{2916 + 64}}{2}
x = \frac{-54 \± \sqrt{2980}}{2}
x = \frac{-54 \± 59.6}{2}
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x = -27 + 29.8    or    x = -27 - 29.8
x = 2.8    or    x = 56.8

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             or

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+ 56.8        + 56.8
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        (x, y) = (-56.8, 2.8)

The two numbers that add up to -54 and multiply to -16 are -56.8 and 2.8.
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