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slega
2 months ago
11

G Find non-zero vectors x and y that are both orthogonaland orthogonal to each other.

Mathematics
1 answer:
babunello [11.8K]2 months ago
4 0

Answer:

x = <1,5>

y = <-5,1>

Step-by-step explanation:

Let x, y represent two vectors in 2D

defined as:

x = <a, b>

y = <c, d>

for x, y to be orthogonal, we understand that their dot or scalar product should equal zero, therefore

x. y = 0

<a, b>. <c, d> = 0

a*c + b*d = 0 ------ (A)

Now, select any four values for a, b, c, and d that fulfill equation (A)

for example, I choose (this is subject to personal choice):

a=1, b=5, c=-5, d=1

Verifying Equation (A):

(1)*(5) + (-5)*(1) =0

5 - 5 = 0

0 = 0 (True)

Thus,

The identified vectors x, y are:

x = <1, 5>

y = <-5, 1>

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The minimum number of batches Joyce can produce is one.
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Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.
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A="b is situated in the center"

B="c lies to the right of b"

C="The letters def occur sequentially in that arrangement"

a) b can occupy 7 positions; however, only one of these is the center. Therefore, P(A)=1/7

b) Let X=i; "b holds the i-th position"

Y=j; "c occupies the j-th position"

P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}

P(B)=1/2

c) Let X=i; "d holds the i-th position"

Y=j; "e occupies the j-th position"

Let Z=k; "f is in the i-th position"

P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

Events A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

Events A and C are not independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

Events B and C are not independent

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2 months ago
Hi! I'm confused with this question, It's about tessellations. Can someone give a complete example of a possible answer? It woul
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Yes, this reflects the correct answer as [[TAG_10]]
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