Answer:
one unit vector is ur=(-1/√3 ,1/√3 ,1/√3 )
Step-by-step explanation:
first we need to find a vector that is ortogonal to u and v . This vector r can be generated through the vectorial product of u and v , u X v :![r=u X v =\left[\begin{array}{ccc}i&j&k\\1&0&1\\0&1&1\end{array}\right] = \left[\begin{array}{ccc}0&1\\1&1\end{array}\right]*i + \left[\begin{array}{ccc}1&0\\1&1\end{array}\right]*j + \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]*k = -1 * i + 1*j + 1*k = (-1,1,1)](https://tex.z-dn.net/?f=r%3Du%20X%20v%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%260%261%5C%5C0%261%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%2Ai%20%2B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%2Aj%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%2Ak%20%3D%20-1%20%2A%20i%20%2B%201%2Aj%20%2B%201%2Ak%20%3D%20%28-1%2C1%2C1%29)
then the unit vector ur can be found through r and its modulus |r| :
ur=r/|r| = 1/[√[(-1)²+(1)²+(1)²]] * (-1,1,1)/√3 =(-1/√3 ,1/√3 ,1/√3 )
ur=(-1/√3 ,1/√3 ,1/√3 )
Factors 5x are absolutely correct
Explanation
Answer:
<h2>15°</h2>
Step-by-step explanation:
We know that the central angle of a circle is 360°.
In this case, we have the circle divided into 12 equal parts, so each angle would be also the 12th part of 360°.

Each circular sector has an angle of 30°, and each arc is equal to 30°.
Now, notice that tha angle formed by the two blue cables is an external angle formed by two secants, that means we can use the following relation

Where
is the angle formed by the two blue cables.

Therefore, the angle measures 15°.
Step-by-step explanation:
fifth row=19+4=23
sixth row= 23+4=27