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 )
A+b= 1246
a-b=1234 => a= b+1234
2b+1234= 1246/-1234
2b= 12/:2
b= 6
a=b+1234
a= 6+ 1234
a= 1240
Answer:
The area of the square is greater than the area of the circle.
I hope this helps you
w = 2l-7
2 (2l-7+l)=43,6
3l-7=21,8
3l=28,8l=9,6
w=2.9,6-7=12,2
Area=9,6.12,2=117,12