Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
![\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%261%262%5C%5C1%26-2%263%5Cend%7Barray%7D%5Cright%5D)
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
![\frac{[7,2,-1]}{3\sqrt{6} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B7%2C2%2C-1%5D%7D%7B3%5Csqrt%7B6%7D%20%7D)
or
[ 
, 
, 
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ 
, 
, 
]
In conclusion, the two unit vectors are;
[ , , ]
and
[ , , ]
<em>Hope this helps!</em>
we;retk;tykh'tk';rekw'kr'e
Answer:
Step-by-step explanation:
Given the explicit function as
f(n) = 15n+4
The first term of the sequence is at when n= 1
f(1) = 15(1)+4
f(1) = 19
a = 19
Common difference d = f(2)-f(1)
f(2) = 15(2)+4
f(2) = 34
d = 34-19
d = 15
Sum of nth term of an AP = n/2{2a+(n-1)d}
S20 = 20/2{2(19)+(20-1)15)
S20 = 10(38+19(15))
S20 = 10(38+285)
S20 = 10(323)
S20 = 3230.
Sum of the 20th term is 3230
For the explicit function
f(n) = 4n+15
f(1) = 4(1)+15
f(1) = 19
a = 19
Common difference d = f(2)-f(1)
f(2) = 4(2)+15
f(2) = 23
d = 23-19
d = 4
Sum of nth term of an AP = n/2{2a+(n-1)d}
S20 = 20/2{2(19)+(20-1)4)
S20 = 10(38+19(4))
S20 = 10(38+76)
S20 = 10(114)
S20 = 1140
Sum of the 20th terms is 1140
Answer:
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Step-by-step explanation:
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