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
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>
(x+1)(x+2)
this is what i got hope it helps
The tree is 52.8 feet, since the hypotenuse is only part of the tree, you have to add 12, since then it would equal the total height of the tree.
a^2+b^2=c^2
12^2+ 39^2=c^2
144+1521= 1665
√1665
= 40.8
40.8 + 12
= 52.8 feet
The answer is 2. Here's why:
First if all, anything to the power of 0 is 1.
So n^0=1.
You need to plug in 3 for m, so then 6(3)^-1.
3^-1=1/3.
6 × 1/3= 6/3=2
2×1=2.