Given that OA = 2i + 3j and OB = 3i — 2j, find the magnitude of AB to one decimal place.
a•b = 0, so OA⊥OB
Just use the Pythagorean theorem. AB = √26
Or, since you are showing that you work with vectors ....
vector AB = vector AO + vector OB
= <-2,-3> + <3,-2> = <1,-5>
|<1,-5>| = √26 = appr 5.1
Well, looks like OA and OB are vectors, and you want to find the magnitude of AB. Let's see, if we subtract the coordinates of point O from point B, we get AB = (3-2)i + (-2-3)j.
So, AB simplifies to AB = i - 5j. Now, to find the magnitude of AB, we use the Pythagorean theorem. It's like finding the length of a hypotenuse in a right-angled triangle.
The formula for the magnitude of a vector is √(x² + y²), where x and y are the components of the vector. In this case, x = 1 and y = -5.
Calculating, we have AB = √(1² + (-5)²) = √(1 + 25) = √26, which is approximately 5.1.
So, the magnitude of AB to one decimal place is 5.1. It seems like AB is on its way to becoming a superstar!
To find the magnitude of AB, we need to calculate the distance between points A and B. The distance between two points can be found using the distance formula:
d = √((x2 - x1)² + (y2 - y1)²)
In this case, we have two points A and B, with coordinates OA = 2i + 3j and OB = 3i - 2j.
Let's find the coordinates of point B first. Subtracting the coordinates of point A from point B, we get:
OB - OA = (3i - 2j) - (2i + 3j)
= (3i - 2i) + (-2j - 3j)
= i - 5j
Now that we have the coordinates of AB, which are i - 5j, we can substitute these into the distance formula:
d = √((x2 - x1)² + (y2 - y1)²)
= √((1 - 0)² + (-5 - 0)²)
= √(1² + (-5)²)
= √(1 + 25)
= √(26)
≈ 5.099
Therefore, the magnitude of AB to one decimal place is approximately 5.1.