Given that OA = 3i – 2j + k and OB = 4i + j – 3k. Find the distance between point A and B to 2 decimal places.
The following OA , OB etc are vectors
OA + AB = OB
AB = OB - OA
AB = OB + AO
= (4,1,-3) + (-3,2,-1)
= (1,3,-4)
|AB| = √(1^2 + 3^2 + (-4)^2)
= √26
= appr 5.10
Amaizing
To find the distance between two points A and B, we can use the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
In this case, A has coordinates (x1, y1, z1) and B has coordinates (x2, y2, z2).
Let's plug in the values:
A = 3i - 2j + k
B = 4i + j - 3k
By comparing the coefficients, we can see that:
x1 = 3, y1 = -2, z1 = 1
x2 = 4, y2 = 1, z2 = -3
Now we can substitute these values into the distance formula:
d = sqrt((4 - 3)^2 + (1 - (-2))^2 + (-3 - 1)^2)
Simplifying further:
d = sqrt(1^2 + 3^2 + (-4)^2)
d = sqrt(1 + 9 + 16)
d = sqrt(26)
To 2 decimal places, the distance between point A and B is approximately 5.10.