Relative to an origin O, the position vector of A and B are (9) and (k) respectively.
(-4) (1)
A) express AB as a column vector.
B) given that |AB| = 13 units find the possible values of k
C) the point C is such that AC = (8)
(-3). Find the coordinates of C
A) The position vector AB can be found by subtracting the position vector of A from the position vector of B:
AB = B - A
= (k) - (9)
(1) (-4)
= (k - 9)
(1 + 4)
Therefore, the column vector AB is (k - 9) in the first row and (5) in the second row.
B) The magnitude (or length) of AB is given as |AB| = 13 units. Using the column vector AB above, we can calculate the magnitude as follows:
|AB| = sqrt((k - 9)^2 + 5^2) = 13
Squaring both sides:
(k - 9)^2 + 25 = 169
Expanding and simplifying:
k^2 - 18k + 81 + 25 = 169
k^2 - 18k - 63 = 0
Factoring or using the quadratic formula, we find that the possible values of k are k = -3 and k = 21.
C) We are given that AC = (8)
(-3)
To find the coordinates of C, we can add the position vector of A to the vector AC:
C = A + AC
= (9) + (8)
(-4) (-3)
= (9 + 8)
(-4 - 3)
= (17)
(-7)
Therefore, the coordinates of C are (17, -7).
A) To express AB as a column vector, we subtract the position vector of point A from the position vector of point B:
AB = (k - 9)
(1 - (-4))
So, the column vector AB is:
AB = (k - 9)
(5)
B) Given that |AB| = 13, we can use the distance formula to find the possible values of k. The distance between two points A and B in 2D space can be calculated using the formula:
|AB| = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the distance |AB| = 13.
Substituting the coordinates of A and B:
|AB| = √((k - 9)^2 + (5 - (-4))^2)
Simplifying the equation:
13 = √((k - 9)^2 + 9^2)
Squaring both sides of the equation:
169 = (k - 9)^2 + 81
Now, let's solve for k:
(k - 9)^2 = 169 - 81
(k - 9)^2 = 88
Taking the square root of both sides:
k - 9 = ±√88
k - 9 = ±2√22
Now, we solve for k in two separate equations:
k = 9 ± 2√22
So, the possible values of k are:
k = 9 + 2√22
k = 9 - 2√22
C) The point C is such that AC = (8)
(-3)
To find the coordinates of point C, we add the position vector of A to the given vector AC:
C = A + AC
Using column vector addition:
C = (9, -4) + (8, -3)
Adding corresponding entries:
C = (9 + 8, -4 + (-3))
Simplifying:
C = (17, -7)
Therefore, the coordinates of point C are (17, -7).
A) To express the vector AB as a column vector, subtract the position vector of point A from the position vector of point B:
AB = (9 - (-4), k - 1) = (13, k - 1)
B) To find the possible values of k given that |AB| = 13 units, we can use the magnitude formula for a vector:
|AB| = √((13)^2 + (k - 1)^2) = 13
Squaring both sides of the equation, we have:
(13)^2 = (13)^2 + (k - 1)^2
169 = 169 + (k - 1)^2
(k - 1)^2 = 0
Taking the square root of both sides, we have:
k - 1 = ±√0 = 0
Therefore, the possible values of k are:
k = 1 (when we take the positive square root)
k = 1 (when we take the negative square root)
C) Given that AC = (8, -3), we can find the coordinates of point C by adding the position vector of point A to the vector AC:
C = A + AC = (9, -4) + (8, -3) = (9 + 8, -4 + (-3)) = (17, -7)
Therefore, the coordinates of point C are (17, -7).