Question

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

Answer 1

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).