Question

Relative to a fixed origin O, the point A has position vector (2i + 3j - 4k), the point B has position vector (4i - 2j + 3k), and the point C has position vector (ai + 5j – 2k), where a is a constant and a < 0.

D is the point such that AB = BD.
(a) Find the position vector of D.
Given AC| = 4,
(b) find the value of a.​

Answer 1

Answer: (a) To find the position vector of point D, we need to find the midpoint of the line segment AB. We can do this by taking the average of the position vectors of points A and B:

D = (A + B) / 2 = ((2i + 3j - 4k) + (4i - 2j + 3k)) / 2 = (3i + 0.5j - 0.5k)

So, the position vector of point D is (3i + 0.5j - 0.5k).

(b) To find the value of a, we use the fact that the distance between points A and C is 4. We can find the distance between two points by finding the magnitude of the difference between their position vectors. The position vector of point C is (ai + 5j – 2k), so the difference between the position vectors of points A and C is (ai + 5j – 2k) - (2i + 3j - 4k) = (a-2)i + 2j - (-2k).

The magnitude of this difference is given by:

|(ai + 5j - 2k) - (2i + 3j - 4k)| = √((a-2)^2 + 2^2 + 2^2) = 4

So, we have:

√((a-2)^2 + 2^2 + 2^2) = 4

(a-2)^2 + 2^2 + 2^2 = 16

(a-2)^2 = 16 - 4 - 4

(a-2)^2 = 8

a-2 = ±2√2

Since a < 0, we have:

a = 2 - 2√2

So, the value of a is 2 - 2√2.

Explanation: