Question

The point A, B, and C lie on a straight line. The position vectors of A, B, and C are 2i+3j+9k and 5i-3j+4k, respectively. B divides AC internally in the ratio 2:1. Find the following:

(a.) The position vector of B.
(b.) Distance of B from the origin.

Answer 1

The position vector of B is 3i - 3j + 7k, and the distance of B from the origin is approximately 8.19 units.

Step-by-step explanation:

The question involves finding the position vector of point B, which divides the line segment AC in the ratio 2:1, and the distance of point B from the origin. Let's denote point A's position vector as ℝ₀, point B's as ℝ, and point C's as ℝ₁. Since B divides AC in the ratio 2:1, the position vector of B (ℝ) can be found using the section formula:

ℝ = (ℝ₁ + 2ℝ₀) / (1 + 2)

Substitute the given vectors:

ℝ = ((5î - 3ï + 4ı) + 2(2î + 3ï + 9ı)) / 3

Now, calculate the components:

ℝ = (5 + 4î - 3 - 6ï + 4 + 18ı) / 3

ℝ = (9î - 9ï + 22ı) / 3

ℝ = 3î - 3ï + ≈ı

To find the distance of B from the origin, calculate the magnitude of the position vector ℝ:

|B| = √(3² + (-3)² + 7²)

|B| = √(9 + 9 + 49)

|B| = √67 ≈ 8.19 units