Section Formula in 3d

Here you will learn section formula in 3d for both internal and external division with examples.

For distance formula and section formula in 2d for internal and external division see this What is the Section Formula | Distance Formula in 2d.

Let’s begin –

Section Formula in 3d

(a) Internal Division

Let P\((x_1, y_1, z_1)\) and Q\((x_2, y_2, z_2)\) be two points. Let R be a point on the line segment joining P and Q such that it divides the join of P and Q internally in the ratio \(m_1\) : \(m_2\). Then, the coordinates of R are

(\(m_1x_2 + m_2x_1\over m_1 + m_2\), \(m_1y_2 + m_2y_1\over m_1 + m_2\), \(m_1z_2 + m_2z_1\over m_1 + m_2\))

corollary : If R is the mid point of the segment joining P\((x_1, y_1, z_1)\) and Q\((x_2, y_2, z_2)\), then \(m_1\) = \(m_2\) = 1 and the coordinates of R are given by

(\(x_1 + x_2\over 2\), \(y_1 + y_2\over 2\), \(z_1 + z_2\over 2\))

(b) External Division

Let P\((x_1, y_1, z_1)\) and Q\((x_2, y_2, z_2)\) be two points. Let R be a point on PQ produced dividing it externally in the ratio \(m_1\) : \(m_2\) (\(m_1\) \(\ne\) \(m_2\)). Then, the coordinates of R are

(\(m_1x_2 – m_2x_1\over m_1 – m_2\), \(m_1y_2 – m_2y_1\over m_1 – m_2\), \(m_1z_2 – m_2z_1\over m_1 – m_2\))

Example : Find the coordinates of the point which divides the join of P(2, -1, 4) and Q(4, 3, 2) in the ratio 2 : 3 (i) internally (ii) externally.

Solution : Let R (x, y, z) be the required point. then,

(i) Internally,

x = \(2\times 4 + 3\times 2\over 2 + 3\), y = \(2\times 3 + 3\times -1\over 2 + 3\), z = \(2\times 2 + 3\times 4\over 2 + 3\)

\(\implies\) x = \(14\over 5\), y = \(3\over 5\), z = \(16\over 5\)

So, the coordinates of the required point R are (\(14\over 5\), \(3\over 5\), \(16\over 5\)).

(ii) Externally,

x = \(2\times 4 – 3\times 2\over 2 – 3\), y = \(2\times 3 – 3\times -1\over 2 – 3\), z = \(2\times 2 – 3\times 4\over 2 – 3\)

\(\implies\) x = -2, y = -9, z = 8

So, the coordinates of the required point R are (-2, -9, 8).

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