Question

Prove the Section Formula for internal division in vectors.

Hint:

In internal division, multiply each endpoint vector by the opposite ratio and divide by the sum of ratios.

Answer 1

Concept: The section formula gives the position vector of a point dividing a line segment internally in a given ratio. If a point \( P \) divides the line joining points \( A \) and \( B \) internally in the ratio \( m:n \), then its position vector is a weighted average of the vectors of \( A \) and \( B \). Let:

Position vector of \( A = \vec{a} \)
Position vector of \( B = \vec{b} \)
Position vector of \( P = \vec{p} \)
Suppose \( P \) divides \( AB \) internally in the ratio \( m:n \), i.e., \[ AP : PB = m : n \]
Step 1: Express vectors along the line Since \( P \) lies on line \( AB \), \[ \vec{AP} = \vec{p} - \vec{a}, \quad \vec{PB} = \vec{b} - \vec{p} \] Given ratio: \[ \frac{|\vec{AP}|}{|\vec{PB}|} = \frac{m}{n} \] So in vector form: \[ \vec{p} - \vec{a} = \frac{m}{n} (\vec{b} - \vec{p}) \]
Step 2: Solve algebraically \[ n(\vec{p} - \vec{a}) = m(\vec{b} - \vec{p}) \] \[ n\vec{p} - n\vec{a} = m\vec{b} - m\vec{p} \] Bring like terms together: \[ n\vec{p} + m\vec{p} = m\vec{b} + n\vec{a} \] \[ (m+n)\vec{p} = m\vec{b} + n\vec{a} \]
Step 3: Find position vector of \( P \) \[ \vec{p} = \frac{m\vec{b} + n\vec{a}}{m+n} \] Final Result (Section Formula): If a point divides a line segment internally in the ratio \( m:n \), then: \[ \boxed{\vec{p} = \frac{m\vec{b} + n\vec{a}}{m+n}} \] Explanation: The formula shows that the point dividing the segment lies between the two endpoints and is a weighted average of their position vectors. The weights depend on the opposite segments of the ratio.