Section formula, Polygon law

Lesson 7 of 15 • 9 upvotes • 15:00mins

Jagat Singh

Section formula, polygon law and tyoes of vectors Let us consider that the line segment connecting P and Q is divided by a point R lying on PQ. The point R can divide the line segment PQ in two ways: internally and externally. Let us consider both these cases individually. Case 1: Line segment PQ is divided by R internally Let us consider that the point R divides the line segment PQ in the ratio m: n, given that m and n are positive scalar quantities we can say that, mRQ¯¯¯¯¯¯¯¯ = nPR¯¯¯¯¯¯¯¯ Consider the triangles, ∆ORQ and ∆OPR. RQ¯¯¯¯¯¯¯¯ = OQ¯¯¯¯¯¯¯¯ – OR¯¯¯¯¯¯¯¯ = b⃗  – r⃗  PR¯¯¯¯¯¯¯¯ = OR¯¯¯¯¯¯¯¯ – OP¯¯¯¯¯¯¯¯ = r⃗  – a⃗  Therefore, m(b⃗  – r⃗ ) = n(r⃗  – a⃗ ) Rearranging this equation we get: r⃗  = mb⃗ +na⃗ m+n Therefore the position vector of point R dividing P and Q internally in the ratio m:n is given by: OR→ = mb⃗ +na⃗ m+n Case 2: Line segment PQ is divided by R externally Let us consider that the point R divides the line segment PQ in the ratio m: n, given that m and n are positive scalar quantities we can say that, mRQ¯¯¯¯¯¯¯¯ = -nPR¯¯¯¯¯¯¯¯ Consider the triangles, ∆ORQ and ∆OPR. RQ¯¯¯¯¯¯¯¯ = OQ¯¯¯¯¯¯¯¯ – OR¯¯¯¯¯¯¯¯ = b⃗  – r⃗  PR¯¯¯¯¯¯¯¯ = OR¯¯¯¯¯¯¯¯ – OP¯¯¯¯¯¯¯¯ = r⃗  – a⃗  Therefore, m(b⃗  – r⃗ ) = -n(r⃗  – a⃗ ) Rearranging this equation we get: r⃗  = mb⃗ –na⃗ m–n Therefore the position vector of point R dividing P and Q internally in the ratio m:n is given by: OR→ = mb⃗ –na⃗ m–n What if the point R dividing the line segment joining points P and Q is the midpoint of line segment AB? In that case, if R is the midpoint, then R divides the line segment PQ in the ratio 1:1, i.e. m = n = 1.The position vector of point R dividing will be given as: OR→ = b⃗ +a⃗ 2