Vector Geometry

Vector Geometry

Vector Geometry: We can study different properties and relations relating to geometry with the help of vectors. Such a study is known as the Vector Geometry.

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Here are 12 geometrical theorems and their proofs by vector method.

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Theorem 1: Midpoint Formula

If ( "a" ) ⃗ and ( "b" ) ⃗ are position vectors of two points A and B respectively and M is the middle point of the line segment AB, then the position vector of M is "1" /"2" (( "a" ) ⃗+( "b" ) ⃗ ). Proof: Let AB be a line segment and O be the origin. Position vectors of points A and B, ("OA" ) ⃗ = ( "a" ) ⃗ and ("OB" ) ⃗ = ( "b" ) ⃗. Suppose, M be the middle point of the line segment AB. ∴ ("AM" ) ⃗ = ("MB" ) ⃗ = "1" /"2" ("AB" ) ⃗ Now, ("OM" ) ⃗ = ("OA" ) ⃗ + ("AM" ) ⃗ = ("OA" ) ⃗ + "1" /"2" ("AB" ) ⃗ = ("OA" ) ⃗ + "1" /"2" (("AO" ) ⃗+("OB" ) ⃗ ) = ("OA" ) ⃗ + "1" /"2" (–("OA" ) ⃗+("OB" ) ⃗ ) = (" a " ) ⃗ + "1" /"2" (–(" a " ) ⃗+(" b " ) ⃗ ) = ("2" ( "a" ) ⃗ - ( "a" ) ⃗ + ( "b" ) ⃗)/"2" = (( "a" ) ⃗ + ( "b" ) ⃗)/"2" = "1" /"2" (( "a" ) ⃗+( "b" ) ⃗ ) Hence, position vector of M = ("OM" ) ⃗ = "1" /"2" (( "a" ) ⃗+( "b" ) ⃗ ). Proved.

Theorem 2: Section Formula for Internal Division

If ( "a" ) ⃗ and ( "b" ) ⃗ be the position vectors of two points A and B respectively and the point M divides the line segment AB internally in the ratio m : n, then the position vector of M is ("m" ( "b" ) ⃗" + n" ( "a" ) ⃗)/"m + n" . Proof: Let AB be a line segment and O be the origin. Position vectors of A and B are ("OA" ) ⃗ = ( "a" ) ⃗ and ("OB" ) ⃗ = ( "b" ) ⃗. Suppose, M divides the line segment AB internally in the ratio of m : n. ∴ ("AM" ) ⃗/("MB" ) ⃗ = "m" /"n" or, n("AM" ) ⃗ = m("MB" ) ⃗ or, n(("AO" ) ⃗+("OM" ) ⃗) = m(("MO" ) ⃗+("OB" ) ⃗) or, n(-(" a " ) ⃗+("OM" ) ⃗) = m(-("OM" ) ⃗+(" b " ) ⃗) or, –n( "a" ) ⃗ + n("OM" ) ⃗ = –m("OM" ) ⃗ + m( "b" ) ⃗ or, n("OM" ) ⃗ + m("OM" ) ⃗ = m( "b" ) ⃗ + n( "a" ) ⃗ or, ("OM" ) ⃗ (m + n) = m( "b" ) ⃗ + n( "a" ) ⃗ or, ("OM" ) ⃗ = ("m" ( "b" ) ⃗" + n" ( "a" ) ⃗)/"m + n" ∴ Position vector of M = ("OM" ) ⃗ = ("m" ( "b" ) ⃗" + n" ( "a" ) ⃗)/"m + n" Proved.

Theorem 3: Section Formula: External Division

If ( "a" ) ⃗ and ( "b" ) ⃗ be the position vectors of two points A and B respectively and the point P divides the line segment AB externally in the ratio m : n, then the position vector of P is ("m" ( "b" ) ⃗" " -" n" ( "a" ) ⃗)/("m " -" n" ). Proof: Let AB be a line segment and O be the origin. Position vectors of A and B are ("OA" ) ⃗ = ( "a" ) ⃗ and ("OB" ) ⃗ = ( "b" ) ⃗. Suppose, P divides the line segment AB externally in the ratio of m : n. ∴ ("AP" ) ⃗/("BP" ) ⃗ = "m" /"n" or, m("BP" ) ⃗ = n("AP" ) ⃗ or, m(("BO" ) ⃗ + ("OP" ) ⃗) = n(("AO" ) ⃗ + ("OP" ) ⃗) or, m(-("OB" ) ⃗ + ("OP" ) ⃗) = n(-("OA" ) ⃗ + ("OP" ) ⃗) or, m(-(" b " ) ⃗ + ("OP" ) ⃗) = n(-(" a " ) ⃗ + ("OP" ) ⃗) or, –m(" b " ) ⃗ + m("OP" ) ⃗ = –n(" a " ) ⃗ + n("OP" ) ⃗ or, m("OP" ) ⃗ – n("OP" ) ⃗ = m(" b " ) ⃗ – n(" a " ) ⃗ or, ("OP" ) ⃗(m – n) = m(" b " ) ⃗ – n(" a " ) ⃗ or, ("OP" ) ⃗ = ("m" ( "b" ) ⃗" " -" n" ( "a" ) ⃗)/("m " -" n" ) ∴ Position vector of P = ("OP" ) ⃗ = ("m" ( "b" ) ⃗" " -" n" ( "a" ) ⃗)/("m " -" n" ) Proved.

Theorem 4: Centroid Formula

The position vector of the centroid of a triangle is given by ( "g" ) ⃗ = "1" /"3" ((" a " ) ⃗" + " (" b " ) ⃗" + " (" c " ) ⃗ ) where ( "a" ) ⃗, ( "b" ) ⃗ and ( "c" ) ⃗ are the position vectors of the vertices and ( "g" ) ⃗ is the position vector of the centroid. Proof: Let ABC be a triangle and O be the origin. Here, ("OA" ) ⃗ = ( "a" ) ⃗, ("OB" ) ⃗ = ( "b" ) ⃗ and ("OC" ) ⃗ = ( "c" ) ⃗. Let D be the midpoint of AC. Therefore, by midpoint theorem, ("OD" ) ⃗ = "1" /"2" (( "a" ) ⃗ +( "c" ) ⃗ ) Let G be the centroid of the triangle ABC. Then, G divides BD internally in the ratio 2 : 1. Therefore, by section formula, ("OG" ) ⃗ = ("2" ("OD" ) ⃗" + 1" ("OB" ) ⃗)/"2 + 1" = ("2 × " "1" /"2" " " ((" a " ) ⃗" + " (" c " ) ⃗ )" + " (" b " ) ⃗)/"3" = ((" a " ) ⃗" + " (" c " ) ⃗" + " (" b " ) ⃗)/"3" Therefore, position vector of Centroid G = ("OG" ) ⃗ = (" g " ) ⃗ = ((" a " ) ⃗" + " (" b " ) ⃗" + " (" c " ) ⃗)/"3" proved.

Theorem 5:

Prove by vector method that the line joining midpoints of two sides of a triangle is parallel to the third side and half of it. Proof: Let ABC be a triangle and P and Q be the midpoints of the sides AB and AC respectively. ∴ ("PA" ) ⃗ = "1" /"2" ("BA" ) ⃗ and ("AQ" ) ⃗ = "1" /"2" ("AC" ) ⃗ By triangle law of vector addition, ("PQ" ) ⃗ = ("PA" ) ⃗ + ("AQ" ) ⃗ = "1" /"2" ("BA" ) ⃗ + "1" /"2" ("AC" ) ⃗ = "1" /"2" (("BA" ) ⃗ + ("AC" ) ⃗) = "1" /"2" ("BC" ) ⃗ ∴ PQ ∥ BC and PQ = "1" /"2" BC. Proved.

Theorem 6:

Prove by vector method that the line joining vertex and midpoint of the base of an isosceles triangle is perpendicular to the base. Proof: Let ABC be an isosceles triangle where AB = AC and AM is the median to the base BC. Let ("AB" ) ⃗ = ( "a" ) ⃗ and ("AC" ) ⃗ = ( "b" ) ⃗. Then, |( "a" ) ⃗| = |( "b" ) ⃗| i.e. a = b. And, ("BM" ) ⃗ = "1" /"2" ("BC" ) ⃗ By triangle law of vector addition, ("BC" ) ⃗ = ("BA" ) ⃗ + ("AC" ) ⃗ = – ("AB" ) ⃗ + ("AC" ) ⃗ = – (" a " ) ⃗ + (" b " ) ⃗ = ((" b " ) ⃗ – (" a " ) ⃗) ("AM" ) ⃗ = ("AB" ) ⃗ + ("BM" ) ⃗ = ("AB" ) ⃗ + "1" /"2" ("BC" ) ⃗ = (" a " ) ⃗ + "1" /"2" (– (" a " ) ⃗ + (" b " ) ⃗) = ("2" ( "a" ) ⃗ - ( "a" ) ⃗ + ( "b" ) ⃗)/"2" = (( "a" ) ⃗ + ( "b" ) ⃗)/"2" = "1" /"2" ((" a " ) ⃗" +" (" b " ) ⃗ ) Now, ("AM" ) ⃗ . ("BC" ) ⃗ = "1" /"2" ((" a " ) ⃗" +" (" b " ) ⃗ ) . ((" b " ) ⃗ – (" a " ) ⃗) = "1" /"2" (b2 – a2) = 0 [∵ a = b] ∴ AM ⊥ BC proved.

Theorem 7:

Prove by vector method that the quadrilateral joining the midpoints of the adjacent sides of any quadrilateral is a parallelogram. Proof: Let ABCD be a quadrilateral and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Join BD. From the figure, ("BA" ) ⃗ + ("AD" ) ⃗ = ("BD" ) ⃗ or, 2("PA" ) ⃗ + 2("AS" ) ⃗ = ("BD" ) ⃗ or, 2(("PA" ) ⃗ + ("AS" ) ⃗) = ("BD" ) ⃗ or, ("PA" ) ⃗ + ("AS" ) ⃗ = "1" /"2" ("BD" ) ⃗ or, ("PS" ) ⃗ = "1" /"2" ("BD" ) ⃗ ∴ PS ∥ BD ……….. (i) Again, ("BC" ) ⃗ + ("CD" ) ⃗ = ("BD" ) ⃗ or, 2("QC" ) ⃗ + 2("CR" ) ⃗ = ("BD" ) ⃗ or, 2(("QC" ) ⃗ + ("CR" ) ⃗) = ("BD" ) ⃗ or, ("QC" ) ⃗ + ("CR" ) ⃗ = "1" /"2" ("BD" ) ⃗ or, ("QR" ) ⃗ = "1" /"2" ("BD" ) ⃗ ∴ QR ∥ BD ………. (ii) From (i) and (ii), PS ∥ QR. Similarly by joining AC, we can prove that PQ ∥ SR. Hence PQRS is a parallelogram. Proved.

Theorem 8:

Prove by vector method that the diagonals of a parallelogram bisect each other. Proof: Let OACB be a parallelogram and O be the origin. Let, ("OA" ) ⃗ = ( "a" ) ⃗ and ("OB" ) ⃗ = ( "b" ) ⃗. Suppose, M be the middle point of AB. Then by midpoint formula, ("OM" ) ⃗ = "1" /"2" (( "a" ) ⃗+( "b" ) ⃗ ) …………….. (i) Again, let N be the middle point of OC. Then by parallelogram law of vector addition, ("ON" ) ⃗ = "1" /"2" ("OC" ) ⃗ = "1" /"2" (("OA" ) ⃗ + ("OB" ) ⃗) = "1" /"2" (( "a" ) ⃗+( "b" ) ⃗ ) …………….. (ii) From (i) and (ii), M and N have the same position vector. So, they are same point. Hence, the diagonals of a parallelogram bisect each other. Proved.

Theorem 9:

Prove by vector method that the diagonals of a rectangle are equal. Proof: Let ABCD be a rectangle and AC and BD are diagonals. Then, ("AB" ) ⃗ = ("DC" ) ⃗, ("AD" ) ⃗ = ("BC" ) ⃗, ∠A = ∠B = ∠C = ∠D = 90° and hence, ("AB" ) ⃗ . ("BC" ) ⃗ = 0 and ("BA" ) ⃗ . ("AD" ) ⃗ = 0. Now, by triangle law of vector addition, ("AC" ) ⃗ = ("AB" ) ⃗ + ("BC" ) ⃗ or, (("AC" ) ⃗)2 = (("AB" ) ⃗ + ("BC" ) ⃗)2 or, AC2 = (("AB" ) ⃗)2 + 2 ("AB" ) ⃗ . ("BC" ) ⃗ + (("BC" ) ⃗)2 or, AC2 = AB2 + 2 × 0 + BC2 or, AC2 = AB2 + BC2 ……………… (i) Again, ("BD" ) ⃗ = ("BA" ) ⃗ + ("AD" ) ⃗ or, (("BD" ) ⃗)2 = (("BA" ) ⃗ + ("AD" ) ⃗)2 or, BD2 = (("BA" ) ⃗)2 + 2 ("BA" ) ⃗ . ("AD" ) ⃗ + (("AD" ) ⃗)2 or, BD2 = BA2 + 2 × 0 + AD2 or, BD2 = BA2 + AD2 or, BD2 = AB2 + BC2 ……………… (ii) From (i) and (ii), AC2 = BD2 ∴ AC = BD Hence the diagonals of a rectangle are equal. Proved.

Theorem 10:

Prove by vector method that the diagonals of a rhombus bisect each other at right angles. Proof: Let OABC is a rhombus, ("OA" ) ⃗ = ("CB" ) ⃗ = (" a " ) ⃗ and ("OC" ) ⃗ = ("AB" ) ⃗ = (" b " ) ⃗ where |(" a " ) ⃗| = |(" b " ) ⃗| i.e. a = b. Let M be the midpoint of diagonal OB and N is the midpoint of diagonal AC. Now, by triangle law of vector addition, ("OB" ) ⃗ = ("OA" ) ⃗ + ("AB" ) ⃗ = (" a " ) ⃗ + (" b " ) ⃗. Since, M is the midpoint of OB, ("OM" ) ⃗ = "1" /"2" ("OB" ) ⃗ = "1" /"2" ((" a " ) ⃗ + (" b " ) ⃗) ………….. (i) Again, since N is the midpoint of AC, by midpoint theorem, ("ON" ) ⃗ = "1" /"2" (("OA" ) ⃗ + ("OC" ) ⃗) = "1" /"2" ((" a " ) ⃗ + (" b " ) ⃗) ………….. (ii) From (i) and (ii), the midpoints M and N of diagonals OB and AC coincide each other, i.e. they are the same point. Hence, the diagonals of a rhombus bisect each other. Now, ("AC" ) ⃗ . ("OB" ) ⃗ = (–( "a" ) ⃗ + ( "b" ) ⃗ ) . (( "a" ) ⃗ + ( "b" ) ⃗) = ((" b " ) ⃗)2 – ((" a " ) ⃗)2 = b2 – a2 = 0 [∵ a = b] ∴ ("AC" ) ⃗ ⊥ ("OB" ) ⃗. Hence diagonals of a rhombus bisect each other at right angle. Proved.

Theorem 11:

Prove by vector method that the angle in the semi-circle is a right angle. Proof: Let O be the centre of a circle and AB be a diameter. ∴ ("AO" ) ⃗ = ("OB" ) ⃗. ∠ACB is an angle in the semi-circle. Join OC. AO = OB = OC (Radii of same circle). By triangle law of vector addition, ("AC" ) ⃗ = ("AO" ) ⃗ + ("OC" ) ⃗ ("CB" ) ⃗ = ("CO" ) ⃗ + ("OB" ) ⃗ = ("CO" ) ⃗ + ("AO" ) ⃗ = ("AO" ) ⃗ – ("OC" ) ⃗ Now, ("AC" ) ⃗ . ("CB" ) ⃗ = ((AO) ⃗+(OC) ⃗ ) . (("AO" ) ⃗ – ("OC" ) ⃗) = (("AO" ) ⃗)2 – (("OC" ) ⃗)2 = AO2 – OC2 = 0 [∵ AO = OC] ∴ ∠ACB = 90°. Proved.

Theorem 12:

Prove by vector method that the midpoint of the hypotenuse of a right angled triangle is equidistant from each vertex. Proof: Let AOC be a right angled triangle and O be the origin. Let ∠AOC = 90° and M is the mid-point of AC. Then, By midpoint theorem, ("OM" ) ⃗ = "1" /"2" (("OA" ) ⃗ + ("OC" ) ⃗) or, 2("OM" ) ⃗ = ("OA" ) ⃗ + ("OC" ) ⃗ or, (2("OM" ) ⃗)2 = (("OA" ) ⃗ + ("OC" ) ⃗)2 or, 4OM2 = (("OA" ) ⃗)2 + 2 ("OA" ) ⃗ . ("OC" ) ⃗ + (("OC" ) ⃗)2 or, 4OM2 = OA2 + 2 × 0 + OC2 or, 4OM2 = OA2 + OC2 = AC2 or, OM2 = "1" /"4" AC2 or, OM = "1" /"2" AC Also, AM = MC = "1" /"2" AC ∴ AM = MC = OM Hence, the midpoint of the hypotenuse of a right angled triangle is equidistant from each vertex. Proved.

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