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.

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

 

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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