Product of Vectors

Product of Vectors

Product of Vectors: Product of vectors or multiplication of vectors can be performed in the following two ways:

- Scalar Product or Dot Product

- Vector Product or Cross Product

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Scalar Product or Dot Product

The scalar product or dot product of two vectors (" a " ) ⃗ and (" b " ) ⃗, denoted by (" a " ) ⃗. (" b " ) ⃗ and read as (" a " ) ⃗ dot (" b " ) ⃗, is defined as the product of the magnitude of two vectors multiplied by the cosine of the angle θ between their directions. Thus, (" a " ) ⃗. (" b " ) ⃗ = |(" a " ) ⃗| |(" b " ) ⃗| cosθ = ab cosθ Where, |(" a " ) ⃗| = a and |(" b " ) ⃗| = b. Let, ("OA" ) ⃗ = (" a " ) ⃗ and ("OB" ) ⃗ = (" b " ) ⃗ be to co-initial vectors as given in the figure alongside. Draw perpendicular BM from B to OA. Now, (" a " ) ⃗. (" b " ) ⃗ = |(" a " ) ⃗| |(" b " ) ⃗| cosθ = ab cosθ = (OA) (OB) cosθ = (OA) (OB cosθ) = (OA) (OM) = (magnitude of (" a " ) ⃗) (component of (" b " ) ⃗ in the direction of (" a " ) ⃗) So, the scalar product of two vectors is equivalent to the product of the magnitude of one vector with the component of the other vector in the direction of this vector. Note: If we write (" a " ) ⃗. (" b " ) ⃗, the rotation of (" a " ) ⃗ toward (" b " ) ⃗ is anticlockwise and the angle θ is taken to be positive. ∴ (" a " ) ⃗. (" b " ) ⃗ = ab cosθ If we write (" b " ) ⃗. (" a " ) ⃗, the rotation of (" b " ) ⃗ toward (" a " ) ⃗ is clockwise and the angle θ is taken to be negative. ∴ (" b " ) ⃗. (" a " ) ⃗ = ba cos(–θ) = ab cosθ Thus, (" a " ) ⃗. (" b " ) ⃗ = (" b " ) ⃗. (" a " ) ⃗, i.e. scalar product is commutative.

Scalar Product in Component Form

Let us consider two points A(a1, a2) and B(b1, b2) in the plane. Then, Position vector of A = ("OA" ) ⃗ = (" a " ) ⃗ = (■("a" _"1" @"a" _"2" )) Position vector of B = ("OB" ) ⃗ = (" b " ) ⃗ = (■("b" _"1" @"b" _"2" )) Magnitude of (" a " ) ⃗ and (" b " ) ⃗ are |("OA" ) ⃗| = OA = a = |(" a " ) ⃗| |("OB" ) ⃗| = OB = b = |(" b " ) ⃗| Let ∠XOA = 𝛽, ∠XOB = 𝛼 and ∠AOB = θ. Then, 𝛼 – 𝛽 = θ. Draw perpendiculars AM and BN from A and B to the X-axis. Then, OM = a1, MA = a2, ON = b1 and NB = b2. Now, from the right angled ΔOMA, cos𝛽 = "OM" /"OA" = "a" _"1" /"a" ⟹ a1 = a cos𝛽 sin𝛽 = "MA" /"OA" = "a" _"2" /"a" ⟹ a2 = a sin𝛽 Similarly, from the right angled ΔONB, b1 = b cos𝛼 b2 = b sin𝛼 Now, a1b1 + a2b2 = a cos𝛽 b cos𝛼 + a sin𝛽 b sin𝛼 = ab (cos𝛽 cos𝛼 + sin𝛽 sin𝛼) = ab cos(𝛼 – 𝛽) = |(" a " ) ⃗| |(" b " ) ⃗| cosθ ………………... (i) But, by the definition of scalar product of two vectors (" a " ) ⃗ and (" b " ) ⃗ , (" a " ) ⃗. (" b " ) ⃗ = |(" a " ) ⃗| |(" b " ) ⃗| cosθ ……………..... (ii) From (i) and (ii), (" a " ) ⃗. (" b " ) ⃗ = a1b1 + a2b2 This result gives us the scalar product of two vectors in component form. Thus, if (" a " ) ⃗ = (■("a" _"1" @"a" _"2" )) and (" b " ) ⃗ = (■("b" _"1" @"b" _"2" )) be two vectors in component form, then the scalar product of (" a " ) ⃗ and (" b " ) ⃗ is defined by (" a " ) ⃗. (" b " ) ⃗ = a1b1 + a2b2.

The angle between two vectors

From (i), |(" a " ) ⃗| |(" b " ) ⃗| cosθ = a1b1 + a2b2 or, cosθ = ("a" _"1" "b" _"1" +"a" _"2" "b" _"2" )/(|( "a" ) ⃗| |( "b" ) ⃗|) This result gives us angle between two vectors (" a " ) ⃗ and (" b " ) ⃗. θ = cos-1(("a" _"1" "b" _"1" +"a" _"2" "b" _"2" )/(|( "a" ) ⃗| |( "b" ) ⃗|)).

Properties of the scalar product

Let (" a " ) ⃗, (" b " ) ⃗ and (" c " ) ⃗ be three vectors, then the following properties are satisfied by the scalar product of vectors: Commutative property: (" a " ) ⃗. (" b " ) ⃗ = (" b " ) ⃗. (" a " ) ⃗ Distributive property: (" a " ) ⃗. ( (" b " ) ⃗+(" c " ) ⃗) = (" a " ) ⃗. (" b " ) ⃗ + (" a " ) ⃗. (" c " ) ⃗ Associative property: m(" a " ) ⃗. n(" b " ) ⃗ = mn ((" a " ) ⃗. (" b " ) ⃗)

Perpendicular vectors

Let (" a " ) ⃗ = (■("a" _"1" @"a" _"2" )) and (" b " ) ⃗ = (■("b" _"1" @"b" _"2" )) be two vectors. If (" a " ) ⃗ and (" b " ) ⃗ are perpendicular to each other, then the angle between (" a " ) ⃗ and (" b " ) ⃗ is θ = 90°. Now, (" a " ) ⃗. (" b " ) ⃗ = |(" a " ) ⃗| |(" b " ) ⃗| cosθ = ab cos90° = 0 Conversely, let (" a " ) ⃗. (" b " ) ⃗ = 0 Then, |(" a " ) ⃗| |(" b " ) ⃗| cosθ = 0 or, ab cosθ = 0 or, cosθ = 0 = cos90° ∴ θ = 90° Thus, two vectors (" a " ) ⃗ and (" b " ) ⃗ are perpendicular to each other (or orthogonal) if (" a " ) ⃗. (" b " ) ⃗ = 0.

Parallel Vectors

Let (" a " ) ⃗ and (" b " ) ⃗ be two vectors. If (" a " ) ⃗ and (" b " ) ⃗ are parallel to each other, then the angle between (" a " ) ⃗ and (" b " ) ⃗ is θ = 0° or 180°. Now, If θ = 0°, (" a " ) ⃗. (" b " ) ⃗ = |(" a " ) ⃗| |(" b " ) ⃗| cosθ = ab cos0° = ab If θ = 180°, (" a " ) ⃗. (" b " ) ⃗ = |(" a " ) ⃗| |(" b " ) ⃗| cosθ = ab cos180° = – ab Thus, two vectors (" a " ) ⃗ and (" b " ) ⃗ are parallel to each other if (" a " ) ⃗. (" b " ) ⃗ = ab or, (" a " ) ⃗. (" b " ) ⃗ = – ab.

Mutually Perpendicular Unit Vectors

(" i " ) ⃗ = (■("1" @"0" )) and (" j " ) ⃗ = (■("0" @"1" )) are the unit vectors along X-axis and Y-axis respectively. Since two axes on co-ordinate plane are perpendiculars, the unit vectors (" i " ) ⃗ and (" j " ) ⃗ are mutually perpendicular to each other. Therefore, (" i " ) ⃗.(" j " ) ⃗ = 0 and (" j " ) ⃗.(" i " ) ⃗ = 0. Scalar products of (" i " ) ⃗ and (" j " ) ⃗, (" i " ) ⃗.(" i " ) ⃗ = (■("1" @"0" )) . (■("1" @"0" )) = 1 + 0 = 1 (" j " ) ⃗.(" j " ) ⃗ = (■("0" @"1" )) . (■("0" @"1" )) = 0 + 1 = 1 (" i " ) ⃗.(" j " ) ⃗ = (■("1" @"0" )) . (■("0" @"1" )) = 0 + 0 = 0 (" j " ) ⃗.(" i " ) ⃗ = (■("0" @"1" )) . (■("1" @"0" )) = 0 + 0 = 0 The value of scalar product of (" i " ) ⃗ and (" j " ) ⃗ can be remembered from the table given below. . (" i " ) ⃗ (" j " ) ⃗ (" i " ) ⃗ 1 0 (" j " ) ⃗ 0 1

Scalar Product in Terms of Unit Vectors

Let (" a " ) ⃗ = a1(" i " ) ⃗ + a2(" j " ) ⃗ and (" b " ) ⃗ = b1(" i " ) ⃗ + b2(" j " ) ⃗ be two vectors, then (" a " ) ⃗.(" b " ) ⃗ = (a1(" i " ) ⃗ + a2(" j " ) ⃗).( b1(" i " ) ⃗ + b2(" j " ) ⃗) = a1b1 (" i " ) ⃗.(" i " ) ⃗ + a1b2 (" i " ) ⃗.(" j " ) ⃗ + a2b1 (" j " ) ⃗.(" i " ) ⃗ + a2b2 (" j " ) ⃗.(" j " ) ⃗ = a1b1 + 0 + 0 + a2b2 [∵(" i " ) ⃗.(" j " ) ⃗ = (" j " ) ⃗.(" i " ) ⃗ = 0] = a1b1 + a2b2

Worked Out Examples:

Example 1: If (" a " ) ⃗ = 3(" i " ) ⃗ + 4(" j " ) ⃗ and (" b " ) ⃗ = (" i " ) ⃗ + (" j " ) ⃗, find (" a " ) ⃗.(" b " ) ⃗ angle between (" a " ) ⃗ and (" b " ) ⃗ unit vector along (" a " ) ⃗. Solution: (" a " ) ⃗.(" b " ) ⃗ = (3(" i " ) ⃗ + 4(" j " ) ⃗).( (" i " ) ⃗ + (" j " ) ⃗) = 3(" i " ) ⃗.(" i " ) ⃗ + 3(" i " ) ⃗.(" j " ) ⃗ + 4(" j " ) ⃗.(" i " ) ⃗ + 4(" j " ) ⃗.(" j " ) ⃗ = 3 + 0 + 0 + 4 = 7 |(" a " ) ⃗| = √("3" ^"2" " +" 〖" 4" 〗^"2" ) = √("25" ) = 5 |(" b " ) ⃗| = √("1" ^"2" " +" 〖" 1" 〗^"2" ) = √("2" ) ∴ Angle, θ = cos-1((( "a" ) ⃗.( "b" ) ⃗)/(|( "a" ) ⃗| |( "b" ) ⃗|)) = cos-1("7" /("5" √("2" ))) = 8.13° Unit vector along (" a " ) ⃗ = (" a " ) ⃗ = ( "a" ) ⃗/(|( "a" ) ⃗|) = "1" /"5" (3(" i " ) ⃗ + 4(" j " ) ⃗) = "3" /"5" (" i " ) ⃗ + "4" /"5" (" j " ) ⃗

Example 2: If (" a " ) ⃗ = 6(" i " ) ⃗ – 8(" j " ) ⃗ and (" b " ) ⃗ = 8(" i " ) ⃗ + 6(" j " ) ⃗, show that (" a " ) ⃗ and (" b " ) ⃗ are perpendicular to each other. Solution: Here, (" a " ) ⃗ = 6(" i " ) ⃗ – 8(" j " ) ⃗ and (" b " ) ⃗ = 8(" i " ) ⃗ + 6(" j " ) ⃗ ∴ (" a " ) ⃗.(" b " ) ⃗ = (6(" i " ) ⃗ – 8(" j " ) ⃗).( 8(" i " ) ⃗ + 6(" j " ) ⃗) = 48(" i " ) ⃗.(" i " ) ⃗ + 36(" i " ) ⃗.(" j " ) ⃗ – 64(" j " ) ⃗.(" i " ) ⃗ – 48(" j " ) ⃗.(" j " ) ⃗ = 48 + 0 – 0 – 48 = 0 Since (" a " ) ⃗.(" b " ) ⃗ = 0, (" a " ) ⃗ and (" b " ) ⃗ are perpendicular to each other.

Example 3: Show that (" a " ) ⃗ = (■("3" @"6" )) and (" b " ) ⃗ = (■("6" @"12" )) are parallel vectors. Solution: Here, (" a " ) ⃗ = (■("3" @"6" )) and (" b " ) ⃗ = (■("6" @"12" )) ∴ (" a " ) ⃗.(" b " ) ⃗ = (■("3" @"6" )) . (■("6" @"12" )) = 18 + 72 = 90 a = |(" a " ) ⃗| = √("3" ^"2" " +" 〖" 6" 〗^"2" ) = √("9 +36" ) = √("45" ) = 3√("5" ) b = |(" b " ) ⃗| = √("6" ^"2" " +" 〖" 12" 〗^"2" ) = √("36 +144" ) = √("180" ) = 6√("5" ) ∴ ab = 3√("5" ) × 6√("5" ) = 90 ∴ (" a " ) ⃗.(" b " ) ⃗ = ab So (" a " ) ⃗ and (" b " ) ⃗ are parallel to each other.

Example 4: Find the angle between (" a " ) ⃗ = 3(" i " ) ⃗ – 4(" j " ) ⃗ and X-axis. Solution: Here, (" a " ) ⃗ = 3(" i " ) ⃗ – 4(" j " ) ⃗ = (■("3" @–"4" )) ∴ |(" a " ) ⃗| = √("3" ^"2" " +" 〖" (–4)" 〗^"2" ) = √("9 + 16" ) = √("25" ) = 5 Let (" b " ) ⃗ = unit vector along X-axis = (■("1" @0)) ∴ |(" b " ) ⃗| = √("1" ^"2" " +" 〖" 0" 〗^"2" ) = 1 Now, (" a " ) ⃗.(" b " ) ⃗ = (■("3" @–"4" )) . (■("1" @0)) = 3 + 0 = 3 ∴ θ = cos-1((( "a" ) ⃗.( "b" ) ⃗)/(|( "a" ) ⃗| |( "b" ) ⃗|)) = cos-1("3" /"5" ) = 53.13°

Example 5: If ("AB" ) ⃗ displaces apoint A(5, -4) to the point B(6, 1), express ("AB" ) ⃗ in the form of x(" i " ) ⃗ + y(" j " ) ⃗. Solution: Here, Position vector of A = ("OA" ) ⃗ = (■("5" @–"4" )) Position vector of B = ("OB" ) ⃗ = (■("6" @1)) ∴ ("AB" ) ⃗ = ("OB" ) ⃗ – ("OA" ) ⃗ = (■("6" @1)) – (■("5" @–"4" )) = (■("1" @5)) = (" i " ) ⃗ + 5(" j " ) ⃗

Example 6: In ΔABC, if ("AB" ) ⃗ = 3(" i " ) ⃗ + 2(" j " ) ⃗ and ("BC" ) ⃗ = –(" i " ) ⃗ – 5(" j " ) ⃗, show that the triangle is right angled triangle. Which angle is a right angle ? Solution: Here, ("AB" ) ⃗ = 3(" i " ) ⃗ + 2(" j " ) ⃗ and ("BC" ) ⃗ = –(" i " ) ⃗ – 5(" j " ) ⃗ Now, ("AC" ) ⃗ = ("AB" ) ⃗ + ("BC" ) ⃗ = 3(" i " ) ⃗ + 2(" j " ) ⃗ – (" i " ) ⃗ – 5(" j " ) ⃗ = 2(" i " ) ⃗ – 3(" j " ) ⃗ Again, ("AB" ) ⃗ . ("AC" ) ⃗ = (3(" i " ) ⃗ + 2(" j " ) ⃗) . (2(" i " ) ⃗ – 3(" j " ) ⃗) = 6(" i " ) ⃗.(" i " ) ⃗ – 9(" i " ) ⃗.(" j " ) ⃗ + 4(" j " ) ⃗.(" i " ) ⃗ – 6(" j " ) ⃗.(" j " ) ⃗ = 6 – 0 + 0 – 6 = 0 ∴ ("AB" ) ⃗ ⊥ ("AC" ) ⃗ Hence, ΔABC is a right angled triangle in which ∠A = 90°.

Example 7: If A(1, 1), B(2, 3) and C(5, -1) are the vertices of a triangle ABC, prove by vector method that ABC is a right angled triangle. Solution: Here, the vertices of triangle ABC are A(1, 1), B(2, 3) and C(5, -1). Let O be the origin. Then, ("OA" ) ⃗ = (■("1" @1)) , ("OB" ) ⃗ = (■("2" @3)) and ("OC" ) ⃗ = (■("5" @"-" 1)) Now, ("AB" ) ⃗ = ("OB" ) ⃗ – ("OA" ) ⃗ = (■("2" @3)) – (■("1" @1)) = (■("1" @2)) = (" i " ) ⃗ + 2(" j " ) ⃗ ("BC" ) ⃗ = ("OC" ) ⃗ – ("OB" ) ⃗ = (■("5" @–1)) – (■("2" @3)) = (■("3" @"-4" )) = 3(" i " ) ⃗ – 4(" j " ) ⃗ ("CA" ) ⃗ = ("OA" ) ⃗ – ("OC" ) ⃗ = (■("1" @1)) – (■("5" @–1)) = (■(–"4" @2)) = –4(" i " ) ⃗ + 2(" j " ) ⃗ Now, ("CA" ) ⃗ . ("AB" ) ⃗ = (–4(" i " ) ⃗ + 2(" j " ) ⃗) . ((" i " ) ⃗ + 2(" j " ) ⃗) = -4(" i " ) ⃗.(" i " ) ⃗ – 8(" i " ) ⃗.(" j " ) ⃗ + 2(" j " ) ⃗.(" i " ) ⃗ + 4(" j " ) ⃗.(" j " ) ⃗ = -4 – 0 + 0 + 4 = 0 ∴ ("CA" ) ⃗ ⊥ ("AB" ) ⃗ Hence ABC is a right angled triangle.

Example 8: If ("AB" ) ⃗ = 3(" i " ) ⃗ – 5(" j " ) ⃗ and ("BC" ) ⃗ = 2(" i " ) ⃗ + 8(" j " ) ⃗, prove that ABC is an isosceles right angled triangle. Solution: Here, ("AB" ) ⃗ = 3(" i " ) ⃗ – 5(" j " ) ⃗ and ("BC" ) ⃗ = 2(" i " ) ⃗ + 8(" j " ) ⃗ Then, ("AC" ) ⃗ = ("AB" ) ⃗ – ("BC" ) ⃗ = 3(" i " ) ⃗ – 5(" j " ) ⃗ + 2(" i " ) ⃗ + 8(" j " ) ⃗ = 5(" i " ) ⃗ + 3(" j " ) ⃗ Now, ("AB" ) ⃗ . ("AC" ) ⃗ = (3(" i " ) ⃗ – 5(" j " ) ⃗) . (5(" i " ) ⃗ + 3(" j " ) ⃗) = 15(" i " ) ⃗.(" i " ) ⃗ + 9(" i " ) ⃗.(" j " ) ⃗ – 25(" j " ) ⃗.(" i " ) ⃗ – 15(" j " ) ⃗.(" j " ) ⃗ = 15 + 0 – 0 – 15 = 0 ∴ ("AB" ) ⃗ ⊥ ("AC" ) ⃗ and hence ∠A = 90°. Again, AB = |("AB" ) ⃗| = √("3" ^"2" " +" 〖" (–5)" 〗^"2" ) = √("9 + 25" ) = √("34" ) AC = |("AC" ) ⃗| = √("5" ^"2" " +" 〖" 3" 〗^"2" ) = √("25 + 9" ) = √("34" ) Since ∠A = 90° and AB = AC, ΔABC is an isosceles right angled triangle. Proved.

Example 9: In the given figure, ABC is a triangle in which ∠ABC = 120° and AB = AC. Show that AB = "1" /√("3" ) AC. Solution: We have, ("AB" ) ⃗ + ("BC" ) ⃗ = ("AC" ) ⃗ [By triangle law] or, (("AB" ) ⃗ + ("BC" ) ⃗)2 = (("AC" ) ⃗)2 [Squaring] or, (("AB" ) ⃗)2 + 2("AB" ) ⃗ . ("BC" ) ⃗ + (("BC" ) ⃗)2 = (("AC" ) ⃗)2 or, AB2 + 2|("AB" ) ⃗|.|("BC" ) ⃗| cos60° + BC2 = AC2 [█("∵ Angle made by " ("AB" ) ⃗" and " ("BC" ) ⃗" is 180° – 120° " @"= 60° and AB = BC" @)] or, AB2 + 2AB.AB × "1" /"2" + AB2 = AC2 or, 2AB2 + 2 AB2 × "1" /"2" = AC2 or, 3AB2 = AC2 or, AB2 = "1" /"3" AC2 or, AB = "1" /√("3" ) AC Proved.

Example 10: If ABC be a right angled triangle with ∠BAC = 90°, show by vector method that AB2 + AC2 = BC2. Solution: By triangle law of vector addition, ("BA" ) ⃗ + ("AC" ) ⃗ = ("BC" ) ⃗ i.e. (("BA" ) ⃗ + ("AC" ) ⃗)2 = (("BC" ) ⃗)2 [Squaring] or, (("BA" ) ⃗)2 + 2 ("BA" ) ⃗ . ("AC" ) ⃗ + (("AC" ) ⃗)2 = (("BC" ) ⃗)2 [Since ("BA" ) ⃗ ⊥ ("AC" ) ⃗, ("BA" ) ⃗ . ("AC" ) ⃗ = 0] or, BA2 + 0 + AC2 = BC2 [∵ BA2 = AB2] or, AB2 + AC2 = BC2 Proved.

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