Inter 1st Year Maths 1A Products of Vectors Formulas

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 5 Products of Vectors to solve questions creatively.

Intermediate 1st Year Maths 1A Products of Vectors Formulas

Scalar or Dot Product of Two Vectors:
The scalar or dot product of two non – zero vectors \(\bar{a}\) and \(\bar{b}\), denoted by \(\bar{a} \cdot \bar{b}\) is defined as \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}|\) cos \((\bar{a}, \bar{b})\). This is a scalar, either \(\bar{a}\) = 0 (or) \(\bar{b}\) = 0, then we define \(\bar{a} \cdot \bar{b}\) = 0. If we write \((\bar{a}, \bar{b})\) = 0, then \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}|\) cos θ, if a ≠ 0, b ≠ 0, since 0 ≤ (a, b) = θ ≤ 7 80°, we get

  • 0 ≤ θ < 90° ⇒ \(\bar{a}\). b > 0.
  • θ = 90° ⇒ \(\bar{a} \cdot \bar{b}\) = 0 and the vectors \(\bar{a}\) and \(\bar{b}\) are perpendicular.
  • 90° < θ ≤ 180° ⇒ \(\bar{a} \cdot \bar{b}\) < 0
  • \(\bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}\)
  • a̅ (b̅ + c̅) = a̅ .b̅ + a̅ .c̅
  • If a̅, b̅ are parallel, a̅.b̅ = ± |a̅ | |b̅ |.
  • If l, m ∈ R, (la̅).(mb̅) = lm(a̅. b̅)
  • Projection of b̅ on a̅ (or) length of the projection a̅ = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\)
  • Orthogonal projection of b̅ on a̅ = \(\frac{(\bar{a} \cdot \bar{b})_{\bar{a}}}{|\bar{a}|^{2}}\); a̅ ≠ 0
    or
    The projection vector b̅ on a̅ = \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^{2}}\right)\) a̅ and it is magnitude = \(\frac{|\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}|}{|\overline{\mathrm{a}}|}\)
  • The component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^{2}}\right)\)a̅

Component vector of a̅ along b̅ = \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\), component vector of a̅ perpendicular to b̅ = a̅ – \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\)

Inter 1st Year Maths 1A Products of Vectors Formulas

Orthogonal unit vectors :
Ifi, j, k are orthogonal unit vector triad in a right handed system, then

  • i̅ .j̅ = j̅.k̅ = k̅.i̅ = 0
  • i̅ .i̅ = j̅.j̅ = k̅.k̅ = 1
  • If r is any vector, r̅ = (r̅.i̅)i̅ +(r̅.j̅)j̅ ≠ (r̅.k̅)k̅

Some identities :
If a̅, b̅, c̅ are three vectors, then

  • (a̅ + b̅)2 = |a̅|2 + |b̅|2 + 2(a̅ . b̅)
  • (a̅ – b̅)2 = |a̅|2 + |b̅|2 – 2(a̅.b̅)
  • (a̅ + b̅)2 + (a̅ – b̅)2 = 2(|a̅|2 + |b̅|2)
  • (a̅ + b̅)2 – (a̅ – b̅)2 = 4(a̅. b̅)
  • (a̅ + b̅). (a̅ – b̅) = |a̅|2 – |b̅|2
  • (a̅ + b̅ + c̅)2 = |a̅|2 + |b̅|2 + |c̅|2 + 2(a̅ . b̅) + 2(b̅ . c̅) + 2(c̅ .a̅)

→ If a̅ = a1 i̅ +a2j̅ + a3k̅ and b̅ = b1i̅ + b2j̅ + b3k̅, then
a̅.b̅ = a1b1 + a2b2 + a3b3
a̅ is perpendicular to b̅
⇔ a1b1 + a2b2 + a3b3 = 0

→ |a̅| = \(\), |b̅| = \(\)

→ If (a̅, b̅) = then cos θ = \(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\) and sin θ = \(\sqrt{\frac{\sum\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}}{\left(\sum a_{1}^{2}\right)\left(\sum b_{1}^{2}\right)}}\)

→ a̅ is parallel to b̅ ⇔ a1: b1 = a2 : b2 = a3: b3

→ a̅.a̅ >0; |a̅.b̅| < |a̅| |b̅|
|a̅ + b̅| ≤ |a̅| + |b̅|; |a̅ – b̅| ≤ |a̅| + |b̅| ;
|a̅ – b̅| ≥ |a̅| – |b̅|

Vector equations of a plane :

  • The equation of the plane, whose perpendicular distance from the origin is p and whose unit normal drawn from the origin towards the plane is h is n̂ is r̅.n̂ = p.
  • Equation of a plane passing through the origin and perpendicular to the unit vector n̅, is r̅.n̅ = 0
  • Vector equation of a plane passing through a point A with position vector a and perpendicular to a vector n̅ is (r – a̅). n̅ = 0.

Perpendicular distance from the origin to the plane (r̅ – a̅).h = 0 is a̅. n̅ . where ‘a̅’ is the position vector of A in the plane and ‘n̅’ is a unit vector perpendicular to the plane.

Angle between two planes :
If π1 and π22 be two planes and \(\bar{M}_{1}, \bar{M}_{2}\) are normals drawn to them, we define the angle between M1 and M2 as the angle between π1 and π2. If the angle between \(\) and \(\) is θ, the angle between the given planes θ = cos-1\(\left[\frac{\bar{M}_{1} \cdot \bar{M}_{2}}{\left|\bar{M}_{1}\right|\left|\bar{M}_{2}\right|}\right]\)

Work done by a constant force F:

  • If a constant force F̅ acting on a particle displaces it from a position ‘A’ to the position B, then the work done ‘W by this constant force T is the dot product of the vectors
    representing the force F̅ and displacement \(\overline{A B}\), i.e., W = F̅.\(\overline{A B}\).
  • If F is the resultant of the forces F̅1, F̅2, ……………….F̅n, then workdone in displacing the particle from A to B is
    \(\bar{W}=\bar{F}_{1} \cdot \overline{A B}+\bar{F}_{2} \cdot \overline{A B}+\ldots \ldots+F_{n} \cdot \overline{A B}\)

Cross Product or Vector Product of two vectors :
The vector product or cross product of two non-parallel non – zero vectors ‘a̅’ and ‘b̅’ is defined as a̅ × b̅ = |a̅||b̅| sin θ n̂, where ‘ n̂’ is a unit vector perpendicular to the plane containing ‘a̅’ and ‘b̅’ such that a̅, b̅ and ‘n̂’ form a vector triad in the right handed system and (a̅, b̅) = θ, this is a vector. If either of a̅, b̅ is a zero vector or ‘a̅’ is parallel to ‘b̅’, we define a̅ × b̅ = 0.

Some important results on vector product:

  • |a̅ × b̅| = |a̅||b̅|sinθ ≤ |a̅||b̅| ;
  • |a̅ × b̅| = |b̅ × a̅|
  • a̅ × b̅ = -(b̅ × a̅):
  • -a̅ × -b̅ = a̅ × b̅
  • (-a̅) × b̅ = a̅ × (-b̅) – (a̅ × b̅)
  • la̅ × mb̅ = lm(a̅ × b̅) ;
  • a̅ × (b̅ + c̅) = a̅ × b̅ + a̅ × c̅
  • a̅ ≠ 0, b̅ ≠ 0 and a̅ × b̅ = 0 ⇔ ‘a̅’ and ‘b̅’ are parallel vectors.
  • a̅, b̅ , c̅ are non-zero vectors and a̅ × c̅ = b̅ × c̅ ⇒ either a̅ = b̅ or a̅ – b̅ is parallel to c̅.

Vector product among i. i and k:
If i̅, j̅ and k̅ are orthogonal unit vectors triad in the right handed system then

  • i̅ × j̅ = j̅ × j = k̅ × k̅ = 0
  • i̅ × j̅ = k̅ =-j̅ × i̅ ; j̅ × k̅ = k̅ × j̅ = i̅ ; k̅ × i̅ = -i̅ × k̅ = j̅
  • If a̅ = a1 i̅ + a2 j + a3k ; b̅ = b1i̅ + b2 j̅ + b3k̅, then
    a̅ × b̅ = a2b3 – a3b2)i̅ + (a3b1 – a1b3)j̅ + (a1b2 – a2b1)k̅

This may be represented in the form of a determinants as a̅ × b̅ = \(\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

  • Unit vectors perpendicular to both ‘a̅’and ‘b̅’ are ± \(\frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}\)
  • If a̅ = a1i̅ + a2j̅ + a3k̅ ; b̅ = b1i̅ + b2 j̅ + b3k̅ and (a̅, b̅) = θ, then
    sin θ = \(\frac{\sqrt{\sum\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\) cos θ = \(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\)

Inter 1st Year Maths 1A Products of Vectors Formulas

Vector areas:

  • If \(\overline{A B}=\bar{c}\) and \(\overline{A C}=\bar{b}\) are two adjacent sides of a triangle ABC, then vector area of ΔABC = \(\frac{1}{2}\)(c̅ × b̅) and the area of the ΔABC = \(\frac{1}{2}\)|c̅ × b̅| $q. units.
  • If a̅, b̅, c̅ are the position vectors of A, B, C respectively then the vector area of
    ΔABC = \(\frac{1}{2}\)[(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)]
    Area of ΔABC = \(\frac{1}{2}\)|(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)|sq. units.
  • If \(\) and \(\) are the diagonals of a parallelogram ABCD, then the vector area of the parallelogram = \(\frac{1}{2}\)|a̅ × b̅| and area = \(\frac{1}{2}\)|a̅ × b̅|sq. units.
  • If AB = a̅ and AD = b̅ are two adjacent sides of a parallelogram ABCD, then its vector area = a̅ × b̅ and area = |a̅ × b̅| sq. units.
  • Vector area of the quadrilateral ABCD = \(\frac{1}{2}\)\((A C \times B D)\) and area of the quadrilateral ABCD = \(\frac{1}{2}\)\(|\overline{A C} \times \overline{B D}|\)sq. units.

Some useful formulas :

  • If a̅, b̅ are two non-zero and non-parallel vectors then
    (a̅ × b̅)2 = a2b2 – (a̅.b̅)2 = \(\left|\begin{array}{cc}
    a \cdot \bar{a} & \bar{a} \cdot \bar{b} \\
    \bar{a} \cdot b & b \cdot \bar{b}
    \end{array}\right|\)
  • For any vector a̅,(a̅ × i̅)2 + (a̅ × j̅)2 + (a̅ × k̅)2 = 2|a|2
  • If a̅, b̅, c̅ are the position vectors of the points A, B, C respectively, then the perpendicular distance from c to the line AB is \(\frac{|\overline{A C} \times \overline{A B}|}{|\overline{A B}|}=\frac{|(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})+(\bar{a} \times \bar{b})|}{|b-\bar{a}|}\)

Moment of a force :
Let 0 be the point of reference (origin) and \(\overline{o p}=\bar{r}\) be the position vector of a point p on the line of action of a force F̅. Then the moment of the force F about 0 is given by r̅ × F̅.

Scalar triple product:
Let a̅, b̅, c̅ he three vectors. We call (a̅ × b̅). c̅ the scalar product of a̅, b and c. This is a scalar (real number). It is written as [a̅ b̅ c̅]

  • If (a̅ × b̅). c̅ = 0, then one or more of the vectors a̅, b̅ and c̅ should be zero vectors. If a ≠ 0, b ≠ 0, c ≠ 0, then c is perpendicular to a̅ × b̅. Hence the vector c̅ lies on the plane determined by a̅ and b̅. Hence a̅, b̅ and c̅ are coplanar.
  • If in a scalar triple product, any two vectors are parallel (equal), then the scalar triple product is zero i.e., [a̅ a̅ b̅] = [a̅ b̅ b̅] = [c̅ b̅ c̅] = 0.
  • In a scalar triple product remains unaltered if the vectors are permutted cyclically i.e., [a̅ b̅ c̅] = [b̅ c̅ a̅] = [c̅ a̅ b̅].
    However [a̅ b̅ c̅] = -[b̅ a̅ c̅] = -[c̅ b̅ a̅] = -[a̅ c̅ b̅].
  • In a scalar triple product, the dot and cross are interchangeable i.e., a̅.b̅ × c̅ = a̅ × b̅.c̅

→ If i̅ , j̅ , k̅ are orthogonal unit vector triad in the right handed system, then

  • [i̅ j̅ k̅ ] = [j̅ k̅ i̅ ] = [k̅ i̅ j̅ ] = 1
  • [i̅ k̅ j̅ ] = [j̅ i̅ k̅ ] = [k̅ j̅ i̅ ] = -1
  • If a̅ = a1 i̅ + a2j̅ + a3 k̅ ; b̅ = b1i̅ + b2 j̅ + b3k̅ and c̅ = c1i̅ + c2 j̅ + c3k̅ then [a̅ b̅ c̅] = \(\left|\begin{array}{lll}
    a_{1} & a_{2} & a_{3} \\
    b_{1} & b_{2} & b_{3} \\
    c_{1} & c_{2} & c_{3}
    \end{array}\right|\)

Inter 1st Year Maths 1A Products of Vectors Formulas

→ A necessary and sufficient condition that three non-parallel (non-collinear) and non-zero vectors a, b and c to be coplanar is [a̅ b̅ c̅] = 0. If [a̅ b̅ c̅] ≠ 0, then the three vectors are non-coplanar.

→ If a̅, b̅, c̅ are three non-zero, non-coplanar vectors and V is the volume of the parallelopiped with co-terminus edges a̅, b̅ and c̅, then v = |(a × b). c|. = |[a b c]|

→ The volume of the parallelopiped formed with A, B, C, D as vertices is \(|[A B A C A D]|\) cubic units.

→ If a̅, b̅, c̅ represent the co-terminus edges of a tetrahedron, then its volume = \(\frac{1}{6}\)[a̅, b̅, c̅] cubic units.

→ If A(x1 y1 z1], B(x2, y2, z2] C(x3 y3 z3) and D(x4, y4 z4] are the vertices of a tetrahedron = \(\frac{1}{6}\)\(|[A B A C A D]|\)

  • Vector equation of a plane containing three non-collinear points a̅, b̅, c̅ is r̅ .[(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)] = [a̅ b̅ c̅]
  • A unit vector perpendicular to the plane containing three non-collinear points a̅, b̅, c̅ is \(\frac{(\bar{a} \times b)+(b \times \bar{c})+(\bar{c} \times \bar{a})}{|(\bar{a} \times \bar{b})+(b \times \bar{c})+(\bar{c} \times a)|}\)
  • Length of the perpendicular from the origin to the plane containing three non-collinear points a̅, b̅, c̅ is \(\frac{|[\bar{a} b c]|}{|(\bar{a} \times \bar{b})+(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})|}\)

→ Vector equation of the plane passing through three non-collinear points a̅, b̅ and c̅ is [r̅ – a̅ b̅ – a̅ c̅ – a̅] = 0

→ Vector equation of the plane passing through a given point a̅ and parallel to the vectors b̅ and c̅ is [r b̅ c̅] = [a̅ b̅ c̅]

→ Vector equation of the line passing through the point a̅ and parallel to the vector b̅ is (r – a̅) × b̅ = 0

→ Distance of the point p(c) from a line joining the points A(a) and B(b) = |(c̅ – a̅) × b̅|
(25) i) Equation of the plane passing through the point p(x1, y1, z1) and perpendicular to the vector ai̅ + bj̅ + ck̅ is a (x – x1) + b(y – y1) + c(z – z1) = 0.

→ The equation of the plane passing through the points (x1 y1 z1), (x2, y2, z2) and
(x3, y3, z3) is \(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0

Skew lines:
Two lines l and m are called skew lines if there is no plane passing through these lines.

Shortest distance between the skew lines:
Shortest distance between the skew lines r̅ = a̅ + tb̅ and r̅ = c̅ + sd̅ is

Inter 1st Year Maths 1A Products of Vectors Formulas

Vector triple product:
If a̅, b̅ and c̅ are three vectors, products of the type (a̅ × b̅) × c̅, a̅ × (b̅ × c̅) from the vector triple products. From this definition.

  • If any one of a̅, b̅ and c̅ is a zero vector, a̅ × (b̅ × c̅) or (a̅ × b̅) c̅ = 0
  • If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0 and a̅ is parallel to b, then (a̅ × b̅) × c̅ = 0
  • If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0 and c̅ is perpendicular to the plane of a and b, then (a̅ × b̅) × c̅ = 0.

→ If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0, a̅ and b̅ are non-parallel vectors and c̅ is not perpendicular to the plane passing through a̅ and b, then
(a̅ × b̅) × c̅ = (a̅.c̅)b̅ – (b̅.c)a̅
a̅ × (b̅ × c̅) = (a̅.c̅)b̅ – (a.b̅)c̅

  • In general, vector triple product of three vectors need not satisfy the associative law. i.e., (a̅ × b) × c ≠ a̅ × (b × c)
    For any three vectors a̅, b̅ and c
  • a̅ × (b̅ × c̅) + b̅ × (c̅ × a̅) + c̅ × (a̅ × b̅) = 0
  • [a̅ × b̅ b̅ × c̅ c̅ × a̅] = [a̅ b̅ c̅]2

Scalar product of four vectors :
Scalar product of a̅, b̅, c̅ and d̅ is () = (a̅. c̅) (b̅.d̅) – (a̅. d̅) (b̅ .c̅) = \(\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\)

Vector product of four vectors:
If a̅, b̅, c̅ and d̅ are four vectors,
(a̅ × b̅) × (c̅ × d̅) = [a̅ c̅ d̅] b̅ – [b̅ c̅ d̅]a̅ – [a b̅ d̅]c̅ – [a̅ b̅ c̅]d̅

Some important results:

  • [a̅ + b̅ b̅ + c̅ c̅ + a̅] = 2[a̅ b̅ c̅]
  • i̅ × (j̅ × k̅) + i̅ × (k̅ × i̅) + k̅ × (i̅ × j̅) = 0
  • [a̅ × b̅ b̅ × c̅ c̅ × a̅] = [a̅ b̅ c̅]2
    [a̅ b̅ c̅] [l̅ m̅ n̅] = \(\)
  • If a̅, b̅, c̅ be such that a is perpendicular to (b̅ + c̅), bis perpendicular to (c̅ + a̅), c̅ is perpendicular to (a̅ + b̅), then |a̅ + b̅ + c̅| = \(\sqrt{a^{2}+b^{2}+c^{2}}\)
  • If a line makes angles α, β, γ and δ with the diagonals of a cube, then
    cos2α + cos2β + cos2γ + cos2δ = \(\frac{4}{3}\)
  • i̅ × (a̅ × i̅) + j̅ × (a̅ × j̅) + k̅ × (a̅ × k̅) – 2a̅

→ Equation of the sphere with centre at c and radius ‘a’ is r2 – 2r̅. c̅ + c2 = a2