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 – \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\)

**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 = a_{1} i̅ +a_{2}j̅ + a_{3}k̅ and b̅ = b_{1}i̅ + b_{2}j̅ + b_{3}k̅, then

a̅.b̅ = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}

a̅ is perpendicular to b̅

⇔ a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3} = 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 ⇔ a_{1}: b_{1} = a_{2} : b_{2} = a_{3}: b_{3}

→ 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 π_{1} and π_{2}2 be two planes and \(\bar{M}_{1}, \bar{M}_{2}\) are normals drawn to them, we define the angle between M_{1} and M_{2} as the angle between π_{1} and π_{2}. 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 F̅
_{1}, F̅_{2}, ……………….F̅_{n}, 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 = a
_{1}i̅ + a_{2}j + a_{3}k ; b̅ = b_{1}i̅ + b_{2}j̅ + b_{3}k̅, then

a̅ × b̅ = a_{2}b_{3}– a_{3}b_{2})i̅ + (a_{3}b_{1}– a_{1}b_{3})j̅ + (a_{1}b_{2}– a_{2}b_{1})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 = a
_{1}i̅ + a_{2}j̅ + a_{3}k̅ ; b̅ = b_{1}i̅ + b_{2}j̅ + b_{3}k̅ 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 = \(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\)

**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}= a^{2}b^{2}– (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 = a
_{1}i̅ + a_{2}j̅ + a_{3}k̅ ; b̅ = b_{1}i̅ + b_{2}j̅ + b_{3}k̅ and c̅ = c_{1}i̅ + c_{2}j̅ + c_{3}k̅ 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|\)

→ 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(x_{1} y_{1} z_{1}], B(x_{2}, y_{2}, z_{2}] C(x_{3} y_{3} z_{3}) and D(x_{4}, y_{4} z_{4}] 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(x_{1}, y_{1}, z_{1}) and perpendicular to the vector ai̅ + bj̅ + ck̅ is a (x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0.

→ The equation of the plane passing through the points (x_{1} y_{1} z_{1}), (x_{2}, y_{2}, z_{2}) and

(x_{3}, y_{3}, z_{3}) 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

**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}

**Scala****r 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

cos^{2}α + cos^{2}β + cos^{2}γ + cos^{2}δ = \(\frac{4}{3}\) - i̅ × (a̅ × i̅) + j̅ × (a̅ × j̅) + k̅ × (a̅ × k̅) – 2a̅

→ Equation of the sphere with centre at c and radius ‘a’ is r^{2} – 2r̅. c̅ + c^{2} = a^{2}

**Scalar Product**

Def: Let \(\vec{a}, \vec{b}\) be two vectors dot product (or) scalar product (or) direct product (or) inner product denoted by \(\vec{a}, \vec{b}\). Which is defined as \(|\vec{a}||\vec{b}|\)cos θ where cos θ = \((\vec{a}, \vec{b})\)

* The product \(\vec{a}, \vec{b}\) is zero when \(|\vec{a}|\) = 0 (or) \(|\vec{b}|\) = 0 (or) θ = 90°.

**Sign of the scalar product :**

Let \(\vec{a}, \vec{b}\) are two non-zero vectors

- If θ is acute then \(\vec{a}.\vec{b}\)> 0 (i.e 0 < θ < 90°).
- If θ is obtuse then \(\vec{a}.\vec{b}\) < 0 (i.e 90° < θ < 180°).
- If θ = 90° then\(\vec{a} \cdot \vec{b}\) = o.
- If θ = 0° then \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\)
- If θ = 180° then \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\)

Note:

- The dot product of two vectors is always scalar.
- \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) i.e dot product of two vectors is commutative.
- If \(\vec{a} \cdot \vec{b}\) are two vectors then \(\vec{a} \cdot(-\vec{b})=(-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b})\)
- \((-\vec{a}) \cdot(-\bar{b})=\vec{a} \cdot \vec{b}\)
- If l,m are two scalars and \(\vec{a} \cdot \vec{b}\) are two vectors then \((l \bar{a}) \cdot(m \bar{b})={lm}(\vec{a} \cdot \vec{b})\)
- If \(\vec{a}\) and \(\vec{b}\) are two vectors then \(\vec{a} \cdot \vec{b}=\pm|\vec{a}||\vec{b}|\)
- If \(\vec{a}\) is a vector then \(\vec{a} \cdot \vec{a}=|\vec{a}|^{2}\)
- If \(\vec{a}\) is a vector \(\vec{a}\).\(\vec{a}\) is denoted by \(\overline{(a)^{2}}\) hence \(\overline{(a)^{2}}=|\vec{a}|^{2}\)

**Components and orthogonal projection:**

Def: Let \(\vec{a}=\overline{O A} \quad \vec{b}=\overline{O B}\) be two non zero vectors let the plane passing through B and perpendicular to a intersect \(\overline{O A}\) ln M.

- If \((\vec{a}, \vec{b})\) is acute then OM is called component of \(\vec{b}\) on \(\vec{a}\).
- If \((\vec{a}, \vec{b})\) is obtuse then -(OM) is called the component of \(\vec{b}\) on \(\vec{a}\).
- The vector \(\overline{O M}\) is called component vector of \(\vec{b}\) on \(\vec{a}\).

Def: Let \(\vec{a}=\overline{O A}\); \(\vec{b}=\overrightarrow{P Q}\) be two vectors let the planes passing through P, Q and perpendicular to a intersect \([latex]\)[/latex] in L, M respectively then \(\overline{L M}\) is called orthogonal projection of \(\vec{b}\) on \(\vec{a}\)

Note:

- The orthogonal projection of a vector b on a is equal tb component vector of b on a .
- Component of a vector \(\vec{b}\) on \(\vec{a}\) is also called projection of \(\vec{b}\) on \(\vec{a}\)
- If A< B, C, D are four points in the space then the component of \(\overline{A B}\) on \(\overline{C D}\) is same as the projection of \(\overline{A B}\) on the ray \(\overline{C D}\).

→ If \(\vec{a}, \vec{b}\) be two vectors (\(\vec{a} \neq \vec{o}\)) then

- The component of \(\vec{b}\) on \(\vec{a}\) is \(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}\)
- The orthogonal projection of \(\vec{b}\) on \(\vec{a}\) is \(\frac{(\vec{b} \cdot \vec{a}) \vec{a}}{|\vec{a}|^{2}}\)

→ If \(\vec{i}, \vec{j}, \vec{k}\) form a right handed system of Ortho normal triad then

- \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{j}=\vec{k} \cdot \vec{k}\) = 1
- \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{i}=0 ; \vec{j} \cdot \vec{k}=\vec{k} \cdot \vec{j}=0 ; \vec{k} \cdot \vec{i}=\vec{i} \cdot \vec{k}=0\)

→ If \(\vec{a}=a_{1} \vec{i}+a_{2} \vec{j}+a_{3} \vec{k}\) \(\vec{b}=b_{1} \vec{i}+b_{2} \vec{j}+b_{3} \vec{k}\) then \(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\)

→ If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors then

→ If \(\vec{r}\) is vector then \(\vec{r}=(\vec{r} \cdot \vec{i}) i+(\vec{r}+\vec{j}) \vec{j}+(\vec{r} \cdot \vec{k}) \vec{k}\)

**Angle between the planes:**

The angle between the planes is defined as the angle between the normals to the planes drawn from any point in the space.

**Sphere:**

The vector equation of a sphere with centre C having position vector \(\vec{c}\) and radius a is \((\vec{r}-\vec{c})^{2}\) = a^{2} i.e. \(\vec{r}^{2}-2 \vec{r} \cdot \vec{c}+c^{2}\) = a^{2}

The vector equation of a sphere with A(a) and B(b) as the end points of a diameter is \((\vec{r}-\vec{a}) \cdot(\vec{r}-\vec{b})\) = 0 (or) \((\vec{r})^{2}-\vec{r} \cdot(\vec{a}+\vec{b})+\vec{a} \cdot \vec{b}\) = 0

**Work done by a force :**

If a force \(\vec{F}\) acting on a particle displaces it from a position A to the position B then work done W by this force \(\vec{F}\) is \(\vec{F} \cdot \overline{A B}\)

- The vector equation of the plane which is at a distance of p from the origin along the unit vector \(\vec{n}\) is \(\vec{r} \cdot \vec{n}\) = p.
- The vector equation of the plane passing through the origin and perpendicular to the vector m is r.m =0
- The Cartesian equation of the plane which is at a distance of p from the origin along the unit vector n = li + mj + nk of the plane is n = lx + my + nz
- The vector equation of the plane passing through the point a having position vector \(\vec{a}\) and perpendicular to the vector \(\vec{m}\) is \((\vec{r}-\vec{a}) \cdot \vec{m}\) = 0.
- The vector equation of the plane passing through the point a having position vector \(\vec{a}\) and parallel to the plane r.m=q is \((\vec{r}-\vec{a}) \cdot \vec{m}\) = 0.

**Cross( Vector) Product of Vectors:**

Let \(\vec{a}, \vec{b}\) be two vectors. The cross product or vector product or skew product of vectors \(\vec{a}, \vec{b}\) is denoted by \(\vec{a} \times \vec{b}\) and is defined as follows

- If \(\vec{a}\) = 0 or \(\vec{b}\) = 0 or \(\vec{a}, \vec{b}\) are parallel then \(\vec{a} \times \vec{b}\) = 0
- If \(\vec{a}\) ≠ 0, \(\vec{b}\) ≠ 0, \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}|(\sin \theta) \vec{n}\) where \(\vec{n}\) is a unit vector perpendicular to a and b so that a, b, n form a right handed system.

Note:

- \(\vec{a} \times \vec{b}\) is a vector
- If \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a} \times \vec{b}\) is perpendicular to both a and b
- If \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a}, \vec{b}\) , \(\vec{a} \times \vec{b}\) form a right handed system .
- For any vector \(\vec{a}\) \(\vec{a} \times \vec{b}\) = o

2. If \(\vec{a}, \vec{b}\) are two vectors \(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\) this is called “anti commutative law”

3. If \(\vec{a}, \vec{b}\) are two vectors then \(\vec{a} \times(-\vec{b})=(-\vec{a}) \times \vec{b}=-(\vec{a} \times \vec{b})\)

4. If \(\vec{a}, \vec{b}\) are two vectors then \((-\vec{a}) \times(-\vec{b})=\vec{a} \times \vec{b}\)

5. If \(\vec{a}, \vec{b}\) are two vectors l,m are two scalars then (la) x (mb) = lm(a x b)

6. If \(\vec{a}, \vec{b}, \vec{b}\) are three vectos, then

- \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\)
- \((\vec{b}+\vec{c}) \times \vec{a}=\vec{b} \times \vec{a}+\vec{c} \times \vec{a}\)

7. If \(\vec{l}, \vec{l}, \vec{k}\) from a right handed system of orthonormal triad then

- \(\vec{l} \times \vec{l}=\vec{j} \times \vec{j}=\vec{k} \times \vec{k}=\vec{o}\)
- \(\vec{i} \times \vec{j}=\vec{k}=-\vec{j} \times \vec{l} ; \vec{j} \times \vec{k}=\vec{l}=-\vec{k} \times \vec{j} ; \vec{k} \times \vec{l}=\vec{j}=-\vec{l} \times \vec{k}\)

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{j}+a_{3} \vec{k}, \vec{b}=b_{1} \vec{l}+b_{2} \vec{j}+b_{3} \vec{k}\) then \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}

\vec{l} & \vec{j} & \vec{k} \\

a_{1} & a_{2} & a_{3} \\

b_{1} & b_{2} & b_{3}

\end{array}\right|\)

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{m}+a_{3} \vec{n}, \quad \vec{b}=b_{1} \vec{l}+b_{2} \vec{m}+b_{3} \vec{n}\) where \(\vec{l}, \vec{m}, \vec{n}\) form a right system of non coplanar vectors then \(\)

→ If \(\vec{a}, \vec{b}\) are two vectors then \((\vec{a} \times \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2}\) = a^{2}b^{2}.

Vector Area:

If A is the area of the region bounded by a plane curve and \(\vec{n}\) is the unit vector perpendicular to the plane of the curve such that the direction of curve drawn can be considered anti clock wise then A\(\vec{n}\) is called vector area of the plane region bounded by the curve.

- The vector area of triangle ABC is \(\frac{1}{2} \overline{A B} \times \overrightarrow{A C}=\frac{1}{2} \overrightarrow{B C} \times \overrightarrow{B A}=\frac{1}{2} \overline{C A} \times \overrightarrow{C B}\)
- If \(\vec{a}, \vec{b}, \vec{c}\)are the position vectors of the vertices of a triangle then the vector area of the triangle is \(\frac{1}{2}(\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\)
- If ABCD is a parallelogram and \(\overrightarrow{A B}=\vec{a}, \quad \overrightarrow{B C}=\vec{b}\) then the vector area of ABCD is \(\vec{a} \times \vec{b}\).
- If ABCD is a parallelogram and \(\overrightarrow{A C}=\vec{a}, \overrightarrow{B C}=\vec{b}\) then vector area of parallelogram ABCD is \(\frac{1}{2}(\vec{a} \times \vec{b})\)
- The vector equation of a line passing through the point A with position vector a and perpendicular to the vectors \(\vec{b} \times \vec{c}\) is \(\vec{r}=\vec{a}+t(\vec{b} \times \vec{c})\).

**Scalar Triple Product:**

- If \(\vec{a}, \vec{b}, \vec{c}\) are the three vectors, then the real numbers \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is called scalar triple product denoted by \([\vec{a} \vec{b} \vec{c}]\). This is read as ‘box’ \(\vec{a}, \vec{b}, \vec{c}\)
- If V is the volume of the parallelepiped with coterminous edges a, b, c then V = |\([\vec{a} \vec{b} \vec{c}]\)|
- If \(\vec{a}, \vec{b}, \vec{c}\) form the right handed system of vectors then V = \([\vec{a} \vec{b} \vec{c}]\)
- If \(\vec{a}, \vec{b}, \vec{c}\) form left handed system of vectors then -V = \([\vec{a} \vec{b} \vec{c}]\)

Note:

- The scalar triple product is independent of the position of dot and cross. i.e. \(\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c}\)
- The value of the scalar triple product is unaltered so long as the cyclic order remains unchanged

\([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\) - The value of a scalar triple product is zero if two of its vectors are equal

\([\vec{a} \vec{a} \vec{b}]\)= 0 \([\vec{b} \vec{b} \vec{c}]\) = 0 - If a, b, c are coplanar then \([\vec{a} \vec{b} \vec{c}]\) = 0
- If a,b,c form right handed system then \([\vec{a} \vec{b} \vec{c}]\) > 0
- If a,b,c form left handed system then \([\vec{a} \vec{b} \vec{c}]\) < 0
- The value of the triple product changes its sign when two vectors are interchanged

\([\vec{a} \vec{b} \vec{c}]\) = –\([\vec{a} \vec{c} \vec{b}]\) - If l,m, n are three scalars \(\vec{a}, \vec{b}, \vec{c}\) are three vectors then \(\)

→ Three non zero non parallel vectors abc nare coplanar iff \([l \vec{a} m \vec{b} \quad n \vec{c}]={lmn}\left[\begin{array}{lll}

\vec{a} & \vec{b} & \vec{c}

\end{array}\right]\)= 0

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{m}+a_{3} \vec{n}, \vec{b}=b_{1} \vec{l}+b_{2} \vec{m}+b_{3} \vec{n}, \vec{c}=c_{1} \vec{l}+c_{2} \vec{m}+c_{3} \vec{n}\) where \(\vec{l}, \vec{m}, \vec{n}\) form a right handed system of non coplanar vectors, then \([\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}

\vec{m} \times \vec{n} & \vec{n} \times \vec{l} & \vec{l} \times \vec{m} \\

b_{1} & b_{2} & b_{3} \\

c_{1} & c_{2} & c_{3}

\end{array}\right|\)

→ The vectors equation of plane passing through the points A, B with position vectors \(\vec{a}, \vec{b}\) and parallel to the vector \(\vec{c}\) is \([\vec{r}-\vec{a} \vec{b}-\vec{a} \vec{c}]\) (or) \([\vec{r} \vec{b} \vec{c}]+[\vec{r} \vec{c} \vec{a}]=\left[\begin{array}{lll}

\vec{a} & \vec{b} & \vec{c}

\end{array}\right]\)

→ The vector equation of the plane passing through the point A with position vector \(\vec{a}\) and parallel to \(\vec{b}, \vec{c}\) is \([\vec{r}-\vec{a} \vec{b} \vec{c}]\) = 0 i.e. \(\left[\begin{array}{lll}

\vec{r} & \vec{b} & \vec{c}

\end{array}\right]=\left[\begin{array}{lll}

\vec{a} & \vec{b} & \vec{c}

\end{array}\right]\)

**Skew lines:**

Two lines are said to be skew lines if there exist no plane passing through them i.e. the lines lie on two difference planes

Def:- l_{1} and l_{2} are two skew lines. If P is a point on l_{1} and Q is a point on l_{2} such that \(\overleftarrow{P Q}\) ⊥ l_{1} and PQ ⊥ l_{2} then \(\overleftarrow{P Q}\) is called shortest distance and \(\overleftarrow{P Q}\) is called shortest distance line between the lines l_{1} and l_{2}.

The shortest distance between the skew lines \(\vec{r}=\vec{a}+t \vec{b}\) and \(\vec{r}=\vec{c}+t \vec{d}\) is \(\frac{|[\vec{a}-\vec{c} \vec{b} \vec{d}]|}{|\vec{b} \times \vec{d}|}\)

**Vector Triple Product:**

Cross Product of Three vectors : For any three vectors \(\bar{a}, \bar{b}\) or \(\bar{c}\) then cross product or vector product of these vectors are given as \(\bar{a} \times(\bar{b} \times \bar{c}),(\bar{a} \times \bar{b}) \times \bar{c}\) or \((\bar{b} \times \bar{c}) \times \bar{a}\) etc.

vi. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are non zero vectors and \(\) then b and c are parallel (or collinear) vectors.

vii. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are non zero and non parallel vectors then \(\bar{a} \times(\bar{b} \times \bar{c}), \quad \bar{b} \times(\bar{c} \times\bar{a})\) and \(\bar{c} \times(\bar{a} \times \bar{b})\) are non collinear vectors.

viii. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are any three vectors then \(\bar{a}(\bar{b} \times \bar{c})+\bar{b} \times(\bar{c} \times \bar{a})+\bar{c} \times(\bar{a} \times \bar{b})=\overline{\mathrm{O}}\)

ix. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are any three vectors then \(\bar{a}(\bar{b} \times \bar{c})+\bar{b} \times(\bar{c} \times \bar{a})+\bar{c} \times(\bar{a} \times \bar{b})\) are coplanar. [since sum of these vectors is zero]

x. \(\bar{a}(\bar{b} \times \bar{c})\) is vector lies in the plane of \(\bar{b}\) and \(\bar{c}\) or parallel to the plane of \(\bar{b}\) and \(\bar{c}\).

**Product of Four Vectors:
**Dot product of four vectors : The dot product of four vectors a̅, b̅, c̅ and d̅ is given as \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})=(\bar{a} \cdot \bar{c})(\bar{b} \cdot \bar{d})-(\bar{a} \cdot \bar{d})(\bar{b} \cdot \bar{c})=\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|\)

→ Cross product of four vectors : If a̅, b̅, c̅ and d̅ are any four vectors then

→ The vectorial equation of the plane passing through the point a and parallel to the vectors b̅, c̅is \([\overline{\mathrm{r}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ The vectorial equation of the plane passing through the points a̅, b̅ and parallel to the vector c̅ is \([\overline{\mathrm{rb}} \overline{\mathrm{c}}]+[\overline{\mathrm{r}} \overline{\mathrm{c}} \overline{\mathrm{a}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ The vectorial equation of the plane passing through the points a̅, b̅, c̅ is \([\overline{\mathrm{r}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+[\overline{\mathrm{r}} \overline{\mathrm{c}} \overline{\mathrm{a}}]+[\overline{\mathrm{ra}} \overline{\mathrm{b}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ If the points with the position vectorsa̅, b̅, c̅, d̅ are coplanar, then the condition is \([\overline{\mathrm{a}} \overline{\mathrm{bd}}]+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{d}}]+[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{d}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ Length of the perpendicular from the origin to the plane passing through the points a̅, b̅, c̅ is \(\frac{|[\overline{\mathrm{a} b} \overline{\mathrm{c}}]|}{|\overline{\mathrm{b}} \times \overline{\mathrm{c}}+\overline{\mathrm{c}} \times \overline{\mathrm{a}}+\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}\)

→ Length of the perpendicular from the point c̅ on to the line joining the points a̅, b̅ is \(\frac{\mid(\overline{\mathrm{a}}-\overline{\mathrm{c}}) \times(\overline{\mathrm{c}}-\overline{\mathrm{b}})}{|\overline{\mathrm{a}}-\overline{\mathrm{b}}|}\)

→ P, Q, R are non collinear points. Then distance of P to the plane OQR is OP. \(\left|\frac{\overline{\mathrm{OP}} \cdot(\overline{\mathrm{OQ}} \times \overline{\mathrm{OR}})}{|\overline{\mathrm{OQ}} \times \overline{\mathrm{OR}}|}\right|\)

→ Perpendicular distance from P(α̅ ) to the plane passing through A(a̅) and parallel to the vectors b and c is

→ Length of the perpendicular from the point c̅ to the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{tb}\) is \(\frac{|(\overline{\mathrm{c}}-\overline{\mathrm{a}}) \times \overline{\mathrm{b}}|}{|\overline{\mathrm{b}}|}\)