CBSE Class 12 Mathematics Vector Algebra Notes Set 02

Important Notes and Formulas

Introduction

Vectors constitute one of the several mathematical systems which can be usefully employed to provide mathematical handling for certain types of problems in Geometry, Mechanics and other branches of Applied Mathematics.
Vectors facilitate mathematical study of such physical quantities as possess direction in addition to magnitude. Velocity of a particle is one such quantity.

Classification of Physical Quantities

Physical quantities are broadly divided in two categories: (a) Vector quantities    (b) Scalar quantities.
(a) Vector quantities: Any quantity that has both magnitude and direction and for which vector addition is defined and meaningful is treated as a vector quantity, such as velocity, momentum or force.
Note: Quantities having magnitude and direction but not obeying the vector law of addition will not be treated as vectors. For example, the rotations of a rigid body through finite angles have both magnitude and direction but do not satisfy the law of vector addition, therefore they are not vectors.
(b) Scalar quantities: A quantity that has size or magnitude but does not involve the concept of direction is called a scalar quantity, such as mass, length, time, density or energy.

Mathematical Description of Vector

To understand vectors mathematically we will first understand a directed line segment.
Directed line segment: Any given portion of a given straight line where the two end points are distinguished as Initial and Terminal is called a Directed Line Segment.
The directed line segment with initial point A and terminal point B is denoted by the symbol \( \overrightarrow{AB} \). The two end points of a directed line segment are not interchangeable and the directed line segments \( \overrightarrow{AB} \) and \( \overrightarrow{BA} \) must be thought of as different.
Vector: A directed line segment is called a vector. Every directed line segment has three essential characteristics.

(i) Length: The length of \( \overrightarrow{AB} \) will be denoted by the symbol \( |\overrightarrow{AB}| \). Clearly, we have \( |\overrightarrow{AB}| = |\overrightarrow{BA}| \).
(ii) Support: The line of unlimited length of which a directed line segment is a part is called its line of support or simply the Support.
(iii) Sense: The sense of \( \overrightarrow{AB} \) is from A to B and that of \( \overrightarrow{BA} \) from B to A, so that the sense of a directed line segment is from its initial to the terminal point.

Position Vector

Let O be a fixed origin, then the position vector of a point P is the vector \( \overrightarrow{OP} \). If \( \vec{a} \) and \( \vec{b} \) are position vectors of two points A and B, then
\( \overrightarrow{AB} = \vec{b} - \vec{a} \) = P.V. of B − P.V. of A.

Displacement Vector

The distance between the end points A and B of the directed line segment AB is called the displacement vector.
Displacement vector = P.V. of final point − P.V. of initial point.

Component of a Vector in a Plane

Consider a plane equipped with a two-dimensional coordinate system with O as origin; X′OX and Y′OY two perpendicular lines through O as axes of coordinates as shown in the figure.

Let \( \hat{i} \) and \( \hat{j} \) denote the unit vectors respectively along the positive directions of X-axis and Y-axis.
Let \( \vec{v} = \overrightarrow{AB} \) be any vector in the plane under consideration.
Let P be the point such that \( \overrightarrow{OP} = \overrightarrow{AB} \). Let M be the foot of the perpendicular from P on X′OX and let OM = x and MP = y, then
\( \overrightarrow{OM} = x\hat{i} \) and \( \overrightarrow{MP} = y\hat{j} \)
Now, \( \vec{v} = \overrightarrow{AB} = \overrightarrow{OP} = \overrightarrow{OM} + \overrightarrow{MP} = x\hat{i} + y\hat{j} \)
The ordered pair (x, y) are called the components of \( \vec{v} \); x being the component along X-axis and y the component along Y-axis. \( x\hat{i} \) and \( y\hat{j} \) are called the component vectors of \( \vec{v} \) along X-axis and Y-axis respectively.

Some Observations:
(i) If the components of a vector \( \vec{v} \) are x and y, then
\( |\vec{v}| = |\overrightarrow{AB}| = |\overrightarrow{OP}| = \sqrt{OP^2} = \sqrt{OM^2 + MP^2} \), i.e. \( |\vec{v}| = \sqrt{x^2 + y^2} \)

This means that the length (or modulus) of a vector (written in component form) is the non-negative square root of the sum of the squares of its components.
(ii) A point P in the XOY-plane has the coordinate pair (x, y) if the position vector of P is \( x\hat{i} + y\hat{j} \). (∵ P.V. of P = \( \overrightarrow{OP} = x\hat{i} + y\hat{j} \))
(iii) If \( A(x_1, y_1) \) and \( B(x_2, y_2) \) are the initial and final points of a vector \( \vec{v} = \overrightarrow{AB} \), then
\( \vec{v} \) = P.V. of B − P.V. of A
\( = (x_2\hat{i} + y_2\hat{j}) - (x_1\hat{i} + y_1\hat{j}) \)
\( = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} \)
⇒ Components of \( \vec{v} \) are \( x_2 - x_1 \) and \( y_2 - y_1 \).

Let M and N be the feet of the perpendiculars drawn from A and B respectively on X′OX, then
OM = abscissa of A = \( x_1 \) and ON = abscissa of B = \( x_2 \)
∴ MN = ON − OM = \( x_2 - x_1 \) = component of \( \vec{v} \) along X-axis.
⇒ Component of a vector along X-axis is equal to its projection (directed) on X-axis.
Similarly, we can show that the component of a vector along Y-axis is equal to its projection on Y-axis.
Distance between the points \( A(x_1, y_1) \) and \( B(x_2, y_2) \)
\( = AB = |\overrightarrow{AB}| = |(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Note: If a vector \( \vec{v} = x\hat{i} + y\hat{j} \), then it is conventional to write \( \vec{v} \) = (x, y).

Vector Joining Two Points

If \( P_1(x_1, y_1, z_1) \) and \( P_2(x_2, y_2, z_2) \) are any two points, then the vector joining \( P_1 \) and \( P_2 \) is the vector \( \overrightarrow{P_1P_2} \) (figure).

Joining the points \( P_1 \) and \( P_2 \) with the origin O, and applying the triangle law, from the triangle \( OP_1P_2 \),
\( \overrightarrow{OP_1} + \overrightarrow{P_1P_2} = \overrightarrow{OP_2} \)
Using the properties of vector addition, the above equation becomes
\( \overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} \)
i.e. \( \overrightarrow{P_1P_2} = (x_2\hat{i} + y_2\hat{j} + z_2\hat{k}) - (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) \)
\( = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k} \)
The magnitude of the vector \( \overrightarrow{P_1P_2} \) is given by \( |\overrightarrow{P_1P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).

Direction Cosines of a Line

If α, β, γ are the angles made by a line with the x-axis, y-axis and z-axis respectively, then cos α, cos β and cos γ are called the direction cosines of the line, denoted by l, m and n respectively.

Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), then \( \cos\alpha = \frac{x}{|\vec{r}|} \), \( \cos\beta = \frac{y}{|\vec{r}|} \) and \( \cos\gamma = \frac{z}{|\vec{r}|} \).

Direction Ratios

Any three numbers a, b, c proportional to the direction cosines l, m, n are called direction ratios of the line, i.e. \( \frac{l}{a} = \frac{m}{b} = \frac{n}{c} \).
It is easy to see that there can be infinitely many sets of direction ratios for a given line.
Note:
1. The relation between l, m, n is given by \( l^2 + m^2 + n^2 = 1 \) but \( a^2 + b^2 + c^2 \neq 1 \), in general.
2. The direction ratios a, b, c of a vector \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) are just the respective components x, y and z of the vector.

Types of Vectors

(a) Zero vector or Null vector: A vector of zero magnitude, i.e. which has the same initial and terminal point (\( \overrightarrow{AA}, \overrightarrow{BB} \)), is called a zero vector. It is denoted by \( \vec{0} \). It can have any arbitrary direction and any line as its line of support.

(b) Unit vector: A vector of unit magnitude in the direction of a non-zero vector \( \vec{a} \) is called the unit vector along \( \vec{a} \) and is denoted by \( \hat{a} \). Symbolically \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \).

(c) Co-initial vectors: Vectors having the same initial point are called co-initial vectors.

(d) Collinear vectors: Two vectors are said to be collinear if their supports are parallel, disregarding their direction and magnitude. Collinear vectors are also called parallel vectors. If they have the same direction, they are named as like vectors; otherwise unlike vectors.

(e) Equality of two vectors: Two vectors are said to be equal if they have
(a) the same length, (b) the same or parallel supports and (c) the same sense.
Note: Components of two equal vectors taken in any arbitrary direction are equal, i.e.
if \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then \( \vec{a} = \vec{b} \Leftrightarrow a_1 = b_1,\ a_2 = b_2,\ a_3 = b_3 \).

(f) Negative of a vector: A vector whose magnitude is the same as that of a given vector (say \( \overrightarrow{AB} \)) but whose direction is opposite to it, is called the negative of the given vector. For example, the vector \( \overrightarrow{BA} \) is negative to the vector \( \overrightarrow{AB} \) and is written as \( \overrightarrow{BA} = -\overrightarrow{AB} \).

(g) Free vectors: If a vector can be translated anywhere in space without changing its magnitude and direction, then such a vector is called a free vector. In other words, the initial point of a free vector can be taken anywhere in space keeping its magnitude and direction the same.

(h) Localised vectors: For a vector of given magnitude and direction, if its initial point is fixed in space, then such a vector is called a localised vector.
Examples: Force in rotational motion, moment of inertia etc.

Test of Collinearity of Three Points

(a) Three points A, B, C will be collinear if \( \overrightarrow{AB} = \lambda \overrightarrow{BC} \), where λ is a scalar.
(b) If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) are two collinear vectors, then \( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \).
(c) Three points A, B, C with position vectors \( \vec{a}, \vec{b}, \vec{c} \) respectively are collinear if and only if there exist scalars x, y, z, not all zero simultaneously, such that \( x\vec{a} + y\vec{b} + z\vec{c} = \vec{0} \), where x + y + z = 0.

Algebra of Vectors

It is possible to develop an algebra of vectors which proves useful in the study of Geometry, Mechanics and other branches of Applied Mathematics.
(a) Addition of two vectors: The vectors have magnitude as well as direction, therefore their addition is different from the addition of scalars.
Let \( \vec{a} \) and \( \vec{b} \) be two vectors in a plane, which are represented by \( \overrightarrow{AB} \) and \( \overrightarrow{CD} \). Their addition can be performed in the following two ways:

(i) Triangle law of addition of vectors: If two vectors can be represented by the two sides of a triangle, taken in order, then their sum will be represented by the third side in reverse order.
Let O be a fixed point in the plane of the vectors. Draw a line segment \( \overrightarrow{OE} \) from O, equal and parallel to \( \overrightarrow{AB} \), which represents the vector \( \vec{a} \). Now from E, draw a line segment \( \overrightarrow{EF} \), equal and parallel to \( \overrightarrow{CD} \), which represents the vector \( \vec{b} \). The line segment \( \overrightarrow{OF} \) obtained by joining O and F represents the sum of the vectors \( \vec{a} \) and \( \vec{b} \), i.e. \( \overrightarrow{OE} + \overrightarrow{EF} = \overrightarrow{OF} \) or \( \vec{a} + \vec{b} = \overrightarrow{OF} \).

This method of addition of two vectors is called the Triangle law of addition of vectors.

(ii) Parallelogram law of addition of vectors: If two vectors are represented by the two adjacent sides of a parallelogram, then their sum will be represented by the diagonal through the co-initial point.
Let \( \vec{a} \) and \( \vec{b} \) be vectors drawn from the point O, denoted by the line segments \( \overrightarrow{OP} \) and \( \overrightarrow{OQ} \). Now complete the parallelogram OPRQ. Then the vector represented by the diagonal \( \overrightarrow{OR} \) will represent the sum of the vectors \( \vec{a} \) and \( \vec{b} \), i.e. \( \overrightarrow{OP} + \overrightarrow{OQ} = \overrightarrow{OR} \) or \( \vec{a} + \vec{b} = \overrightarrow{OR} \).

This method of addition of two vectors is called the Parallelogram law of addition of vectors.

(iii) Properties of vector addition:
(1) \( \vec{a} + \vec{b} = \vec{b} + \vec{a} \) (commutative)
(2) \( (\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c}) \) (associative)
(3) \( \vec{a} + \vec{0} = \vec{a} = \vec{0} + \vec{a} \) (additive identity)
(4) \( \vec{a} + (-\vec{a}) = \vec{0} = (-\vec{a}) + \vec{a} \) (additive inverse)

Subtraction of Vectors

Vector \( -\vec{b} \) has length equal to that of vector \( \vec{b} \) but its direction is opposite.
Subtraction of vectors \( \vec{a} \) and \( \vec{b} \) is defined as the addition of \( \vec{a} \) and \( (-\vec{b}) \). It is written as follows:
\( \vec{a} - \vec{b} = \vec{a} + (-\vec{b}) \)

Geometrical representation: In the given diagram, \( \vec{a} \) and \( \vec{b} \) are represented by \( \overrightarrow{OA} \) and \( \overrightarrow{AB} \). We extend the line AB in the opposite direction up to C, where AB = AC. The line segment \( \overrightarrow{AC} \) will represent the vector \( -\vec{b} \). By joining the points O and C, the vector represented by \( \overrightarrow{OC} \) is \( \vec{a} + (-\vec{b}) \), i.e. it denotes the vector \( \vec{a} - \vec{b} \).
In the parallelogram law, the diagonal not passing through the co-initial point is represented by \( \vec{a} - \vec{b} \).

Note:
(i) \( \vec{a} - \vec{a} = \vec{a} + (-\vec{a}) = \vec{0} \)
(ii) \( \vec{a} - \vec{b} \neq \vec{b} - \vec{a} \). Hence, subtraction of vectors does not obey the commutative law.
(iii) \( \vec{a} - (\vec{b} - \vec{c}) \neq (\vec{a} - \vec{b}) - \vec{c} \), i.e. subtraction of vectors does not obey the associative law.

Multiplication of a Vector by a Scalar

Let \( \vec{a} \) be a given vector and λ a scalar. Then the product of the vector \( \vec{a} \) by the scalar λ, denoted by \( \lambda\vec{a} \), is called the multiplication of the vector \( \vec{a} \) by the scalar λ. Note that \( \lambda\vec{a} \) is also a vector, collinear to the vector \( \vec{a} \). The vector \( \lambda\vec{a} \) has the direction same as (or opposite to) that of the vector \( \vec{a} \) according as the value of λ is positive (or negative). Also, the magnitude of the vector \( \lambda\vec{a} \) is |λ| times the magnitude of the vector \( \vec{a} \), i.e. \( |\lambda\vec{a}| = |\lambda|\,|\vec{a}| \).
A geometric visualisation of multiplication of a vector by a scalar is given in the figure.

When λ = −1, then \( \lambda\vec{a} = -\vec{a} \), which is a vector having magnitude equal to the magnitude of \( \vec{a} \) and direction opposite to that of \( \vec{a} \). The vector \( -\vec{a} \) is called the negative (or additive inverse) of the vector \( \vec{a} \)
⇒ \( \vec{a} + (-\vec{a}) = (-\vec{a}) + \vec{a} = \vec{0} \)
Also, if \( \lambda = \frac{1}{|\vec{a}|} \), provided \( \vec{a} \) is not a null vector, then
\( |\lambda\vec{a}| = |\lambda|\,|\vec{a}| = \frac{1}{|\vec{a}|}|\vec{a}| = 1 \)
So, \( \lambda\vec{a} \) represents the unit vector in the direction of \( \vec{a} \).
⇒ \( \hat{a} = \frac{1}{|\vec{a}|}\vec{a} \)
Note: For any scalar k, \( k\vec{0} = \vec{0} \).

Section Formula

If \( \vec{a} \) and \( \vec{b} \) are the position vectors of two points A and B, then the p.v. of a point \( C(\vec{r}) \) which divides AB in the ratio m : n is given by:
(a) Internal division: \( \overrightarrow{OC} = \vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \)

Note: Position vector of the mid-point of AB = \( \frac{\vec{a} + \vec{b}}{2} \)
(b) External division: \( \overrightarrow{OC} = \vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n} \)

Multiplication of Two Vectors

There are two types of vector multiplication.

(i) Scalar product of two vectors (dot product)
Definition: Let \( \vec{a} \) and \( \vec{b} \) be two non-zero vectors inclined at an angle θ. Then the scalar product of \( \vec{a} \) with \( \vec{b} \) is denoted by \( \vec{a} \cdot \vec{b} \) and is defined as \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \), \( 0 \leq \theta \leq \pi \).

Geometrical interpretation of scalar product:
\( |\overrightarrow{OA}| = |\vec{a}| \), \( |\overrightarrow{OB}| = |\vec{b}| \)

Now \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \)
\( = |\vec{a}|(OB\cos\theta) \)
= (magnitude of \( \vec{a} \)) (projection of \( \vec{b} \) on \( \vec{a} \))
Again, \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \)
\( = |\vec{b}|(|\vec{a}|\cos\theta) \)
= (magnitude of \( \vec{b} \)) (projection of \( \vec{a} \) on \( \vec{b} \))

Properties:
(a) \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \) \( (0 \leq \theta \leq \pi) \)
Note that if θ is acute then \( \vec{a} \cdot \vec{b} \gt 0 \), if θ is obtuse then \( \vec{a} \cdot \vec{b} \lt 0 \) and if θ is a right angle then \( \vec{a} \cdot \vec{b} = 0 \).
(b) (i) \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = \vec{a}^2 \)    (ii) \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \) (commutative)
(c) \( \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \) (distributive)
(d) Let \( \vec{a} \) and \( \vec{b} \) be any two vectors and λ be any scalar, then \( (\lambda\vec{a}) \cdot \vec{b} = \lambda(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (\lambda\vec{b}) \)
(e) \( \vec{a} \cdot \vec{b} = 0 \Leftrightarrow \vec{a} \perp \vec{b} \); \( (\vec{a}, \vec{b} \neq \vec{0}) \)
(f) \( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \); \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0 \)

Note:
(i) The vector component of \( \vec{a} \) along \( \vec{b} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} \) and perpendicular to \( \vec{b} = \vec{a} - \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} \) [by triangle law of vector addition]

(ii) The angle φ between \( \vec{a} \) and \( \vec{b} \) is given by \( \cos\phi = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \), \( 0 \leq \phi \leq \pi \)
(iii) If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \) and \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \), \( |\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2} \)
(iv) Maximum value of \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \)
(v) Minimum value of \( \vec{a} \cdot \vec{b} = -|\vec{a}||\vec{b}| \)
(vi) Any vector \( \vec{a} \) can be written as \( \vec{a} = (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k} \)

(ii) Vector product of two vectors (cross product)
If \( \vec{a} \) and \( \vec{b} \) are two vectors and θ is the angle between them, then \( \vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n} \), where \( \hat{n} \) is the unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) such that \( \vec{a} \), \( \vec{b} \) and \( \hat{n} \) form a right handed thumb system.

Sign convention
Right handed thumb rule: In a right handed coordinate system, the thumb of the right hand points in the direction of the positive Z-axis when the fingers are curled in the direction away from the positive X-axis toward the positive Y-axis.

Observations:
(a) (i) \( \vec{a} \times \vec{b} = \vec{0} \Leftrightarrow \vec{a} \) and \( \vec{b} \) are parallel (collinear) \( (\vec{a} \neq \vec{0},\ \vec{b} \neq \vec{0}) \), i.e. \( \vec{a} = \lambda\vec{b} \), where λ is a scalar
(ii) \( \vec{a} \times \vec{b} \neq \vec{b} \times \vec{a} \) (not commutative)
(iii) \( (m\vec{a}) \times \vec{b} = \vec{a} \times (m\vec{b}) = m(\vec{a} \times \vec{b}) \), where m is a scalar
(iv) \( \vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \) (distributive over addition)
(v) \( \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0} \)
(vi) \( \hat{i} \times \hat{j} = \hat{k},\ \hat{j} \times \hat{k} = \hat{i},\ \hat{k} \times \hat{i} = \hat{j} \) and \( \hat{j} \times \hat{i} = -\hat{k},\ \hat{k} \times \hat{j} = -\hat{i},\ \hat{i} \times \hat{k} = -\hat{j} \)

(b) If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \)
(c) Geometrically \( |\vec{a} \times \vec{b}| \) = area of the parallelogram whose two adjacent sides are represented by \( \vec{a} \) and \( \vec{b} \).
\( \frac{1}{2}|\vec{a} \times \vec{b}| \) = Area of triangle ABO

(d) (i) Unit vector perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \) is \( \hat{n} = \pm\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \)
(ii) A vector of magnitude 'r' and perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \) is \( \pm\frac{r(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|} \)
(e) Vector area:
(i) If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are the p.v.'s of 3 points A, B and C, then the vector area of triangle ABC = \( \frac{1}{2}\left[\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}\right] \).

(ii) The points A, B and C are collinear if \( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0} \)
(iii) Area of any quadrilateral whose diagonal vectors are \( \vec{d}_1 \) and \( \vec{d}_2 \) is given by \( \frac{1}{2}|\vec{d}_1 \times \vec{d}_2| \).

Projection of a Vector on a Line

Suppose a vector \( \overrightarrow{AB} \) makes an angle θ with a given directed line l (say), in the anticlockwise direction (figure). Then the projection of \( \overrightarrow{AB} \) on l is a vector \( \vec{p} \) (say) with magnitude \( |\overrightarrow{AB}|\cos\theta \), and the direction of \( \vec{p} \) being the same as (or opposite to) that of the line l, depending upon whether cos θ is positive or negative. The vector \( \vec{p} \) can be:

\( \vec{p} \) is called the projection vector, and its magnitude \( |\vec{p}| \) is simply called the projection of the vector \( \overrightarrow{AB} \) on the directed line l.
For example, in each of the above figures (i) to (iv), the projection vector of \( \overrightarrow{AB} \) along the line l is the vector \( \overrightarrow{AC} \).
Observations:
1. If \( \hat{p} \) is the unit vector along a line l, then the projection of a vector \( \vec{a} \) on the line l is given by \( \vec{a} \cdot \hat{p} \).
2. Projection of a vector \( \vec{a} \) on another vector \( \vec{b} \) is given by \( \vec{a} \cdot \hat{b} \), or \( \vec{a} \cdot \left(\frac{\vec{b}}{|\vec{b}|}\right) \), or \( \frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b}) \)
3. If θ = 0, then the projection vector of \( \overrightarrow{AB} \) will be \( \overrightarrow{AB} \) itself and if θ = π, then the projection vector of \( \overrightarrow{AB} \) will be \( \overrightarrow{BA} \).
4. If \( \theta = \frac{\pi}{2} \) or \( \theta = \frac{3\pi}{2} \), then the projection vector of \( \overrightarrow{AB} \) will be the zero vector.

Solved Examples

Illustration 1. Classify the following measures as scalar and vector quantities: [1 Mark]
(i) Mass    (ii) Work    (iii) Force    (iv) Current [¼ × 4]
Answer:
(i) Scalar    (ii) Scalar    (iii) Vector    (iv) Scalar [1]

 

Illustration 2. Represent graphically a displacement of 30 km, 30° west of north after representation of vectors. [1 Mark]
Answer:

\( |\vec{d}| = 30 \) [1]

 

Illustration 3. If A is the point (1, 2) in the X-Y plane and the vector \( \overrightarrow{AB} \) has components 2 and 6, find the point B. [1 Mark]
Answer:
Given that \( \overrightarrow{AB} = 2\hat{i} + 6\hat{j} \) ...(i)
Now P.V. of A = \( \hat{i} + 2\hat{j} \) ...(ii)
⇒ \( \overrightarrow{AB} \) = P.V. of B − P.V. of A,
or P.V. of B = \( \overrightarrow{AB} \) + P.V. of A = \( (2\hat{i} + 6\hat{j}) + (\hat{i} + 2\hat{j}) \) (using (i) and (ii))
\( = 3\hat{i} + 8\hat{j} \)
Hence, B = (3, 8). [1]

 

Illustration 4. Find the direction cosines of the ray from P to Q, where P is the point (1, −2, 2) and Q is the point (3, −5, −4). [1 Mark]
Answer:
\( \overrightarrow{PQ} \) = P.V. of Q − P.V. of P = \( (3\hat{i} - 5\hat{j} - 4\hat{k}) - (\hat{i} - 2\hat{j} + 2\hat{k}) = 2\hat{i} - 3\hat{j} - 6\hat{k} \)
⇒ \( |\overrightarrow{PQ}| = \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = 7 \).
Hence, the unit vector in the direction of \( \overrightarrow{PQ} \)
\( = \frac{1}{7}(2\hat{i} - 3\hat{j} - 6\hat{k}) = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \).
⇒ direction cosines of the ray from P to Q (i.e. of \( \overrightarrow{PQ} \)) are \( \left(\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}\right) \). [1]

 

Illustration 5. Write the direction ratios of the vector \( \vec{a} = 2\hat{i} + \hat{j} - 2\hat{k} \) and hence calculate its direction cosines. [1 Mark]
Answer:
For the given vector, a = 2, b = 1 and c = −2. If l, m and n are the direction cosines of the given vector, then
as \( |\vec{r}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \)
\( l = \frac{a}{|\vec{r}|} = \frac{2}{3},\ m = \frac{b}{|\vec{r}|} = \frac{1}{3},\ n = \frac{c}{|\vec{r}|} = \frac{-2}{3} \)
Thus, the direction cosines are \( \left(\frac{2}{3}, \frac{1}{3}, \frac{-2}{3}\right) \). [1]

 

Illustration 6. Find a vector \( \vec{r} \) of magnitude \( 3\sqrt{2} \) units which makes an angle of \( \frac{\pi}{4} \) and \( \frac{\pi}{2} \) with y and z-axes, respectively. [Exemplar] [1 Mark]
Answer:
Here \( m = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( n = \cos\frac{\pi}{2} = 0 \)
Therefore, \( l^2 + m^2 + n^2 = 1 \) gives
\( l^2 + \frac{1}{2} + 0 = 1 \Rightarrow l = \pm\frac{1}{\sqrt{2}} \)
Hence, the required vector \( \vec{r} = 3\sqrt{2}(l\hat{i} + m\hat{j} + n\hat{k}) \) is given by
\( \vec{r} = 3\sqrt{2}\left(\pm\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}\right) \Rightarrow \vec{r} = \pm 3\hat{i} + 3\hat{j} \). [1]

 

Illustration 7. Find a unit vector in the direction of \( \vec{a} = 12\hat{i} + 5\hat{j} - 3\hat{k} \). [1 Mark]
Answer:
\( |\vec{a}| = |12\hat{i} + 5\hat{j} - 3\hat{k}| = \sqrt{(12)^2 + (5)^2 + (-3)^2} = \sqrt{178} \).
⇒ Unit vector in the direction of \( \vec{a} = \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{12\hat{i} + 5\hat{j} - 3\hat{k}}{\sqrt{178}} \) [1]

 

Illustration 8. If \( \vec{v}_1 \) = (2, −3), \( \vec{v}_2 \) = (0, 1) and \( \vec{v}_3 \) = (−1, 6), find a unit vector [4 Marks]
(i) in the direction of \( \vec{v}_1 + 2\vec{v}_2 - \vec{v}_3 \)    (ii) parallel to \( \vec{v}_1 + 2\vec{v}_2 - \vec{v}_3 \)
Answer:
\( \vec{v}_1 = (2, -3) = 2\hat{i} - 3\hat{j} \)
\( \vec{v}_2 = (0, 1) = \hat{j} \)
\( \vec{v}_3 = (-1, 6) = -\hat{i} + 6\hat{j} \) [1]
Let \( \vec{v} = \vec{v}_1 + 2\vec{v}_2 - \vec{v}_3 \)
\( = (2\hat{i} - 3\hat{j}) + 2(\hat{j}) - (-\hat{i} + 6\hat{j}) \)
\( = 3\hat{i} - 7\hat{j} \)
⇒ \( |\vec{v}| = \sqrt{3^2 + (-7)^2} = \sqrt{58} \) [1]
(i) A unit vector in the direction of \( \vec{v} \), i.e.
\( \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{58}}(3\hat{i} - 7\hat{j}) = \frac{3}{\sqrt{58}}\hat{i} - \frac{7}{\sqrt{58}}\hat{j} \) [1]
(ii) A unit vector parallel to \( \vec{v} \) is \( \hat{v} \) or \( -\hat{v} \)
i.e. \( \frac{3}{\sqrt{58}}\hat{i} - \frac{7}{\sqrt{58}}\hat{j} \) or \( \frac{-3}{\sqrt{58}}\hat{i} + \frac{7}{\sqrt{58}}\hat{j} \) [1]

 

Illustration 9. Show that the points A(3, −4, 5), B(2, 1, 7), C(6, −19, −1) are collinear. [1 Mark]
Answer:
Direction ratios of the line joining A and B are 2 − 3, 1 + 4, 7 − 5, i.e. −1, 5, 2.
Direction ratios of the line joining B and C are 6 − 2, −19 − 1, −1 − 7, i.e. 4, −20, −8.
It is clear that the direction ratios of AB and BC are proportional, hence AB is ∥ to BC. But point B is common to both AB and BC. Therefore A, B, C are collinear points. [1]

 

Illustration 10. Prove that the points with position vectors \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k} \) and \( \vec{c} = -7\hat{j} + 10\hat{k} \) are collinear. [1 Mark]
Answer:
\( \overrightarrow{AB} = \vec{b} - \vec{a} = (2\hat{i} + 3\hat{j} - 4\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \hat{i} + 5\hat{j} - 7\hat{k} \)
\( \overrightarrow{BC} = \vec{c} - \vec{b} = (-7\hat{j} + 10\hat{k}) - (2\hat{i} + 3\hat{j} - 4\hat{k}) = -2\hat{i} - 10\hat{j} + 14\hat{k} = -2(\hat{i} + 5\hat{j} - 7\hat{k}) \)
∴ \( \overrightarrow{BC} = -2\overrightarrow{AB} \) [1]
Hence, the points with position vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \) are collinear.

 

Illustration 11. Find the values of x and y if (x, −1, 3), (3, y, 1) and (−1, 11, 9) are collinear. [4 Marks]
Answer:
Let the given points be A, B and C such that
P.V. of A = \( x\hat{i} - \hat{j} + 3\hat{k} \), P.V. of B = \( 3\hat{i} + y\hat{j} + \hat{k} \) and P.V. of C = \( -\hat{i} + 11\hat{j} + 9\hat{k} \).
∴ A, B and C are collinear iff \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are parallel
⇒ \( \overrightarrow{AB} = \lambda\overrightarrow{AC} \) for a non-zero λ ∈ R. [1]
⇒ P.V. of B − P.V. of A = λ[P.V. of C − P.V. of A]
⇒ \( (3 - x)\hat{i} + (y + 1)\hat{j} - 2\hat{k} = \lambda((-1 - x)\hat{i} + 12\hat{j} + 6\hat{k}) \) [1]
⇒ 3 − x = λ(−1 − x), y + 1 = 12λ
and −2 = 6λ ⇒ \( \lambda = -\frac{1}{3} \) [1]
Now, 3 − x = λ(−x − 1) and y + 1 = 12λ
Hence, \( 3 - x = -\frac{1}{3}(-x - 1) \) and \( y + 1 = 12\left(-\frac{1}{3}\right) \)
⇒ 9 − 3x = x + 1 and y = −4 − 1 ⇒ x = 2 and y = −5 [1]

 

Illustration 12. Find the sum of the vectors \( \vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k} \), \( \vec{b} = -3\hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{c} = 2\hat{i} + \hat{j} + \hat{k} \). [1 Mark]
Answer:
Required sum = \( \vec{a} + \vec{b} + \vec{c} \)
\( = (4\hat{i} - 3\hat{j} + 2\hat{k}) + (-3\hat{i} + 2\hat{j} - 3\hat{k}) + (2\hat{i} + \hat{j} + \hat{k}) \)
\( = (4\hat{i} - 3\hat{i} + 2\hat{i}) + (-3\hat{j} + 2\hat{j} + \hat{j}) + (2\hat{k} - 3\hat{k} + \hat{k}) = 3\hat{i} \) [1]

 

Illustration 13. Find the vector joining the points A(4, 5, 1) and B(−3, −4, −6) directed from A to B. [1 Mark]
Answer:
Since the vector is to be directed from A to B, clearly A is the initial point and B is the terminal point.
So the required vector joining A and B is the vector \( \overrightarrow{AB} \) given by
\( \overrightarrow{AB} = (-3 - 4)\hat{i} + (-4 - 5)\hat{j} + (-6 - 1)\hat{k} \)
\( = -7\hat{i} - 9\hat{j} - 7\hat{k} \) [1]

 

Illustration 14. Find the position vectors of the points which divide the segment joining the points with P.V. \( 2\vec{a} - 3\vec{b} \) and \( 3\vec{a} - 2\vec{b} \) internally and externally in the ratio 2 : 3. [1 Mark]
Answer:
Let A and B be the given points with position vectors \( 2\vec{a} - 3\vec{b} \) and \( 3\vec{a} - 2\vec{b} \) respectively. Let P and Q be the points dividing AB internally and externally in the ratio 2 : 3.
By section formula, P.V. of P = \( \frac{2(\text{P.V. of B}) + 3(\text{P.V. of A})}{2 + 3} = \frac{12\vec{a} - 13\vec{b}}{5} \)
and P.V. of Q = \( \frac{2(\text{P.V. of B}) - 3(\text{P.V. of A})}{2 - 3} = \frac{5\vec{b}}{-1} = -5\vec{b} \) [1]
(∵ Q divides AB in the ratio 2 : −3)

 

Illustration 15. If \( \vec{a} \) and \( \vec{b} \) are the position vectors of points A and B respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA. [Exemplar] [1 Mark]
Answer:
Here the point C divides BA externally in the ratio
BC : AC = BC : (BC − BA) = 1.5 BA : 0.5 BA = 3 : 1

Now from the section formula,
\( \vec{c} \) = position vector of point C = \( \frac{3(\overrightarrow{OA}) - 1(\overrightarrow{OB})}{3 - 1} = \frac{3\vec{a} - \vec{b}}{3 - 1} = \frac{1}{2}(3\vec{a} - \vec{b}) \) [1]

 

Illustration 16. Find the angle between the vectors \( 2\hat{i} - \hat{j} + \hat{k} \) and \( 3\hat{i} + 4\hat{j} - \hat{k} \). [1 Mark]
Answer:
\( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \)
\( \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{(2\hat{i} - \hat{j} + \hat{k}) \cdot (3\hat{i} + 4\hat{j} - \hat{k})}{\sqrt{6} \cdot \sqrt{26}} \)
or \( \cos\theta = \frac{1}{\sqrt{156}} \)
\( \theta = \cos^{-1}\left(\frac{1}{\sqrt{156}}\right) \) [1]

 

Illustration 17. Show that the vectors \( \vec{a} = 3\hat{i} - 2\hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - 3\hat{j} + 5\hat{k} \), \( \vec{c} = 2\hat{i} + \hat{j} - 4\hat{k} \) form a right angled triangle. [1 Mark]
Answer:
\( \vec{b} + \vec{c} = (\hat{i} - 3\hat{j} + 5\hat{k}) + (2\hat{i} + \hat{j} - 4\hat{k}) = 3\hat{i} - 2\hat{j} + \hat{k} = \vec{a} \)

⇒ \( \vec{a}, \vec{b}, \vec{c} \) are coplanar (they form the sides of a triangle).
Further, \( \vec{a} \cdot \vec{c} = (3\hat{i} - 2\hat{j} + \hat{k}) \cdot (2\hat{i} + \hat{j} - 4\hat{k}) = (3)(2) + (-2)(1) + (1)(-4) = 0 \).
Thus, the dot product of the two non-zero vectors \( \vec{a} \) and \( \vec{c} \) is zero, therefore these vectors are perpendicular, i.e. the sides represented by the vectors \( \vec{a} \) and \( \vec{c} \) are perpendicular. [1]
Hence the given vectors form a right angled triangle.

 

Illustration 18. If \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{c} = \hat{i} + 3\hat{j} - \hat{k} \), find λ such that \( \vec{a} \) is perpendicular to \( \lambda\vec{b} + \vec{c} \). [Exemplar] [4 Marks]
Answer:
We have \( \lambda\vec{b} + \vec{c} = \lambda(\hat{i} + \hat{j} - 2\hat{k}) + (\hat{i} + 3\hat{j} - \hat{k}) \)
\( = (\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - (2\lambda + 1)\hat{k} \) [1]
Since \( \vec{a} \perp (\lambda\vec{b} + \vec{c}) \), \( \vec{a} \cdot (\lambda\vec{b} + \vec{c}) = 0 \) [1]
⇒ \( (2\hat{i} - \hat{j} + \hat{k}) \cdot [(\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - (2\lambda + 1)\hat{k}] = 0 \) [1]
⇒ 2(λ + 1) − (λ + 3) − (2λ + 1) = 0 ⇒ −λ − 2 = 0
⇒ λ = −2 [1]

 

Illustration 19. If \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) and \( |\vec{a}| = 3 \), \( |\vec{b}| = 5 \) and \( |\vec{c}| = 7 \), show that the angle between \( \vec{a} \) and \( \vec{b} \) is 60°. [Exemplar] [4 Marks]
Answer:
We have \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \)
⇒ \( |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 \)
or \( |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta = |\vec{c}|^2 \) [2]
⇒ 9 + 25 + 30 cos θ = 49
or \( \cos\theta = \frac{1}{2} \)
⇒ θ = 60° [2]

 

Illustration 20. The two adjacent sides of a parallelogram are \( 2\hat{i} - 4\hat{j} + 5\hat{k} \) and \( \hat{i} - 2\hat{j} - 3\hat{k} \). Find the unit vector parallel to its diagonal. Also, find its area. [4 Marks]
Answer:
Let ABCD be the parallelogram with
\( \overrightarrow{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k} \) and \( \overrightarrow{AD} = \hat{i} - 2\hat{j} - 3\hat{k} \)

Now, \( \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AB} + \overrightarrow{AD} \)
\( = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\hat{i} - 2\hat{j} - 3\hat{k}) = 3\hat{i} - 6\hat{j} + 2\hat{k} \)
Hence, a unit vector parallel to \( \overrightarrow{AC} = \frac{1}{|\overrightarrow{AC}|}(\overrightarrow{AC}) \) [1]
\( = \frac{1}{\sqrt{3^2 + (-6)^2 + 2^2}}(3\hat{i} - 6\hat{j} + 2\hat{k}) \)
\( = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k} \) [1]
Now, area of parallelogram ABCD = \( |\overrightarrow{AB} \times \overrightarrow{AD}| \)
= magnitude of \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} \)
\( = |\hat{i}(12 + 10) - \hat{j}(-6 - 5) + \hat{k}(-4 + 4)| \) [1]
\( = |22\hat{i} + 11\hat{j} + 0\hat{k}| \)
\( = \sqrt{(22)^2 + (11)^2 + 0^2} = \sqrt{(11)^2 + (2^2 \times 11^2)} \)
\( = 11\sqrt{5} \) square units. [1]

 

Illustration 21. Find all vectors of magnitude \( 10\sqrt{3} \) that are perpendicular to the plane of \( \hat{i} + 2\hat{j} + \hat{k} \) and \( -\hat{i} + 3\hat{j} + 4\hat{k} \). [Exemplar] [4 Marks]
Answer:
Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = -\hat{i} + 3\hat{j} + 4\hat{k} \). Then
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -1 & 3 & 4 \end{vmatrix} = \hat{i}(8 - 3) - \hat{j}(4 + 1) + \hat{k}(3 + 2) = 5\hat{i} - 5\hat{j} + 5\hat{k} \) [1]
⇒ \( |\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (-5)^2 + (5)^2} = \sqrt{3(5)^2} = 5\sqrt{3} \). [1]
Therefore, the unit vector perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \) is given by \( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{5\hat{i} - 5\hat{j} + 5\hat{k}}{5\sqrt{3}} \). [1]
Hence, the vector is \( \pm 10\sqrt{3}\left(\frac{5\hat{i} - 5\hat{j} + 5\hat{k}}{5\sqrt{3}}\right) = \pm 10(\hat{i} - \hat{j} + \hat{k}) \) [1]

 

Illustration 22. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{j} - \hat{k} \), find a vector \( \vec{c} \) such that \( \vec{a} \times \vec{c} = \vec{b} \) and \( \vec{a} \cdot \vec{c} = 3 \). [Exemplar] [4 Marks]
Answer:
Let vector \( \vec{c} = a\hat{i} + b\hat{j} + c\hat{k} \)
Given \( \vec{a} \times \vec{c} = \vec{b} \)
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ a & b & c \end{vmatrix} = 0\hat{i} + \hat{j} - \hat{k} \)
⇒ \( \hat{i}(c - b) - \hat{j}(c - a) + \hat{k}(b - a) = 0\hat{i} + \hat{j} - \hat{k} \) [1½]
Equating the coefficients of both the sides, we have
c − b = 0 ...(i),   a − c = 1 ...(ii),   b − a = −1 ...(iii) [½]
Now \( \vec{a} \cdot \vec{c} = 3 \)
⇒ \( (\hat{i} + \hat{j} + \hat{k}) \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = 3 \)
or a + b + c = 3 ...(iv) [1]
On solving equations (i), (ii), (iii) and (iv), we get
\( a = \frac{5}{3},\ b = c = \frac{2}{3} \)
So the required vector \( \vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \) [1]

 

Illustration 23. Find the projection of \( \overrightarrow{PQ} \) on \( \overrightarrow{AB} \), where \( \overrightarrow{PQ} = 5\hat{i} + \hat{j} + 2\hat{k} \) and \( \overrightarrow{AB} = 3\hat{i} - 2\hat{j} - 6\hat{k} \). [1 Mark]
Answer:
\( \frac{\overrightarrow{PQ} \cdot \overrightarrow{AB}{}|\overrightarrow{AB}|} = \frac{15 - 2 - 12}{7} \)
\( = \frac{1}{7} \) [1]

Do Yourself

Do Yourself 1
(i) Find k, if the points (k, −10, 3), (1, −1, 3) and (3, 5, 3) are collinear.
(ii) If a vector \( \vec{v} \) makes angles 60° and 45° respectively with the positive directions of x-axis and y-axis, then find the angle which the vector \( \vec{v} \) makes with the positive direction of z-axis.

Do Yourself 2
(i) Show that the points A, B and C whose position vectors are respectively \( 2\hat{i} + \hat{j} - \hat{k} \), \( 3\hat{i} - 2\hat{j} + \hat{k} \) and \( \hat{i} + 4\hat{j} - 3\hat{k} \) are collinear.
(ii) Find the unit vector in the direction of the sum of the vectors \( -2\hat{i} - 3\hat{j} + 2\hat{k} \), \( 3\hat{i} + 4\hat{j} - 2\hat{k} \) and \( \hat{i} + 2\hat{j} + 6\hat{k} \).
(iii) Prove that the points A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) are the vertices of an equilateral triangle.

Do Yourself 3
(i) Find λ ∈ R such that \( |\lambda\vec{a}| = 1 \); \( \vec{a} \) being a non-zero vector.
(ii) If \( |\vec{a}| = 2 \), find \( |4\vec{a}| \), \( |0\vec{a}| \) and \( |(-5)\vec{a}| \).
(iii) If A, B, C, D are the points with position vectors \( \vec{a} \), \( \vec{b} \), \( 3\vec{a} + 2\vec{b} \) and \( \vec{a} - 2\vec{b} \) respectively, show that \( \overrightarrow{AC} = 2\vec{a} + 2\vec{b} \) and \( \overrightarrow{DB} = 3\vec{b} - \vec{a} \).

Do Yourself 4
(i) Find the value of a for which the vectors \( 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \hat{i} + a\hat{j} + 3\hat{k} \) are (a) perpendicular (b) parallel.
(ii) Find the angles which the vector \( 3\hat{i} - 6\hat{j} + 2\hat{k} \) makes with the coordinate axes.
(iii) If \( \vec{a} \) is any vector in space, then show that \( \vec{a} = (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k} \).

Do Yourself 5
(i) If \( \vec{a} = 3\hat{i} + 4\hat{j} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \), find the value of \( |\vec{a} \times \vec{b}| \).
(ii) Find \( \vec{a} \cdot \vec{b} \) if \( |\vec{a}| = 2 \), \( |\vec{b}| = 5 \) and \( |\vec{a} \times \vec{b}| = 8 \).
(iii) Find the area of the triangle ABC where A, B, C are the points (a, 0, 0), (0, b, 0) and (0, 0, c) respectively, where abc ≠ 0.

Answers for Do Yourself

Do Yourself 1
(i) k = −2    (ii) 60° or 120°

Do Yourself 2
(ii) \( \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k} \)

Do Yourself 3
(i) \( \pm\frac{1}{|\vec{a}|} \)    (ii) 8, 0, 10

Do Yourself 4
(i) (a) −15   (b) \( \frac{2}{3} \)
(ii) \( \cos^{-1}\left(\frac{3}{7}\right), \cos^{-1}\left(-\frac{6}{7}\right), \cos^{-1}\left(\frac{2}{7}\right) \)

Do Yourself 5
(i) \( \sqrt{74} \)    (ii) 6 or −6    (iii) \( \frac{1}{2}\sqrt{a^2b^2 + b^2c^2 + c^2a^2} \)