Scalars and Vectors

In physics and mathematics, physical quantities are broadly classified into two categories:

  • Scalars: Quantities that have only magnitude and no direction. Examples include mass, volume, temperature, distance, and speed. They are completely described by a numerical value along with a unit.
  • Vectors: Quantities that have both magnitude and direction, and obey the triangle law of addition. Examples include displacement, velocity, acceleration, and force.

Geometrically, a vector is represented by a directed line segment. The vector AB⃗\vec{AB} starts at an initial point AA and ends at a terminal point BB. The length of the segment gives the magnitude of the vector, denoted by ∣AB⃗∣|\vec{AB}| or simply ABAB, while the arrow indicates the direction.

A very important distinction is that not every quantity with direction is automatically treated as a vector in mathematics; it must also follow the laws of vector addition. This is why the triangle law is included in the definition.

Position Vector

Consider a point P(x,y,z)P(x, y, z) in a 3D Cartesian coordinate system with origin O(0,0,0)O(0, 0, 0). The vector OP⃗\vec{OP} having OO as its initial point and PP as its terminal point is called the position vector of point PP.

Using unit vectors i^\hat{i}, j^\hat{j}, and k^\hat{k} along the positive x, y, and z axes respectively, the position vector r⃗\vec{r} of point PP is written as: r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

The magnitude of this position vector is the distance of point PP from the origin and is given by: ∣r⃗∣=x2+y2+z2|\vec{r}| = \sqrt{x^2 + y^2 + z^2}

Thus, the position vector not only tells us where the point is located, but also gives a compact algebraic way to describe points in space.

Direction Cosines and Direction Ratios

Let a position vector r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} make angles α\alpha, β\beta, and γ\gamma with the positive x, y, and z-axes respectively. These are called the direction angles of the vector.

The cosines of these angles, namely cos⁡α\cos\alpha, cos⁡β\cos\beta, and cos⁡γ\cos\gamma, are called the direction cosines of the vector and are usually denoted by ll, mm, and nn respectively. l=cos⁡α=x∣r⃗∣,m=cos⁡β=y∣r⃗∣,n=cos⁡γ=z∣r⃗∣l = \cos\alpha = \frac{x}{|\vec{r}|}, \quad m = \cos\beta = \frac{y}{|\vec{r}|}, \quad n = \cos\gamma = \frac{z}{|\vec{r}|}

A fundamental identity satisfied by direction cosines is: l2+m2+n2=1l^2 + m^2 + n^2 = 1 This follows directly from: (x∣r⃗∣)2+(y∣r⃗∣)2+(z∣r⃗∣)2=x2+y2+z2∣r⃗∣2=1\left(\frac{x}{|\vec{r}|}\right)^2 + \left(\frac{y}{|\vec{r}|}\right)^2 + \left(\frac{z}{|\vec{r}|}\right)^2 = \frac{x^2+y^2+z^2}{|\vec{r}|^2} = 1

Any three numbers aa, bb, and cc proportional to ll, mm, and nn are called the direction ratios of the vector. In particular, the scalar components (x,y,z)(x, y, z) of the vector are direction ratios of the vector.

Types of Vectors

  • Zero Vector (Null Vector): A vector whose initial and terminal points coincide. Its magnitude is zero. It is denoted by 0⃗\vec{0}.
  • Unit Vector: A vector whose magnitude is exactly 1. If a⃗\vec{a} is a non-zero vector, then the unit vector in its direction is: a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}
  • Coinitial Vectors: Two or more vectors having the same initial point.
  • Collinear (or Parallel) Vectors: Two or more vectors are collinear if they are parallel to the same line. If a⃗\vec{a} and b⃗\vec{b} are collinear, then one is a scalar multiple of the other: a⃗=λb⃗\vec{a} = \lambda \vec{b} for some scalar λ\lambda.
  • Equal Vectors: Two vectors are equal if they have the same magnitude and the same direction, regardless of where they are placed in space.
  • Negative of a Vector: If a⃗\vec{a} is a vector, then −a⃗-\vec{a} is the vector with the same magnitude but exactly opposite direction.

These basic vector types are used constantly in later topics such as vector addition, dot product, cross product, lines, and planes.

Example 1: Finding Magnitude

Compute the magnitude of the vector a⃗=3i^−4j^+12k^\vec{a} = 3\hat{i} - 4\hat{j} + 12\hat{k}.

Solution: Step 1: Identify the scalar components of the vector. Here, x=3x = 3, y=−4y = -4, and z=12z = 12.

Step 2: Use the magnitude formula: ∣a⃗∣=x2+y2+z2|\vec{a}| = \sqrt{x^2 + y^2 + z^2}

Step 3: Substitute the values: ∣a⃗∣=32+(−4)2+122|\vec{a}| = \sqrt{3^2 + (-4)^2 + 12^2}

Step 4: Compute the squares: ∣a⃗∣=9+16+144=169|\vec{a}| = \sqrt{9 + 16 + 144} = \sqrt{169}

Step 5: Take the square root: ∣a⃗∣=13|\vec{a}| = 13

Answer: 13 units.

Example 2: Finding a Unit Vector

Find the unit vector in the direction of the vector a⃗=2i^+3j^+k^\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}.

Solution: Step 1: Find the magnitude of the vector: ∣a⃗∣=22+32+12=4+9+1=14|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}

Step 2: Use the formula for the unit vector: a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}

Step 3: Substitute the vector and its magnitude: a^=2i^+3j^+k^14\hat{a} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}}

Step 4: Write in component form: a^=214i^+314j^+114k^\hat{a} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}

Answer: 214i^+314j^+114k^\frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}

Example 3: Direction Cosines

Find the direction cosines of the vector r⃗=i^+2j^+3k^\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}.

Solution: Step 1: Identify the scalar components: x=1,y=2,z=3x = 1, \quad y = 2, \quad z = 3

Step 2: Find the magnitude: ∣r⃗∣=12+22+32=1+4+9=14|\vec{r}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

Step 3: Use the formulas for direction cosines: l=x∣r⃗∣,m=y∣r⃗∣,n=z∣r⃗∣l = \frac{x}{|\vec{r}|}, \quad m = \frac{y}{|\vec{r}|}, \quad n = \frac{z}{|\vec{r}|}

Step 4: Substitute the values: l=114,m=214,n=314l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}

Answer: The direction cosines are 114\frac{1}{\sqrt{14}}, 214\frac{2}{\sqrt{14}}, and 314\frac{3}{\sqrt{14}}.

Example 4: Equality of Vectors

Find the values of xx and yy so that the vectors 2i^+3j^2\hat{i} + 3\hat{j} and xi^+yj^x\hat{i} + y\hat{j} are equal.

Solution: Step 1: Two vectors are equal if and only if their corresponding components are equal.

Step 2: Compare the i^\hat{i} components: 2=x  ⟹  x=22 = x \implies x = 2

Step 3: Compare the j^\hat{j} components: 3=y  ⟹  y=33 = y \implies y = 3

Answer: x=2,y=3x = 2, y = 3.

Example 5: Collinear Vectors

Show that the vectors a⃗=2i^−3j^+4k^\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} and b⃗=−4i^+6j^−8k^\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k} are collinear.

Solution: Step 1: Two vectors are collinear if one is a scalar multiple of the other.

Step 2: Factor the vector b⃗\vec{b}: b⃗=−4i^+6j^−8k^=−2(2i^−3j^+4k^)\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k} = -2(2\hat{i} - 3\hat{j} + 4\hat{k})

Step 3: Recognize that the bracketed expression is exactly a⃗\vec{a}: b⃗=−2a⃗\vec{b} = -2\vec{a}

Since one vector is a scalar multiple of the other, the vectors are collinear.

Answer: Proved.

Example 6: Vector with Given Magnitude

Find a vector in the direction of vector a⃗=i^−2j^\vec{a} = \hat{i} - 2\hat{j} that has a magnitude of 7 units.

Solution: Step 1: Find the magnitude of a⃗\vec{a}: ∣a⃗∣=12+(−2)2=1+4=5|\vec{a}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}

Step 2: Find the unit vector in the direction of a⃗\vec{a}: a^=a⃗∣a⃗∣=15i^−25j^\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}

Step 3: Multiply this unit vector by the required magnitude 7: Required vector=7a^=7(15i^−25j^)\text{Required vector} = 7\hat{a} = 7\left(\frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}\right)

Step 4: Simplify: Required vector=75i^−145j^\text{Required vector} = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}

Answer: 75i^−145j^\frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}

Example 7: Direction Cosines Property

If a vector makes angles α\alpha, β\beta, and γ\gamma with the x, y, and z axes respectively, prove that sin⁡2α+sin⁡2β+sin⁡2γ=2\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2.

Solution: Step 1: Use the standard identity for direction cosines: cos⁡2α+cos⁡2β+cos⁡2γ=1\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

Step 2: Use the identity cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta for each angle.

Step 3: Substitute into the direction cosine identity: (1−sin⁡2α)+(1−sin⁡2β)+(1−sin⁡2γ)=1(1 - \sin^2\alpha) + (1 - \sin^2\beta) + (1 - \sin^2\gamma) = 1

Step 4: Simplify: 3−(sin⁡2α+sin⁡2β+sin⁡2γ)=13 - (\sin^2\alpha + \sin^2\beta + \sin^2\gamma) = 1

Step 5: Rearranging, sin⁡2α+sin⁡2β+sin⁡2γ=2\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2

Answer: Proved.

Example 8: Initial and Terminal Points

Find the scalar and vector components of the vector with initial point A(2,1)A(2, 1) and terminal point B(−5,7)B(-5, 7).

Solution: Step 1: The vector from AA to BB is obtained by subtracting coordinates of AA from coordinates of BB: AB⃗=(x2−x1)i^+(y2−y1)j^\vec{AB} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j}

Step 2: Substitute the coordinates: AB⃗=(−5−2)i^+(7−1)j^\vec{AB} = (-5-2)\hat{i} + (7-1)\hat{j}

Step 3: Simplify: AB⃗=−7i^+6j^\vec{AB} = -7\hat{i} + 6\hat{j}

So the scalar components are −7-7 and 66, and the vector components are −7i^-7\hat{i} and 6j^6\hat{j}.

Answer: Scalar components: −7,6-7, 6. Vector components: −7i^,6j^-7\hat{i}, 6\hat{j}.

Example 9: Angles with the Axes

Find the angles at which the vector a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k} is inclined to the coordinate axes.

Solution: Step 1: Find the magnitude of the vector: ∣a⃗∣=12+12+12=3|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}

Step 2: Find the direction cosines: l=13,m=13,n=13l = \frac{1}{\sqrt{3}}, \quad m = \frac{1}{\sqrt{3}}, \quad n = \frac{1}{\sqrt{3}}

Step 3: Relate direction cosines to direction angles: cos⁡α=13,cos⁡β=13,cos⁡γ=13\cos\alpha = \frac{1}{\sqrt{3}}, \quad \cos\beta = \frac{1}{\sqrt{3}}, \quad \cos\gamma = \frac{1}{\sqrt{3}}

Step 4: Hence, α=cos⁡−1(13),β=cos⁡−1(13),γ=cos⁡−1(13)\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right), \quad \beta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right), \quad \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)

Since all three direction cosines are equal, the vector is equally inclined to all three axes.

Answer: cos⁡−1(1/3)\cos^{-1}(1/\sqrt{3}) with all three axes.

Example 10: Checking for a Unit Vector

Determine whether the vector v⃗=12i^+32j^\vec{v} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} is a unit vector.

Solution: Step 1: A vector is a unit vector if its magnitude is 1.

Step 2: Compute the magnitude: ∣v⃗∣=(12)2+(32)2|\vec{v}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}

Step 3: Simplify each square: ∣v⃗∣=14+34|\vec{v}| = \sqrt{\frac{1}{4} + \frac{3}{4}}

Step 4: Add the fractions: ∣v⃗∣=44=1=1|\vec{v}| = \sqrt{\frac{4}{4}} = \sqrt{1} = 1

Since the magnitude is exactly 1, the vector is a unit vector.

Answer: Yes, it is a unit vector.