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Vector Algebra - Vectors and scalars, magnitude and direction of a vector

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Scalars and Vectors: A scalar quantity has only magnitude (e.g., distance, mass, time). A vector quantity has both magnitude and direction (e.g., displacement, velocity, force). Visually, a vector is represented by a directed line segment where the length of the segment denotes the magnitude and an arrowhead indicates the direction.

Position Vector: For a point P(x,y,z)P(x, y, z) in a 3D coordinate system, the vector OP\vec{OP} with the origin O(0,0,0)O(0, 0, 0) as the initial point and PP as the terminal point is called the position vector of PP. It is visually represented as an arrow pointing from the center of the axes to the specific point in space.

Magnitude of a Vector: The magnitude (or length) of a vector a=xi^+yj^+zk^\vec{a} = x\hat{i} + y\hat{j} + z\hat{k} is denoted by a|\vec{a}|. It represents the distance between the initial and terminal points. In a geometric sense, it is the actual length of the arrow representing the vector, calculated using the Pythagorean theorem in three dimensions.

Direction Cosines and Ratios: The angles α,β,\alpha, \beta, and γ\gamma made by a vector with the positive directions of the x,y,x, y, and zz axes respectively are called direction angles. Their cosines, l=cosα,m=cosβ,n=cosγl = \cos\alpha, m = \cos\beta, n = \cos\gamma, are the direction cosines. Direction ratios are any real numbers a,b,ca, b, c such that l/a=m/b=n/cl/a = m/b = n/c. Visually, these define the orientation of the vector in 3D space relative to the coordinate axes.

Unit Vector: A vector whose magnitude is exactly 11 unit is called a unit vector. It is denoted by adding a 'hat' symbol (e.g., a^\hat{a}). A unit vector represents pure direction without scaling the magnitude. Geometrically, it is a vector that lies on the unit sphere centered at the origin.

Types of Vectors: (i) Zero Vector: Initial and terminal points coincide (magnitude 0). (ii) Co-initial Vectors: Vectors having the same initial point. (iii) Collinear/Parallel Vectors: Vectors parallel to the same line, regardless of their magnitudes or directions. (iv) Equal Vectors: Vectors that have the same magnitude and direction, even if they have different initial points (they are translations of each other).

Vector Joining Two Points: If we have two points P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2), the vector P1P2\vec{P_1P_2} is found by subtracting the coordinates of the initial point from the terminal point. Visually, this creates a displacement vector representing the path from P1P_1 to P2P_2.

📐Formulae

Magnitude of a=xi^+yj^+zk^\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}: a=x2+y2+z2|\vec{a}| = \sqrt{x^2 + y^2 + z^2}

Unit vector in the direction of a\vec{a}: a^=aa=xi^+yj^+zk^x2+y2+z2\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{\sqrt{x^2 + y^2 + z^2}}

Direction Cosines (l,m,nl, m, n): l=xr,m=yr,n=zrl = \frac{x}{|\vec{r}|}, m = \frac{y}{|\vec{r}|}, n = \frac{z}{|\vec{r}|}

Identity for Direction Cosines: l2+m2+n2=1l^2 + m^2 + n^2 = 1

Vector joining points P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2): P1P2=(x2x1)i^+(y2y1)j^+(z2z1)k^\vec{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

Magnitude of vector joining two points: P1P2=(x2x1)2+(y2y1)2+(z2z1)2|\vec{P_1P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

💡Examples

Problem 1:

Find the magnitude and direction cosines of the vector a=2i^+3j^6k^\vec{a} = 2\hat{i} + 3\hat{j} - 6\hat{k}.

Solution:

Step 1: Calculate the magnitude a|\vec{a}|. a=22+32+(6)2=4+9+36=49=7|\vec{a}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 Step 2: Calculate direction cosines l,m,nl, m, n. l=27l = \frac{2}{7}, m=37m = \frac{3}{7}, n=67n = \frac{-6}{7}. Therefore, the magnitude is 77 and direction cosines are (27,37,67)(\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}).

Explanation:

To find the magnitude, we use the 3D distance formula components. Direction cosines are then obtained by dividing each component of the vector by its total magnitude.

Problem 2:

Find the unit vector in the direction of the vector v=PQ\vec{v} = \vec{PQ}, where PP and QQ are the points (1,2,3)(1, 2, 3) and (4,5,6)(4, 5, 6) respectively.

Solution:

Step 1: Find the vector PQ\vec{PQ}. PQ=(41)i^+(52)j^+(63)k^=3i^+3j^+3k^\vec{PQ} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} Step 2: Calculate the magnitude of PQ\vec{PQ}. PQ=32+32+32=9+9+9=27=33|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3} Step 3: Find the unit vector v^\hat{v}. v^=PQPQ=3i^+3j^+3k^33=13i^+13j^+13k^\hat{v} = \frac{\vec{PQ}}{|\vec{PQ}|} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}

Explanation:

First, we determine the vector by subtracting coordinates of the initial point from the terminal point. Then, we normalize the vector by dividing it by its magnitude to ensure the resulting vector has a length of 1.