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Vectors - Concept of Vectors and Scalars

Grade 12ICSE

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

🔑Concepts

Scalars and Vectors: A scalar quantity possesses only magnitude (e.g., mass, temperature), whereas a vector quantity possesses both magnitude and direction (e.g., displacement, velocity). Visually, a vector is represented as a directed line segment where the length of the line denotes the magnitude and the arrowhead indicates the direction.

Position Vector: The position vector of a point P(x,y,z)P(x, y, z) with respect to the origin O(0,0,0)O(0, 0, 0) is given by OP=xi^+yj^+zk^\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}. Geometrically, this is an arrow originating at (0,0,0)(0,0,0) and terminating at point PP, showing its spatial location relative to the center.

Direction Cosines and Direction Ratios: If a vector r\vec{r} makes angles α,β,γ\alpha, \beta, \gamma with the positive directions of the x,y,zx, y, z axes respectively, then cosα,cosβ,cosγ\cos \alpha, \cos \beta, \cos \gamma are called its direction cosines (denoted by l,m,nl, m, n). Direction ratios are any numbers proportional to these cosines. Visually, these represent the orientation of the vector relative to the 3D coordinate frame.

Types of Vectors: Key types include the Zero Vector (magnitude zero, arbitrary direction), Unit Vector (magnitude of 1), and Co-initial Vectors (vectors having the same initial point). Equal vectors must have the same magnitude and direction regardless of their starting positions, appearing as parallel arrows of identical length pointing the same way.

Triangle Law of Vector Addition: If two vectors are represented as two sides of a triangle taken in order, their sum is represented by the third side taken in the opposite order. Visually, if you place the tail of vector b\vec{b} at the head of vector a\vec{a}, the vector from the tail of a\vec{a} to the head of b\vec{b} is the resultant a+b\vec{a} + \vec{b}.

Parallelogram Law: If two vectors are represented by the adjacent sides of a parallelogram, their sum is the diagonal passing through their common initial point. This visualizes addition when both vectors act from the same point, with the diagonal representing the combined effect.

Section Formula: This determines the position vector of a point RR which divides a line segment joining two points PP and QQ (with position vectors a\vec{a} and b\vec{b}) in the ratio m:nm:n. If RR divides PQPQ internally, its position vector is mb+nam+n\frac{m\vec{b} + n\vec{a}}{m + n}. Visually, RR is a point located on the line segment between PP and QQ.

📐Formulae

Magnitude of vector 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\hat{a} = \frac{\vec{a}}{|\vec{a}|}

Relationship between direction cosines: l2+m2+n2=1l^2 + m^2 + n^2 = 1

Direction cosines from coordinates: l=xx2+y2+z2,m=yx2+y2+z2,n=zx2+y2+z2l = \frac{x}{\sqrt{x^2+y^2+z^2}}, m = \frac{y}{\sqrt{x^2+y^2+z^2}}, n = \frac{z}{\sqrt{x^2+y^2+z^2}}

Vector joining two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2): PQ=(x2x1)i^+(y2y1)j^+(z2z1)k^\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

Internal Section Formula: r=mb+nam+n\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}

External Section Formula: r=mbnamn\vec{r} = \frac{m\vec{b} - n\vec{a}}{m-n}

💡Examples

Problem 1:

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

Solution:

Step 1: Calculate the magnitude of a\vec{a}. a=22+32+12=4+9+1=14|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} Step 2: Apply the unit vector formula a^=aa\hat{a} = \frac{\vec{a}}{|\vec{a}|}. a^=2i^+3j^+k^14\hat{a} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}} Step 3: Simplify. a^=214i^+314j^+114k^\hat{a} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}

Explanation:

To find a unit vector, we divide the original vector by its magnitude. This scales the vector down so that its length becomes exactly 1 while maintaining the original direction.

Problem 2:

Find the direction cosines of the vector joining the points A(1,2,3)A(1, 2, -3) and B(1,2,1)B(-1, -2, 1) directed from AA to BB.

Solution:

Step 1: Find the vector AB\vec{AB}. AB=(11)i^+(22)j^+(1(3))k^=2i^4j^+4k^\vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k} Step 2: Find the magnitude AB|\vec{AB}|. AB=(2)2+(4)2+42=4+16+16=36=6|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 Step 3: Calculate direction cosines l,m,nl, m, n. l=26=13,m=46=23,n=46=23l = \frac{-2}{6} = -\frac{1}{3}, \quad m = \frac{-4}{6} = -\frac{2}{3}, \quad n = \frac{4}{6} = \frac{2}{3} Direction cosines are (13,23,23)(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}).

Explanation:

The direction cosines are obtained by taking the components of the vector (direction ratios) and dividing them by the total magnitude of the vector.