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Vector Algebra - Types of vectors, position vector of a point, components of a vector

Grade 12CBSE

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

πŸ”‘Concepts

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Vector Definition: A quantity that has both magnitude and direction is called a vector. Visually, it is represented as a directed line segment with an initial point AA and a terminal point BB, denoted as ABβƒ—\vec{AB}. The length of the segment represents the magnitude ∣ABβƒ—βˆ£|\vec{AB}|.

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Types of Vectors: Vectors are classified based on their properties. A 'Zero Vector' has zero magnitude and no specific direction. 'Unit Vectors' have a magnitude of exactly 1 unit. 'Co-initial Vectors' share the same starting point. 'Collinear or Parallel Vectors' lie on the same or parallel lines, regardless of their magnitude or direction. Visually, parallel vectors look like arrows pointing in the same or opposite directions that never intersect.

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Equal and Negative Vectors: Two vectors are equal if they have the same magnitude and direction, even if their starting points differ. A 'Negative Vector' has the same magnitude but the exact opposite direction. Visually, a negative vector βˆ’aβƒ—-\vec{a} is an arrow of the same length as aβƒ—\vec{a} but flipped 180∘180^{\circ}.

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Position Vector: For any point P(x,y,z)P(x, y, z) in space, the vector OP⃗\vec{OP} originating from the origin O(0,0,0)O(0, 0, 0) is called the position vector of PP. Visually, this is an arrow starting at the intersection of the X,Y,ZX, Y, Z axes and ending at point PP. Its magnitude is given by the distance formula from the origin.

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Components of a Vector: A vector r⃗\vec{r} can be decomposed into scalar components x,y,zx, y, z and vector components xi^,yj^,zk^x\hat{i}, y\hat{j}, z\hat{k} along the X,Y,and ZX, Y, \text{and } Z axes respectively. Visually, these components represent the 'shadow' or projection of the vector onto each coordinate axis, forming a rectangular box where the vector is the space diagonal.

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Direction Cosines and Ratios: If a vector makes angles α,β,γ\alpha, \beta, \gamma with the positive X,Y,and ZX, Y, \text{and } Z axes, then cos⁑α,cos⁑β,cos⁑γ\cos \alpha, \cos \beta, \cos \gamma are its direction cosines (often denoted as l,m,nl, m, n). These values define the specific orientation of the vector in 3D space.

πŸ“Formulae

Magnitude of a vector rβƒ—=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}: ∣rβƒ—βˆ£=x2+y2+z2|\vec{r}| = \sqrt{x^2 + y^2 + z^2}

Unit vector in the direction of aβƒ—\vec{a}: a^=aβƒ—βˆ£aβƒ—βˆ£\hat{a} = \frac{\vec{a}}{|\vec{a}|}

Vector joining 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): P1P2βƒ—=(x2βˆ’x1)i^+(y2βˆ’y1)j^+(z2βˆ’z1)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 P1P2βƒ—\vec{P_1P_2}: ∣P1P2βƒ—βˆ£=(x2βˆ’x1)2+(y2βˆ’y1)2+(z2βˆ’z1)2|\vec{P_1P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Direction Cosines: l=x∣rβƒ—βˆ£,m=y∣rβƒ—βˆ£,n=z∣rβƒ—βˆ£l = \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 (or cos⁑2α+cos⁑2β+cos⁑2γ=1\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1)

πŸ’‘Examples

Problem 1:

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:

  1. First, calculate the magnitude of aβƒ—\vec{a}: ∣aβƒ—βˆ£=22+32+12=4+9+1=14|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}
  2. Use the formula for the unit vector: a^=aβƒ—βˆ£aβƒ—βˆ£=2i^+3j^+k^14\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}}
  3. Separate the components: 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 preserves the direction but scales the length to 1.

Problem 2:

Find the vector and its magnitude joining the points A(1,2,βˆ’3)A(1, 2, -3) and B(βˆ’1,βˆ’2,1)B(-1, -2, 1), directed from AA to BB.

Solution:

  1. Identify the coordinates: (x1,y1,z1)=(1,2,βˆ’3)(x_1, y_1, z_1) = (1, 2, -3) and (x2,y2,z2)=(βˆ’1,βˆ’2,1)(x_2, y_2, z_2) = (-1, -2, 1).
  2. Calculate the vector ABβƒ—\vec{AB}: ABβƒ—=(βˆ’1βˆ’1)i^+(βˆ’2βˆ’2)j^+(1βˆ’(βˆ’3))k^\vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} ABβƒ—=βˆ’2i^βˆ’4j^+4k^\vec{AB} = -2\hat{i} - 4\hat{j} + 4\hat{k}
  3. Calculate the magnitude ∣ABβƒ—βˆ£|\vec{AB}|: ∣ABβƒ—βˆ£=(βˆ’2)2+(βˆ’4)2+42=4+16+16=36=6|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6

Explanation:

The vector joining two points is found by subtracting the coordinates of the initial point from the terminal point. The magnitude is the distance between these two points.