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Vectors - Types of Vectors and Vector Operations

Grade 12ICSE

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

🔑Concepts

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Definition of a Vector: A vector is a quantity that possesses both magnitude and direction. Geometrically, it is represented by a directed line segment AB⃗\vec{AB}, where AA is the initial point (tail) and BB is the terminal point (head). The length of the segment represents the magnitude ∣AB⃗∣|\vec{AB}|.

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Position Vector: For any point P(x,y,z)P(x, y, z) in three-dimensional space, the vector OP⃗\vec{OP} originating from the origin O(0,0,0)O(0, 0, 0) to PP is called the position vector. It is expressed as r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, where i^,j^,k^\hat{i}, \hat{j}, \hat{k} are unit vectors along the x,y,zx, y, z axes respectively.

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Types of Vectors: Key types include the Zero Vector (magnitude is 0, direction indeterminate), Unit Vector (magnitude is 1 unit), Co-initial Vectors (vectors having the same starting point), and Collinear Vectors (vectors parallel to the same line, regardless of their magnitudes or directions).

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Vector Addition (Triangle Law): If two vectors a⃗\vec{a} and b⃗\vec{b} are represented by two sides of a triangle taken in order, then their sum a⃗+b⃗\vec{a} + \vec{b} is represented by the third side of the triangle taken in the opposite order. Visually, this follows the 'head-to-tail' rule where the resultant connects the start of the first vector to the end of the second.

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Parallelogram Law of Addition: If two vectors a⃗\vec{a} and b⃗\vec{b} are represented by two adjacent sides of a parallelogram directed away from a common vertex, then their sum a⃗+b⃗\vec{a} + \vec{b} is represented by the diagonal of the parallelogram passing through that same vertex.

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Direction Cosines and Ratios: If a vector makes angles α,β,γ\alpha, \beta, \gamma with the positive directions of the x,y,zx, y, z axes, then cos⁡α,cos⁡β,cos⁡γ\cos \alpha, \cos \beta, \cos \gamma are called direction cosines (denoted as l,m,nl, m, n). The coordinates of a unit vector in the direction of r⃗\vec{r} are exactly its direction cosines.

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Section Formula: If a point RR divides the line segment joining two points PP (with position vector a⃗\vec{a}) and QQ (with position vector b⃗\vec{b}) in the ratio m:nm:n, its position vector is calculated based on whether the division is internal or external. Visually, RR lies on the segment PQPQ for internal division and on the extension of the segment for external division.

📐Formulae

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

Unit vector in direction of a⃗\vec{a}: a^=a⃗∣a⃗∣\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): P1P2⃗=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^\vec{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

Direction Cosines: l=x∣r⃗∣,m=y∣r⃗∣,n=z∣r⃗∣l = \frac{x}{|\vec{r}|}, m = \frac{y}{|\vec{r}|}, n = \frac{z}{|\vec{r}|}

Relation between 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

Internal Section Formula: r⃗=mb⃗+na⃗m+n\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}

External Section Formula: r⃗=mb⃗−na⃗m−n\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}

Midpoint Formula: r⃗=a⃗+b⃗2\vec{r} = \frac{\vec{a} + \vec{b}}{2}

💡Examples

Problem 1:

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

Solution:

  1. Find the magnitude of a⃗\vec{a}: ∣a⃗∣=12+12+22=1+1+4=6|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}. \ 2. Apply the unit vector formula: a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}. \ 3. Substitute values: a^=16(i^+j^+2k^)=16i^+16j^+26k^\hat{a} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k}) = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}.

Explanation:

A unit vector is obtained by dividing a vector by its own magnitude, resulting in a vector of length 1 pointing in the same direction.

Problem 2:

Find the position vector of a point RR which divides the line joining two points PP and QQ whose position vectors are OP⃗=i^+2j^−k^\vec{OP} = \hat{i} + 2\hat{j} - \hat{k} and OQ⃗=−i^+j^+k^\vec{OQ} = -\hat{i} + \hat{j} + \hat{k} respectively, in the ratio 2:12:1 internally.

Solution:

  1. Identify values: a⃗=i^+2j^−k^\vec{a} = \hat{i} + 2\hat{j} - \hat{k}, b⃗=−i^+j^+k^\vec{b} = -\hat{i} + \hat{j} + \hat{k}, m=2m = 2, n=1n = 1. \ 2. Use the internal section formula: OR⃗=mb⃗+na⃗m+n\vec{OR} = \frac{m\vec{b} + n\vec{a}}{m + n}. \ 3. Substitute: OR⃗=2(−i^+j^+k^)+1(i^+2j^−k^)2+1\vec{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1}. \ 4. Simplify numerator: OR⃗=−2i^+2j^+2k^+i^+2j^−k^3\vec{OR} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} - \hat{k}}{3}. \ 5. Result: OR⃗=−i^+4j^+k^3=−13i^+43j^+13k^\vec{OR} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}.

Explanation:

The section formula allows us to find the coordinates of a point on a line segment based on the relative distances from the endpoints.