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Vector Algebra - Some Basic Concepts

Grade 12CBSE

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

🔑Concepts

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A Vector is a quantity that has both magnitude and direction. It is represented geometrically by a directed line segment. The point AA where the vector starts is the initial point, and the point BB where it ends is the terminal point, denoted as AB⃗\vec{AB}.

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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 r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}. The magnitude of this vector is r=∣r⃗∣=x2+y2+z2r = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}.

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Direction Cosines: If a vector r⃗\vec{r} makes angles α,β,γ\alpha, \beta, \gamma with the positive directions of x,y,zx, y, z axes respectively, then cos⁡α,cos⁡β,cos⁡γ\cos \alpha, \cos \beta, \cos \gamma are called direction cosines, usually denoted by l,m,nl, m, n.

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Direction Ratios: Any three numbers a,b,ca, b, c proportional to the direction cosines l,m,nl, m, n are called direction ratios. The relation is given by l=aa2+b2+c2l = \frac{a}{\sqrt{a^2+b^2+c^2}}, m=ba2+b2+c2m = \frac{b}{\sqrt{a^2+b^2+c^2}}, and n=ca2+b2+c2n = \frac{c}{\sqrt{a^2+b^2+c^2}}.

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Types of Vectors:

  1. Zero Vector: Magnitude is zero, direction is indeterminate (0⃗\vec{0}).
  2. Unit Vector: Magnitude is 1 unit, denoted by a^\hat{a}.
  3. Co-initial Vectors: Vectors having the same initial point.
  4. Collinear Vectors: Two or more vectors parallel to the same line, irrespective of their magnitudes and directions.
  5. Equal Vectors: Vectors having the same magnitude and direction.
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Vector Addition: Follows the Triangle Law of vector addition: If two vectors are represented by two sides of a triangle in order, then their sum is represented by the third side taken in the opposite order: AB⃗+BC⃗=AC⃗\vec{AB} + \vec{BC} = \vec{AC}.

📐Formulae

∣a⃗∣=a12+a22+a32\lvert \vec{a} \rvert = \sqrt{a_1^2 + a_2^2 + a_3^2}

a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{\lvert \vec{a} \rvert}

l2+m2+n2=1l^2 + m^2 + n^2 = 1

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}

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

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

💡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:

Given a⃗=i^+j^+2k^\vec{a} = \hat{i} + \hat{j} + 2\hat{k}. Magnitude ∣a⃗∣=12+12+22=1+1+4=6|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}. The unit vector is a^=a⃗∣a⃗∣=i^+j^+2k^6=16i^+16j^+26k^\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}.

Explanation:

To find a unit vector, divide the original vector by its magnitude.

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:

The vector AB⃗\vec{AB} is given by: AB⃗=(−1−1)i^+(−2−2)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} Magnitude ∣AB⃗∣=(−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. Direction Cosines: l=−26=−13l = \frac{-2}{6} = -\frac{1}{3} m=−46=−23m = \frac{-4}{6} = -\frac{2}{3} n=46=23n = \frac{4}{6} = \frac{2}{3}

Explanation:

First, find the vector components by subtracting coordinates, then find the magnitude, and finally calculate l,m,nl, m, n by dividing components by the magnitude.