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Vector Algebra - Addition of vectors, multiplication of a vector by a scalar

Grade 12CBSE

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

πŸ”‘Concepts

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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 OP⃗=xi^+yj^+zk^\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}. Visually, this is represented as a directed arrow starting at the origin and ending at the point PP in three-dimensional space.

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Direction Cosines and Ratios: The coordinates x,y,zx, y, z are the direction ratios of the position vector. The direction cosines cos⁑α,cos⁑β,cos⁑γ\cos \alpha, \cos \beta, \cos \gamma represent the angles the vector makes with the x,y,and zx, y, \text{and } z axes respectively, visually defining the vector's orientation relative to the coordinate frame.

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Triangle Law of Vector Addition: If two vectors are represented by two sides of a triangle taken in order (head-to-tail), their sum is represented by the third side taken in the reverse order. Geometrically, this completes the triangle where the resultant vector acts as a direct path from the start of the first vector to the end of the second.

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Parallelogram Law of Vector Addition: If two vectors a⃗\vec{a} and b⃗\vec{b} are represented by the adjacent sides of a parallelogram, their sum a⃗+b⃗\vec{a} + \vec{b} is represented by the diagonal passing through their common initial point. This visualizes how two concurrent forces or velocities combine into a single resultant.

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Multiplication of a Vector by a Scalar: Multiplying a vector aβƒ—\vec{a} by a scalar Ξ»\lambda results in a vector Ξ»aβƒ—\lambda\vec{a}. Visually, the vector is scaled (stretched or shrunk) by a factor of ∣λ∣|\lambda|. If Ξ»\lambda is positive, the direction remains the same; if Ξ»\lambda is negative, the direction is exactly reversed.

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Unit Vector: A vector with a magnitude of 11 is called a unit vector, denoted by a^\hat{a}. It is calculated by dividing a vector by its own magnitude. Geometrically, it represents a 'pure direction' extending one unit length from the origin along the line of the original vector.

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Components of a Vector: Any vector can be expressed as a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, where a1,a2,a3a_1, a_2, a_3 are scalar components. Visually, these represent the projections of the vector onto the three mutually perpendicular coordinate axes, forming the edges of a rectangular box where the vector is the main diagonal.

πŸ“Formulae

Magnitude of vector aβƒ—=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}: ∣aβƒ—βˆ£=a12+a22+a32|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}

Addition of two vectors: a⃗+b⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}

Scalar Multiplication: λa⃗=(λa1)i^+(λa2)j^+(λa3)k^\lambda\vec{a} = (\lambda a_1)\hat{i} + (\lambda a_2)\hat{j} + (\lambda a_3)\hat{k}

Unit Vector in direction of aβƒ—\vec{a}: a^=aβƒ—βˆ£aβƒ—βˆ£=1a12+a22+a32(a1i^+a2j^+a3k^)\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{a_1^2 + a_2^2 + a_3^2}}(a_1\hat{i} + a_2\hat{j} + a_3\hat{k})

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): PQβƒ—=(x2βˆ’x1)i^+(y2βˆ’y1)j^+(z2βˆ’z1)k^\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

Magnitude of PQβƒ—\vec{PQ}: ∣PQβƒ—βˆ£=(x2βˆ’x1)2+(y2βˆ’y1)2+(z2βˆ’z1)2|\vec{PQ}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

πŸ’‘Examples

Problem 1:

Find the unit vector in the direction of the sum of the vectors aβƒ—=2i^+2j^βˆ’5k^\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k} and bβƒ—=2i^+j^+3k^\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}.

Solution:

Step 1: Find the sum of the vectors cβƒ—=aβƒ—+bβƒ—\vec{c} = \vec{a} + \vec{b}. cβƒ—=(2+2)i^+(2+1)j^+(βˆ’5+3)k^=4i^+3j^βˆ’2k^\vec{c} = (2+2)\hat{i} + (2+1)\hat{j} + (-5+3)\hat{k} = 4\hat{i} + 3\hat{j} - 2\hat{k} Step 2: Calculate the magnitude of the resultant vector ∣cβƒ—βˆ£|\vec{c}|. ∣cβƒ—βˆ£=42+32+(βˆ’2)2=16+9+4=29|\vec{c}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29} Step 3: Find the unit vector c^=cβƒ—βˆ£cβƒ—βˆ£\hat{c} = \frac{\vec{c}}{|\vec{c}|}. c^=129(4i^+3j^βˆ’2k^)=429i^+329j^βˆ’229k^\hat{c} = \frac{1}{\sqrt{29}}(4\hat{i} + 3\hat{j} - 2\hat{k}) = \frac{4}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{2}{\sqrt{29}}\hat{k}

Explanation:

To find a unit vector in a specific direction, we first determine the resultant vector by adding the components of the given vectors, then divide that resultant by its own magnitude.

Problem 2:

Find a vector in the direction of vector aβƒ—=i^βˆ’2j^\vec{a} = \hat{i} - 2\hat{j} that has magnitude 77 units.

Solution:

Step 1: Find the unit vector in the direction of aβƒ—\vec{a}. First, find the magnitude ∣aβƒ—βˆ£=12+(βˆ’2)2=1+4=5|\vec{a}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}. Then, a^=aβƒ—βˆ£aβƒ—βˆ£=15i^βˆ’25j^\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}. Step 2: Multiply the unit vector by the required magnitude (77). vβƒ—=7β‹…a^=7(15i^βˆ’25j^)=75i^βˆ’145j^\vec{v} = 7 \cdot \hat{a} = 7\left(\frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}\right) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}

Explanation:

A vector of a specific magnitude in a given direction is obtained by finding the unit vector (which has magnitude 1) and then scaling it using scalar multiplication.