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Motion in a Plane - Multiplication of Vectors by Real Numbers

Grade 11CBSEPhysics

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

🔑Concepts

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Multiplying a vector A⃗\vec{A} by a positive real number nn results in a new vector nA⃗n\vec{A}. The magnitude of this new vector is nn times the magnitude of A⃗\vec{A}, while the direction remains the same as A⃗\vec{A}.

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Multiplying a vector A⃗\vec{A} by a negative real number −n-n results in a vector −nA⃗-n\vec{A}. The magnitude is nn times the magnitude of A⃗\vec{A}, but the direction is exactly opposite (180∘180^{\circ} reversal) to that of A⃗\vec{A}.

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If the multiplier is a scalar physical quantity with its own dimensions, the resulting vector will have different dimensions and units. For example, multiplying velocity vector v⃗\vec{v} by time Δt\Delta t results in a displacement vector Δr⃗=v⃗Δt\Delta \vec{r} = \vec{v}\Delta t.

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Multiplying any vector by the real number 00 results in a null vector or zero vector, denoted as 0⃗\vec{0}, which has zero magnitude and an indeterminate direction.

📐Formulae

B⃗=nA⃗\vec{B} = n\vec{A}

∣B⃗∣=∣n∣⋅∣A⃗∣|\vec{B}| = |n| \cdot |\vec{A}|

If n>0, direction of B⃗= direction of A⃗\text{If } n > 0, \text{ direction of } \vec{B} = \text{ direction of } \vec{A}

If n<0, direction of B⃗=−(direction of A⃗)\text{If } n < 0, \text{ direction of } \vec{B} = - (\text{direction of } \vec{A})

💡Examples

Problem 1:

A displacement vector d⃗\vec{d} is 5 m5\text{ m} due East. Determine the magnitude and direction of the vectors (i) 2d⃗2\vec{d} and (ii) −1.5d⃗-1.5\vec{d}.

Solution:

(i) 2d⃗=2×5 m=10 m2\vec{d} = 2 \times 5\text{ m} = 10\text{ m} due East. (ii) −1.5d⃗=∣−1.5∣×5 m=7.5 m-1.5\vec{d} = |-1.5| \times 5\text{ m} = 7.5\text{ m} due West.

Explanation:

In the first case, the positive multiplier 22 doubles the magnitude and preserves the direction (East). In the second case, the negative multiplier 1.51.5 increases the magnitude by 1.51.5 times and reverses the direction from East to West.

Problem 2:

If a force vector F⃗=(2i^+3j^) N\vec{F} = (2\hat{i} + 3\hat{j}) \text{ N} is multiplied by a real number λ=4\lambda = 4, what is the resulting force vector?

Solution:

The resulting vector is F⃗′=λF⃗=4(2i^+3j^)=(8i^+12j^) N\vec{F}' = \lambda \vec{F} = 4(2\hat{i} + 3\hat{j}) = (8\hat{i} + 12\hat{j}) \text{ N}.

Explanation:

Multiplication by a real number distributes over the components of the vector in Cartesian form. Both the xx and yy components are scaled by the factor of 44.