Match each vector scenario on the left with its correct resultant magnitude or value on th… — JAMB UTME

Match each vector scenario on the left with its correct resultant magnitude or value on the right.

Resultant of two perpendicular forces of magnitudes $6\text{ N}$ and $8\text{ N}$ $10\text{ N}$
Minimum possible magnitude of the resultant of two forces of $7\text{ N}$ and $12\text{ N}$ $5\text{ N}$
Resultant magnitude of two equal forces of $15\text{ N}$ inclined at an angle of $120^\circ$ to each other $15\text{ N}$
Magnitude of a 3D displacement vector given by $\mathbf{r} = (3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k})\text{ m}$ $13\text{ m}$

Answer

Match perpendicular forces of 6 N and 8 N to 10 N; minimum resultant of 7 N and 12 N forces to 5 N; two 15 N forces at 120 degrees to 15 N; and 3D displacement vector to 13 m.

Each vector calculation correctly applies the geometric or algebraic properties of vectors: orthogonal vector resolution, opposite-direction subtraction, law of cosines for equal magnitudes at 120 degrees, and 3D component magnitude synthesis.

Step-by-Step Solution

Calculate the magnitude of perpendicular vectors

R = \sqrt{6^2 + 8^2} = 10\text{ N}

Perpendicular vectors form a right-angled triangle, so the Pythagorean theorem applies.

Find the minimum resultant magnitude of two vectors

R_{\text{min}} = 12\text{ N} - 7\text{ N} = 5\text{ N}

Minimum resultant occurs when vectors act collinear in opposite directions.

Determine the resultant of two equal vectors at

120^\circ

R = 15\text{ N}

Using the cosine rule

R = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ}

, since

\cos 120^\circ = -0.5

R = F

Compute the magnitude of the 3D displacement vector

|\mathbf{r}| = \sqrt{3^2 + 4^2 + 12^2} = 13\text{ m}

3D magnitude is calculated using the square root of the sum of squared orthogonal components.

Key Concept

Vector Addition, Resolution, and Magnitude Evaluation

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