How to Find Vector Components?

How to Find Vector Components? Is a multiple-choice math challenge from Effortless Math. Read the problem, choose from the answer options below, then open the solution to see the full step-by-step working and check your reasoning against it.

• Trigonometric ratios show the relationship between the magnitude of the vector and the components of the vector.

• Using “Pythagoras theorem” in right triangles with lengths \(v_x\) and \(v_y\):

Vector Components, Example 1:

\(\color{blue}{v_x= v \cos θ}\)

\(\color{blue}{v_y= v \sin θ}\)

\(v_x= v\cos 60° \) → \(v_x= 20×\frac{1}{2}= \frac{20}{2}=10\)

\(v_y= v\sin 60° \) → \(v_y= 20×\frac{\sqrt{3}}{2}= \frac{20\sqrt{3}}{2} =10\sqrt{3}\)

So, the vector \(v⃗\) is \((10, 10\sqrt{3})\).

Vector Components, Example 2:

Find the \(x\) and \(y\) components of a vector having a magnitude of \(10\) and make an angle of \(45\) degrees with the positive \(x\)-axis.

To find the components of a vector use these formulas:

\(\color{blue}{v_x= v \cos θ}\)

\(\color{blue}{v_y= v \sin θ}\)

\(v_x= v\cos 45° \) → \(v_x= 10×\frac{\sqrt{2}}{2}= \frac{10\sqrt{2}}{2} =5\sqrt{2}\)

\(v_y= v\sin 45° \) → \(v_y= 10×\frac{\sqrt{2}}{2}= \frac{10\sqrt{2}}{2} =5\sqrt{2}\)

So, the \(x\)-component and the \(y\)-components of the vector are both equal to \(5\sqrt{2}\).

Exercises for Vector Components

1. Find the value of \( θ \), if \(v_x=15\) and \(v_y=8.66\).
2. Find out the magnitude of a vector \(OA=(-3,4)\).
3. Find the components of the vector, if the magnitude of a vector \(v⃗\) is \(6\) units and the direction of the vector is \(30°\) with the horizontal.
4. Find the direction of \((-4,3)\).

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1. \(\color{blue}{θ=30^\circ}\)
2. \(\color{blue}{|OA|=5}\)
3. \(\color{blue}{v⃗=3, 3\sqrt{3}}\)
4. \(\color{blue}{θ=143.13^\circ}\)

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