Magnitude and direction of a vector

About this lesson

Precalculus › Vectors

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Magnitude and direction of a vector

Warm-up

Three quick questions from earlier lessons.

Discover

How long is the vector ⟨3,4⟩\langle 3, 4 \rangle?

  1. 5
  2. 7

Key idea

A vector's magnitude is its length; its direction angle is measured counterclockwise from the positive xx-axis.

∣⟨a,b⟩∣=a2+b2|\langle a, b \rangle| = \sqrt{a^2 + b^2}. For the direction angle θ\theta: tan⁡θ=ba\tan\theta = \dfrac{b}{a}. The calculator's tan⁡−1\tan^{-1} gives an angle between −90∘-90^\circ and 90∘90^\circ; if the vector points left (a<0a < 0), add 180∘180^\circ.

Worked example

Find the magnitude and direction of ⟨−2,2⟩\langle -2, 2 \rangle.

∣v⃗∣=4+4=22|\vec{v}| = \sqrt{4 + 4} = 2\sqrt{2}
Pythagoras.
tan⁡−1(−1)=−45∘\tan^{-1}(-1) = -45^\circ
tan θ = 2 / (−2) = −1; the calculator gives −45°.
θ=135∘\theta = 135^\circ
The vector points left and up (quadrant II): add 180°.

Your turn, guided

Your turn: ⟨5,−12⟩\langle 5, -12 \rangle.

∣v⃗∣=13|\vec{v}| = 13
√(25 + 144).
θ≈292.6∘\theta \approx 292.6^\circ
Right and down: quadrant IV, so tan⁻¹(−12/5) ≈ −67.4°, or 292.6°.

Practice ★

  1. Find the exact magnitude of v⃗=⟨6,−5⟩\vec{v} = \langle 6, -5 \rangle.
  2. Find the exact magnitude of v⃗=⟨7,−7⟩\vec{v} = \langle 7, -7 \rangle.
  3. Find the exact magnitude of v⃗=⟨3,−7⟩\vec{v} = \langle 3, -7 \rangle.
  4. Find the exact magnitude of v⃗=⟨8,−2⟩\vec{v} = \langle 8, -2 \rangle.

Watch out!

Practice ★★

  1. Find the direction angle of ⟨2,3⟩\langle 2, 3 \rangle, in degrees, to 1 decimal place.
  2. Find the direction angle of ⟨8,9⟩\langle 8, 9 \rangle, in degrees, to 1 decimal place.
  1. Find the direction angle of ⟨−2,3⟩\langle -2, 3 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.
  2. Find the direction angle of ⟨−1,7⟩\langle -1, 7 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.

Exit ticket

Three questions to check you have got it.

  1. Find the direction angle of ⟨−2,3⟩\langle -2, 3 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.
  2. Find the direction angle of ⟨−1,7⟩\langle -1, 7 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.
  3. Find the direction angle of ⟨−8,9⟩\langle -8, 9 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.

Summary card

Magnitude and direction

  • ∣⟨a,b⟩∣=a2+b2|\langle a, b \rangle| = \sqrt{a^2 + b^2}.
  • tan⁡θ=ba\tan\theta = \dfrac{b}{a}; sketch the vector to pick the quadrant.
  • Pointing left: add 180∘180^\circ to the calculator's angle.

Watch out: the magnitude is not a+ba + b.

End of the lesson

Next lesson: Add, subtract and scale vectors Back to the chapter

Answers: Magnitude and direction of a vector

A

  1. 61\sqrt{61}
  2. 727 \sqrt{2}
  3. 58\sqrt{58}
  4. 2172 \sqrt{17}

B

  1. 180tan⁡−1(32)π\frac{180 \tan^{-1}{\left(\frac{3}{2} \right)}}{\pi}
  2. 180tan⁡−1(98)π\frac{180 \tan^{-1}{\left(\frac{9}{8} \right)}}{\pi}

C

  1. −180tan⁡−1(32)π+180- \frac{180 \tan^{-1}{\left(\frac{3}{2} \right)}}{\pi} + 180
  2. −180tan⁡−1(7)π+180- \frac{180 \tan^{-1}{\left(7 \right)}}{\pi} + 180

D

  1. −180tan⁡−1(32)π+180- \frac{180 \tan^{-1}{\left(\frac{3}{2} \right)}}{\pi} + 180
  2. −180tan⁡−1(7)π+180- \frac{180 \tan^{-1}{\left(7 \right)}}{\pi} + 180
  3. −180tan⁡−1(98)π+180- \frac{180 \tan^{-1}{\left(\frac{9}{8} \right)}}{\pi} + 180

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