Multiply a vector by a scalar graphically and component-wise, and find the magnitude and direction of the result

Learning goal: Multiply a vector by a scalar; represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, as c⟨v_x, v_y⟩ = ⟨c v_x, c v_y⟩; compute the magnitude of a scalar multiple cv using ‖cv‖ = |c| ‖v‖; and compute the direction of cv knowing that when |c| ‖v‖ ≠ 0 the direction of cv is either along v (for c > 0) or against v (for c < 0).

Scope note. N.VM.5, with its parts a and b, is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

What a scalar multiple is. A scalar is an ordinary real number: 2, −1.5, ½, 0. Multiplying a vector v by a scalar c produces a new vector cv. Its meaning comes from repeated addition: 2v is v + v, the displacement v done twice, so it points the same way as v and is twice as long. Extending this, 3v is three copies end-to-end, ½v is half the arrow, and (−1)v is the additive inverse −v, the same length pointing the opposite way. In general, cv is the arrow v scaled by the factor |c|, reversed if c is negative.

Component-wise. Since scaling an arrow scales its horizontal and vertical moves by the same factor, scalar multiplication is performed component-wise: c⟨v_x, v_y⟩ = ⟨c v_x, c v_y⟩. Both components are multiplied by c, including the sign of c. Take v = ⟨3, 1⟩. Then 2v = ⟨6, 2⟩ and −1.5v = ⟨−4.5, −1.5⟩. Drawn from the origin, 2v ends at (6, 2), which lies on the same line through the origin as (3, 1), twice as far out; −1.5v ends at (−4.5, −1.5), on the same line but on the other side of the origin.

A fully worked model, magnitudes. ‖v‖ = √(3² + 1²) = √10 ≈ 3.162. ‖2v‖ = √(6² + 2²) = √40 = √(4 × 10) = 2√10 ≈ 6.325. ‖−1.5v‖ = √((−4.5)² + (−1.5)²) = √(20.25 + 2.25) = √22.5 = 1.5√10 ≈ 4.743. In each case the magnitude of the multiple is the absolute value of the scalar times the magnitude of the original: ‖cv‖ = |c| ‖v‖. The absolute value is essential. The scalar −1.5 does not produce a negative length; lengths are never negative. It produces a vector 1.5 times as long, and separately reverses the direction. The algebraic reason: ‖cv‖ = √((c v_x)² + (c v_y)²) = √(c²(v_x² + v_y²)) = √(c²) √(v_x² + v_y²) = |c| ‖v‖, and √(c²) is |c|, not c.

Direction. Provided that |c| ‖v‖ ≠ 0, that is, provided c is not zero and v is not the zero vector, the direction of cv is either along v (when c > 0) or against v (when c < 0). The direction angle of v = ⟨3, 1⟩ is arctan(1/3) ≈ 18.43°. The direction angle of 2v = ⟨6, 2⟩ is arctan(2/6) = arctan(1/3), the same 18.43°: along v. The direction angle of −1.5v = ⟨−4.5, −1.5⟩ is 18.43° + 180° = 198.43°: against v, the arrow pointing down and to the left. Multiplying by a positive scalar never changes direction, however large or small the scalar; multiplying by a negative scalar always reverses it. The condition matters: if c = 0 then cv = ⟨0, 0⟩, the zero vector, which has magnitude 0 and no direction at all, and the same happens when v is itself zero. That is why the rule says "when |c| ‖v‖ ≠ 0".

Graphical representation. To draw cv, keep the tail of v where it is, and stretch or shrink the arrow along its own line to |c| times its length. If c is negative, draw the stretched arrow on the opposite side of the tail. All scalar multiples of a nonzero v lie on one line through its tail; the positive multiples fill one ray and the negative multiples fill the opposite ray. Two nonzero vectors are parallel exactly when one is a scalar multiple of the other, which is a useful test: ⟨4, 2⟩ = 2⟨2, 1⟩ is parallel to ⟨2, 1⟩; ⟨4, 3⟩ is not, because 4/2 = 2 but 3/1 = 3.

Unit vectors. Dividing a nonzero vector by its own magnitude, that is, multiplying by the scalar 1/‖v‖, gives a vector of magnitude exactly 1 pointing along v: ‖(1/‖v‖)v‖ = (1/‖v‖) ‖v‖ = 1. For v = ⟨3, 4⟩ with magnitude 5 the unit vector is ⟨3/5, 4/5⟩ = ⟨0.6, 0.8⟩; check 0.36 + 0.64 = 1. Unit vectors let you build a vector of any wanted magnitude in a given direction: 20 units in the direction of ⟨3, 4⟩ is 20⟨0.6, 0.8⟩ = ⟨12, 16⟩, and 20 units against it is −20⟨0.6, 0.8⟩ = ⟨−12, −16⟩.

Combining with addition. Scalar multiplication and addition mix as you would expect: 2u − 3w means 2u + (−3)w. With u = ⟨1, 2⟩ and w = ⟨−1, 1⟩: 2u = ⟨2, 4⟩, −3w = ⟨3, −3⟩, and 2u − 3w = ⟨5, 1⟩. The distributive rules hold: c(u + w) = cu + cw and (c + d)v = cv + dv, both of which follow from doing the same arithmetic on each component.

The misconceptions. First, "a negative scalar gives a negative magnitude." No: ‖−1.5v‖ = 1.5‖v‖, positive; the sign went into the direction. Second, "multiplying by a scalar between 0 and 1 reverses the vector." No: ½v points along v, just shorter. Reversal is caused only by a negative sign, never by a small size. Third, "cv changes the direction angle by a factor of c." No: the direction angle of 2v is the same as that of v, not twice it. Scalars scale lengths, not angles; the only angle change possible is the 180° flip from a negative c.

Worked, positive scalar: Compute 3v for v = ⟨−2, 5⟩ and verify the magnitude rule. Component-wise: 3⟨−2, 5⟩ = ⟨−6, 15⟩. ‖v‖ = √(4 + 25) = √29 ≈ 5.385. ‖3v‖ = √(36 + 225) = √261 = √(9 × 29) = 3√29 ≈ 16.155, which is 3 × 5.385. Direction: both have components with the same signs and the same ratio 15/(−6) = 5/(−2), so 3v is along v, in the second quadrant.

Worked, negative scalar: Compute −0.5v for v = ⟨8, −6⟩. Component-wise: ⟨−4, 3⟩. ‖v‖ = √(64 + 36) = 10; ‖−0.5v‖ = √(16 + 9) = 5 = |−0.5| × 10. Direction of v: right and down, angle 360° − arctan(6/8) ≈ 323.13°. Direction of −0.5v: left and up, 180° − arctan(3/4) ≈ 143.13°, which is 323.13° − 180°: against v. The magnitude halved and the direction reversed.

Worked, building a vector of given magnitude: Find the vector of magnitude 20 pointing against ⟨3, 4⟩. Unit vector along ⟨3, 4⟩: ⟨3, 4⟩/5 = ⟨0.6, 0.8⟩. Against it with magnitude 20: −20⟨0.6, 0.8⟩ = ⟨−12, −16⟩. Check: √(144 + 256) = √400 = 20, and the components are the negatives of a positive multiple of ⟨3, 4⟩, so the direction is reversed. Equivalently c⟨3, 4⟩ with |c| × 5 = 20 and c < 0 gives c = −4.

Worked, a linear combination: u = ⟨1, 2⟩, w = ⟨−1, 1⟩. 2u − 3w = ⟨2, 4⟩ + ⟨3, −3⟩ = ⟨5, 1⟩. Each scalar multiplies both components of its vector, then the results are added component-wise. Magnitude of the result: √26 ≈ 5.099.

Worked, the zero case: 0 × ⟨3, 1⟩ = ⟨0, 0⟩. The magnitude rule still holds, ‖0v‖ = 0 × √10 = 0, but the direction rule does not apply, because the zero vector has no direction. The condition |c| ‖v‖ ≠ 0 in the direction rule exists precisely to exclude this case.

Recording task: On grid paper draw v = ⟨3, 1⟩ from the origin, then 2v and −1.5v from the same origin, in three colors. Record a table with columns: scalar c, components of cv, ‖cv‖ computed from the components, |c| × ‖v‖, direction angle, along or against v. The rows for c = 2 and c = −1.5 should read ⟨6, 2⟩, √40 ≈ 6.32, 2 × 3.16 = 6.32, 18.43°, along; and ⟨−4.5, −1.5⟩, √22.5 ≈ 4.74, 1.5 × 3.16 = 4.74, 198.43°, against. Add a row for c = 0 and write why the last two columns are blank.

Independent performance: Choose a vector v of your own with both components nonzero and not equal. Compute and draw cv for c = 3, c = ½, c = −1 and c = −2.5. For each, compute the magnitude two ways (from the components and from |c| ‖v‖) and state the direction angle and whether it is along or against v. Then find the unit vector in the direction of v and use it to write the vector of magnitude 13 pointing against v. Finally choose a second vector w and compute 2v − 3w component-wise. Success criteria: every magnitude is nonnegative (the nonzero multiples of v have positive magnitude) and the two computations agree to within rounding; the direction angle of every positive multiple equals that of v and of every negative multiple differs from it by exactly 180°; the unit vector has magnitude 1 when checked; the magnitude-13 vector checks to 13; and a closing sentence explains why the direction rule needs the condition |c| ‖v‖ ≠ 0.