Scale, reverse, and handle the zero vector

Try

Let v=(3,−2). Multiplying by 2 gives (6,−4), which lies twice as far along the same ray from the origin. Multiplying by −2 gives (−6,4), twice as long but pointing in the opposite direction. Multiplying by zero gives (0,0), which has no distinguished direction.

Construct these three endpoints on equal-scale axes. A coordinate description with an arrow from the origin to each nonzero endpoint is equally valid. For the zero result, mark the origin instead of inventing a tiny arrow whose orientation would suggest information that is not present.

Predict

Predict whether the magnitude of −2v is −2 times the magnitude of v. Length cannot be negative. The multiplier for magnitude is the absolute value of the scalar, so both 2v and −2v have magnitude 2√13.

Before using a scalar in a context, ask whether it is dimensionless or has units. Multiplying a velocity by dimensionless 2 produces another velocity. Multiplying it by 2 seconds produces a displacement. Numerically identical coordinates can represent different physical quantities.

Experiment

Complete the endpoint and magnitude table, then explain each direction in words.

Scalar c	cv for v=(3,−2)	Magnitude	Direction compared with v
2	(6,−4)	2√13	Same
−2	(−6,4)	2√13	Opposite
1/2	(3/2,−1)	√13/2	Same
0	(0,0)	0	Undefined

Use the scaling mode in the vector construction lesson to compare positive and negative scalars on a different vector. Its numerical inputs are separate from the fixed table. Test a zero input vector too: no scalar can create a nonzero direction from it.

For a paper or coordinate-list experiment, draw a line through the origin and v’s tip. Every scalar multiple stays on this line. Only positive multiples of this nonzero vector have its same direction; zero is the ray’s endpoint but has no direction.

Understand

For a real scalar c and vector v=(x,y), define cv=(cx,cy). Its magnitude is √((cx)²+(cy)²)=√(c²(x²+y²))=|c|√(x²+y²)=|c| ||v||. The identity √(c²)=|c|, rather than c, is the same principal-root issue seen earlier in this unit.

If v is nonzero and c>0, both coordinates are multiplied by the same positive factor, so the endpoint lies on the ray through v. If c<0, write cv=−(|c|v); the magnitude is scaled positively and the direction reversed. If c=0 or v=0, the result is zero and a direction comparison is not meaningful. A zero vector does not point east simply because a software convention returns angle zero.

Scalar multiplication distributes over vector addition because real multiplication distributes in each coordinate. For u=(a,b) and v=(d,e), c(u+v)=(c(a+d),c(b+e))=(ca+cd,cb+ce)=cu+cv. This proves the rule for every planar pair. A matching drawing illustrates this identity but does not establish it for all inputs.

The context determines units. For velocity v=(3,−2) metres per second, dimensionless −2 gives velocity (−6,4) metres per second, with twice the original speed. By contrast, 2 seconds times v gives displacement (6,−4) metres. A formal time offset of −2 seconds gives a backward extrapolated displacement (−6,4) metres relative to a chosen reference instant; it is not a negative duration someone can wait.

Apply

For a reverse journey of positive duration, reverse the velocity and keep elapsed time positive. This gives a physical interpretation without comparing the magnitude of a velocity directly with the length of a displacement.

In your written proof, include both zero cases. Saying “negative reverses, positive preserves” covers nonzero multiples of nonzero vectors, but leaves out situations in which direction is undefined. Your construction rubric should check that omission just as carefully as an incorrect coordinate.

Continue

The signed constructions and general magnitude proof address N-VM.B.5. They also connect three ideas already developed: absolute value, directed coordinates and the distinction between algebraic expressions and their contextual interpretation.

Next, apply entrywise scalar multiplication to matrices. A matrix holds more than two entries, but each row and column must retain its label. Matching arithmetic positions does not automatically mean the underlying quantities can be combined.