7C, 7E - Vector lengths and unit vectors

7C, 7E - Vector lengths and unit vectors

A displacement has components 33 units east and 44 units north: how long is the direct journey, and how can we keep its direction while changing its length? You will calculate vector magnitudes, build unit vectors and construct vectors of a specified magnitude. In 7C, you find a vector’s magnitude; in 7E, you use that length to make a unit vector.

Magnitude of a vector

Horizontal and vertical components form the perpendicular sides of a right-angled triangle whose hypotenuse is the vector's length.

[DIAGRAM: asset_name: A vector and a unit vector in the same direction; asset_slug: magnitude-unit-direction; caption: A vector with components 3 and 4 has magnitude 5, and dividing it by 5 gives a unit vector; description: A right-angled component triangle has legs 3 and 4 and a blue displacement arrow of length 5. A shorter clay arrow from the same origin lies on the blue arrow and has components three fifths and four fifths, so its length is 1.]
A vector with components 3 and 4 has magnitude 5, and dividing it by 5 gives a unit vector

Magnitude from perpendicular components

For a=xi+yj\mathbf{a}=x\mathbf{i}+y\mathbf{j},

∣a∣=x2+y2|\mathbf{a}|=\sqrt{x^2+y^2}

The magnitude is the nonnegative square root. The vectors i\mathbf{i} and j\mathbf{j} are perpendicular and each has magnitude 11.

This is Pythagoras applied to the component triangle. A component's sign tells you which way to move, but its square contributes positively to the squared length. The magnitude is a scalar; do not attach i\mathbf{i} or j\mathbf{j} to it.

The perpendicular condition matters. For arbitrary nonparallel vectors p\mathbf{p} and q\mathbf{q}, the coefficients of xp+yqx\mathbf{p}+y\mathbf{q} need not be perpendicular lengths, so this formula does not apply just because there are two terms.

Worked example 1: a component triangle

Question. Find the magnitude of a=3i+4j\mathbf{a}=3\mathbf{i}+4\mathbf{j}.

Plan. Use the perpendicular component lengths in Pythagoras.

∣a∣2=32+42=25∣a∣=5\begin{aligned} |\mathbf{a}|^2&=3^2+4^2=25\\ |\mathbf{a}|&=5 \end{aligned}

Answer. ∣a∣=5|\mathbf{a}|=5.

Check. The direct length lies between the larger component 44 and the two-step distance 3+4=73+4=7; 55 fits.

Reverse just the horizontal component. The direction changes, but the reflected right-angled triangle has the same side lengths.

Worked example 2: a negative component

Question. Find the magnitude of b=−3i+4j\mathbf{b}=-3\mathbf{i}+4\mathbf{j}.

Plan. Square the signed components before adding.

∣b∣=(−3)2+42=9+16=5\begin{aligned} |\mathbf{b}|&=\sqrt{(-3)^2+4^2}\\ &=\sqrt{9+16}=5 \end{aligned}

Answer. ∣b∣=5|\mathbf{b}|=5, although b≠a\mathbf{b}\ne\mathbf{a}.

Check. Reflecting the first vector in the vertical axis gives this vector without stretching it, so their magnitudes must agree.

A magnitude need not be an integer. Keep an exact root when necessary.

Unit vectors

Dividing a nonzero vector by its length reduces its magnitude to 11 while keeping its direction.

Unit vector

A unit vector has magnitude 11. The unit vector in the direction of a nonzero vector a\mathbf{a} is

a^=a∣a∣\widehat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}

The hat is a label for this unit vector, not a new operation on its components.

The divisor ∣a∣|\mathbf{a}| is positive, so it does not reverse the arrow. Each component is divided by the same number, preserving the direction. There is no unit vector in the direction of 0\mathbf{0}: the zero vector has no specified direction, and dividing by zero is undefined.

The vectors i\mathbf{i} and j\mathbf{j} are particular unit vectors, but there are infinitely many others. A unit vector need not have both components positive, and its components need not be integers.

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