Basic concepts of vectors

Scalar vs vector. A scalar has magnitude only — mass, temperature, distance, speed. A vector has magnitude and direction — displacement, velocity, force, acceleration. Geometrically a vector is drawn as a directed line segment: if the tail (initial point) is $A$ and the head (terminal point) is $B$, the vector is $\overrightarrow{AB}$, and its magnitude $|\overrightarrow{AB}|$ is the length of segment $AB$. Two arrows of the same length pointing the same way represent the same vector, no matter where in space they are drawn — a vector is defined purely by length and orientation, not by position.

Position vector. To pin a vector to an actual location, fix an origin $O$. For a point $P(x,y,z)$, the vector $\overrightarrow{OP}$ is the position vector of $P$:

where $\hat i,\hat j,\hat k$ are the unit vectors along the $x$, $y$, $z$ axes (each of magnitude 1). The magnitude of this position vector follows from a 3D extension of Pythagoras:

This single formula is used everywhere downstream — it is exactly how you will later compute $|\vec a - \vec b|$, the distance between two points, or the length of a vector obtained by adding two others.

Direction cosines and direction ratios. Suppose $\vec r$ makes angles $\alpha,\beta,\gamma$ with the positive $x$, $y$, $z$ axes. The cosines of these angles, $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$, are the direction cosines of the vector — they encode its orientation independent of its length. Any triple of numbers proportional to $l,m,n$ — most conveniently the components $x,y,z$ themselves — are called direction ratios. Direction ratios are not unique (any nonzero scalar multiple describes the same line direction), but direction cosines are unique for a given direction and always satisfy $l^2+m^2+n^2=1$.

The goal is to show, using only the magnitude formula for a position vector and the definition of a direction cosine as the ratio of a component to the magnitude, that the three direction cosines of any vector always satisfy $l^2+m^2+n^2=1$ — a check that catches almost every arithmetic slip in this topic.

Step 1. Start from a vector $\vec r = x\hat i + y\hat j + z\hat k$ with magnitude $r = |\vec r| = \sqrt{x^2+y^2+z^2}$. By definition, the direction cosines are the components divided by the magnitude: $l = \dfrac{x}{r}$, $m = \dfrac{y}{r}$, $n = \dfrac{z}{r}$. This is exactly saying each direction cosine is "how much of the total length lies along that axis," as a fraction.

Step 2. Square each and add them: $l^2+m^2+n^2 = \dfrac{x^2}{r^2}+\dfrac{y^2}{r^2}+\dfrac{z^2}{r^2} = \dfrac{x^2+y^2+z^2}{r^2}.$

Step 3. Recognize the numerator. Since $r^2 = x^2+y^2+z^2$ by the magnitude formula, the numerator and denominator are identical: $l^2+m^2+n^2 = \dfrac{r^2}{r^2} = 1.$

Step 4. Read the result. This identity holds for every vector, regardless of its actual length — because dividing by $r$ before squaring and adding rescales the vector to a length-1 (unit) vector, and a unit vector's own components must satisfy Pythagoras with total 1. Use $l^2+m^2+n^2=1$ as a built-in error check: after computing direction cosines from any vector, plug them back in — if the sum of squares is not 1, a component or the magnitude was computed wrong.

Find the magnitude and direction cosines of $\vec a = 2\hat i - 3\hat j + 6\hat k$, then verify the identity, and use the same vector to find the angle it makes with the $y$-axis.

Step 1. Magnitude: $|\vec a| = \sqrt{2^2+(-3)^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7.$

Step 2. Direction ratios are just the components themselves: $2, -3, 6$.

Step 3. Direction cosines — divide each component by the magnitude: $l = \dfrac{2}{7}$, $m = \dfrac{-3}{7}$, $n = \dfrac{6}{7}$.

Step 4. Verify the identity: $l^2+m^2+n^2 = \dfrac{4}{49}+\dfrac{9}{49}+\dfrac{36}{49} = \dfrac{49}{49} = 1$. Checks out.

Step 5. Angle with the $y$-axis: $m = \cos\beta = -\dfrac{3}{7}$, so $\beta = \cos^{-1}\!\left(-\dfrac{3}{7}\right) \approx 115.4^\circ$. The negative cosine correctly signals an obtuse angle — $\vec a$ points somewhat "backward" along $y$, matching its negative $y$-component.

$\vec a$ has magnitude 7, direction cosines $\left(\dfrac{2}{7}, -\dfrac{3}{7}, \dfrac{6}{7}\right)$, and makes an angle of about $115.4^\circ$ with the positive $y$-axis.

Competitive stretch

If a vector has direction cosines proportional to $1, 2, -2$ and magnitude $9$, find the vector. First scale to actual direction cosines: $r=\sqrt{1^2+2^2+(-2)^2}=\sqrt{9}=3$, so $l=\tfrac13, m=\tfrac23, n=-\tfrac23$. Multiply by the given magnitude $9$: components are $9l=3$, $9m=6$, $9n=-6$, giving $\vec v = 3\hat i+6\hat j-6\hat k$. Check: $\sqrt{3^2+6^2+6^2}=\sqrt{9+36+36}=\sqrt{81}=9$. ✓

• Treating components as direction cosines directly. $(x,y,z)$ are direction ratios; you must divide by $|\vec r|$ to get true direction cosines, unless the vector already has magnitude 1.
• Forgetting direction ratios are not unique. $(2,-3,6)$ and $(4,-6,12)$ describe the same direction; only direction cosines are unique for a given direction.
• Writing $\vec a = 7$. Magnitude is a scalar; a vector cannot equal a bare number. Write $|\vec a| = 7$.
• Assuming $\cos^{-1}$ must give an acute angle. A negative direction cosine correctly gives an obtuse angle (greater than $90^\circ$) — do not "fix" the sign.

Ask the tutor: want to walk through why dividing by the magnitude always produces a valid set of direction cosines, or practice a few more direction-cosine conversions?