Vector Algebra
Quantities with both size and direction — the algebra of arrows: adding and scaling them, and the three products (dot, cross, and triple) that turn vectors into angles, areas, and volumes
- The difference between scalars and vectors, and the standard types of vectors.
- Components \(a_1\hat i+a_2\hat j+a_3\hat k\), magnitude, and direction cosines.
- Addition and scalar multiplication by the triangle and parallelogram laws, and the section formula.
- The dot product for angles and projections, and the cross product for areas and perpendiculars.
- The scalar triple product as a volume and a test for coplanarity.
- The vector triple product identity.
Scalars & Vectors
A scalar carries only magnitude — mass, time, temperature. A vector carries magnitude and direction — displacement, velocity, force — and is drawn as a directed arrow. Two vectors are equal when they share both magnitude and direction, regardless of where they sit on the page.
The length \(|\vec a|\) is the magnitude; dividing by it gives the unit vector \(\hat a\) pointing the same way. Special cases include the zero vector \(\vec 0\) (no direction), and the position vector \(\overrightarrow{OP}\) of a point relative to the origin.
Components & Direction Cosines
Resolving a vector along the axes turns geometry into arithmetic. Every vector in space is a combination of the unit vectors \(\hat i,\hat j,\hat k\), and its direction cosines record the angles it makes with each axis.
The direction cosines \(l=\tfrac{a_1}{|\vec a|},\ m=\tfrac{a_2}{|\vec a|},\ n=\tfrac{a_3}{|\vec a|}\) are the cosines of the angles with the axes, and they always satisfy \(l^2+m^2+n^2=1\). The position vector of \(P(x,y,z)\) is simply \(x\hat i+y\hat j+z\hat k\).
Addition & Scalar Multiplication
Vectors add nose-to-tail. Slide the second arrow so its tail meets the first arrow's head, and the sum runs from the very start to the very end — the triangle law. Drawn from a common point instead, the sum is the diagonal of the parallelogram. Multiplying by a scalar stretches or shrinks a vector, and a negative scalar reverses it.
In components, add coordinate by coordinate. Addition is commutative and associative. The scalar multiple \(\lambda\vec a\) has magnitude \(|\lambda|\,|\vec a|\); it points the same way for \(\lambda>0\) and the opposite way for \(\lambda<0\). Two vectors are parallel exactly when one is a scalar multiple of the other.
The Section Formula
The position vector of a point that divides a line segment in a given ratio is a weighted average of the endpoints' position vectors — a single formula that handles midpoints, trisection points, and centroids alike.
Here \(\vec a,\vec b\) are the position vectors of the endpoints. The midpoint is the case \(m=n\), giving \(\tfrac12(\vec a+\vec b)\), and the centroid of a triangle is \(\tfrac13(\vec a+\vec b+\vec c)\).
The Dot Product
The scalar (dot) product multiplies two vectors to produce a number that encodes the angle between them. It is the natural tool for testing perpendicularity and for projecting one vector onto another.
From it, \(\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}\), and the two vectors are perpendicular precisely when \(\vec a\cdot\vec b=0\). The projection of \(\vec b\) on \(\vec a\) is \(\dfrac{\vec a\cdot\vec b}{|\vec a|}\), and \(\vec a\cdot\vec a=|\vec a|^2\). The product is commutative.
The Cross Product
The vector (cross) product multiplies two vectors to produce a third vector, perpendicular to both, whose length is the area of the parallelogram they span. Its direction follows the right-hand rule.
The magnitude \(|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta\) is the area of the parallelogram on \(\vec a,\vec b\) (half of it for the triangle). The product vanishes exactly when the vectors are parallel.
The dot product returns a scalar; the cross product returns a vector — never mix them up. And unlike the dot product, the cross product is anti-commutative: \(\vec a\times\vec b=-\,\vec b\times\vec a\), so the order genuinely matters and reversing it flips the resulting vector.
The Scalar Triple Product
Combining a dot and a cross gives a single number that measures the volume of the box spanned by three vectors — and, when that volume is zero, certifies that the three vectors lie in a plane.
Its absolute value is the volume of the parallelepiped with edges \(\vec a,\vec b,\vec c\). The product is cyclic, \([\vec a\,\vec b\,\vec c]=[\vec b\,\vec c\,\vec a]=[\vec c\,\vec a\,\vec b]\), and the three vectors are coplanar exactly when it equals \(0\).
The Vector Triple Product
Crossing a vector with a cross product gives back a vector — and a tidy identity untangles it into ordinary dot products. The result always lies in the plane of the two bracketed vectors.
Each term is a scalar (a dot product) times one of the bracketed vectors, so the answer lives in the plane of \(\vec b\) and \(\vec c\). Beware: the cross product is not associative, so \(\vec a\times(\vec b\times\vec c)\neq(\vec a\times\vec b)\times\vec c\) in general — the brackets are essential.
Putting It to Work
Problem. Find the unit vector along \(\vec a=2\hat i+3\hat j+6\hat k\).
Solution. Its magnitude is \(|\vec a|=\sqrt{4+9+36}=7\), so divide by it:
Problem. Find the angle between \(\vec a=\hat i+\hat j\) and \(\vec b=\hat j+\hat k\).
Solution. Here \(\vec a\cdot\vec b=1\) and \(|\vec a|=|\vec b|=\sqrt2\):
Problem. Find \(\vec a\times\vec b\) for \(\vec a=2\hat i+\hat j+\hat k\), \(\vec b=\hat i-\hat j+\hat k\).
Solution. Expand the determinant:
Problem. Find the point dividing \(A(1,2,3)\) and \(B(4,5,6)\) in the ratio \(2:1\) internally.
Solution. Apply \(\vec r=\tfrac{2\vec b+1\vec a}{3}\):
Problem. Find the volume of the parallelepiped with edges \(\vec a=\hat i+\hat j\), \(\vec b=\hat j+\hat k\), \(\vec c=\hat k+\hat i\).
Solution. Evaluate the scalar triple product as a determinant:
Problem. Evaluate \(\vec a\times(\vec b\times\vec c)\) for \(\vec a=\hat i,\ \vec b=\hat j,\ \vec c=\hat i\).
Solution. Use the identity \((\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c\):
Chapter Summary
Magnitude and direction; \(\hat a=\vec a/|\vec a|\).
\(\vec a=a_1\hat i+a_2\hat j+a_3\hat k\); \(l^2+m^2+n^2=1\).
Triangle/parallelogram law; \(\vec r=\tfrac{m\vec b+n\vec a}{m+n}\).
\(\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta\); \(\perp\Leftrightarrow 0\).
\(|\vec a\times\vec b|\) = area; \(\parallel\Leftrightarrow 0\).
\([\vec a\,\vec b\,\vec c]\) = volume; \(\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c\).
Problems
Decide what each problem asks for — a magnitude, an angle, an area, a volume — then reach for the matching product. Difficulty rises down the list.
- Find the magnitude of \(\vec a=3\hat i-4\hat j\).
- Find the unit vector along \(\vec a=\hat i+2\hat j+2\hat k\).
- For \(\vec a=2\hat i+3\hat j\) and \(\vec b=\hat i-\hat j\), find \(\vec a+\vec b\) and \(\vec a-\vec b\).
- Find \(\vec a\cdot\vec b\) for \(\vec a=\hat i+2\hat j+3\hat k\), \(\vec b=2\hat i-\hat j+\hat k\).
- Find the angle between \(\vec a=\hat i+\hat j\) and \(\vec b=\hat j+\hat k\).
- Find \(\vec a\times\vec b\) for \(\vec a=\hat i+\hat j+\hat k\), \(\vec b=\hat i+2\hat j+3\hat k\).
- Find the area of the parallelogram with sides \(\hat i+\hat j+\hat k\) and \(\hat i+2\hat j+3\hat k\).
- Find the projection of \(\vec a=\hat i+3\hat j+7\hat k\) on \(\vec b=7\hat i-\hat j+8\hat k\).
- Find the position vector of the point dividing \(A(1,-2,3)\) and \(B(3,4,-1)\) in the ratio \(1:2\).
- Find \(\lambda\) so that \(2\hat i+\lambda\hat j+\hat k\) and \(\hat i-2\hat j+3\hat k\) are perpendicular.
- Find the volume of the parallelepiped with edges \(\hat i+\hat j,\ \hat j+\hat k,\ \hat k+\hat i\).
- Show that \(2\hat i-\hat j+\hat k\) and \(-6\hat i+3\hat j-3\hat k\) are parallel.
- Verify \(\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c\) for \(\vec a=\hat i+\hat j,\ \vec b=\hat j,\ \vec c=\hat k\).