Geometrical Interpretation of Vectors

Geometrical Interpretation of Vectors

Junior College 1

Translate the expression into a length, area, direction, line, plane or locus. Name the actual points and lines in your answer. Let A, B and P have position vectors \(\mathbf a,\mathbf b,\mathbf p\). Direction and normal vectors are non-zero; points defining a line are distinct.

Lengths and position vectors

Expression or conditionGeometrical interpretation
\(\mathbf b-\mathbf a\)Displacement vector \(\overrightarrow{AB}\).
\(|\mathbf b-\mathbf a|\)Distance AB.
\(\dfrac{\mathbf a+\mathbf b}{2}\)Position vector of the midpoint of AB.
\(\dfrac{\mathbf b-\mathbf a}{|\mathbf b-\mathbf a|}\)Unit vector pointing from A towards B.

Scalar products and projections

Let \(\widehat{\mathbf b}=\dfrac{\mathbf b}{|\mathbf b|}\). The denominator identifies the direction onto which the other vector is projected.

Expression or conditionGeometrical interpretation
\(\mathbf a\cdot\widehat{\mathbf b}\)Signed scalar projection of OA in the direction of OB.
\(|\mathbf a\cdot\widehat{\mathbf b}|=\dfrac{|\mathbf a\cdot\mathbf b|}{|\mathbf b|}\)Length of the projection of OA onto line OB.
\((\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}\)Vector projection of \(\mathbf a\) onto \(\mathbf b\).
\(\mathbf a\cdot\mathbf b=0\)The non-zero vectors are perpendicular.
\((\mathbf p-\mathbf a)\cdot(\mathbf b-\mathbf a)=0\)AP is perpendicular to AB, provided P differs from A.

Exam check: \(|\mathbf a\cdot\mathbf b|\) is not generally a projection length. If \(|\mathbf a|=1\), it is the projection length of OB onto OA. If \(|\mathbf b|=1\), it is the projection length of OA onto OB. Absolute value removes the sign; a length cannot be negative.

Vector products, areas and perpendicular distances

Expression or conditionGeometrical interpretation
\(\mathbf a\times\mathbf b\)For non-parallel vectors: a vector perpendicular to both, oriented by the right-hand rule.
\(|\mathbf a\times\mathbf b|\)Area of the parallelogram with adjacent sides OA and OB; twice the area of triangle OAB.
\(\dfrac12|(\mathbf b-\mathbf a)\times(\mathbf p-\mathbf a)|\)Area of triangle ABP.
\(|\mathbf a\times\widehat{\mathbf b}|=\dfrac{|\mathbf a\times\mathbf b|}{|\mathbf b|}\)Perpendicular distance from A to line OB.
\(\dfrac{|(\mathbf p-\mathbf a)\times\mathbf d|}{|\mathbf d|}\)Perpendicular distance from P to the line through A parallel to \(\mathbf d\).
\(\mathbf a\times\mathbf b=\mathbf0\)The non-zero vectors are parallel (same or opposite directions).

Why the distance formula works: \(|\mathbf a\times\mathbf b|=|\mathbf b|(|\mathbf a|\sin\theta)\) is base times perpendicular height. Divide by the base length to obtain the height. The foot may lie on the extended line, outside the segment.

Recognising line equations

Expression or conditionGeometrical interpretation
\(\mathbf r=\mathbf a+t\mathbf d,\quad t\in\mathbb R\)Line through A parallel to \(\mathbf d\).
\(\mathbf r=(1-t)\mathbf a+t\mathbf b\)Whole line AB for \(t\in\mathbb R\); segment AB including endpoints for \(0\le t\le1\).
\((\mathbf r-\mathbf a)\times\mathbf d=\mathbf0\)Line through A parallel to \(\mathbf d\).
\(\mathbf r\times\mathbf d=\mathbf a\times\mathbf d\)Rearrange to \((\mathbf r-\mathbf a)\times\mathbf d=\mathbf0\): the same line.
\(\mathbf r\times(\mathbf b-\mathbf a)=\mathbf0\)Line through O parallel to AB; it need not be line AB.
\((\mathbf r-\mathbf a)\times(\mathbf r-\mathbf b)=\mathbf0\)Line AB, including A and B.

Recognising planes and their distances

Expression or conditionGeometrical interpretation
\((\mathbf r-\mathbf a)\cdot\mathbf n=0\)Plane through A with normal vector \(\mathbf n\).
\(\mathbf r\cdot\mathbf n=d\)Plane normal to \(\mathbf n\), at distance \(\dfrac{|d|}{|\mathbf n|}\) from O.
\(\mathbf r\cdot\widehat{\mathbf n}=d\)With a unit normal, d is the signed offset; the distance from O is \(|d|\).
\(\mathbf r=\mathbf a+s\mathbf u+t\mathbf v\)Plane through A parallel to non-parallel directions \(\mathbf u,\mathbf v\), for real s and t.
\(\dfrac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}\)Perpendicular distance from P to the plane \(\mathbf r\cdot\mathbf n=d\).

Interpreting simultaneous solutions

Expression or conditionGeometrical interpretation
Two lines: one common pointIntersecting lines.
Two lines: no common pointsDistinct parallel lines or skew lines.
Two lines: infinitely many common pointsCoincident lines.
Line and plane: one common pointThe line meets the plane at one point.
Line and plane: no common pointsThe line is parallel to and outside the plane.
Line and plane: infinitely many common pointsThe whole line lies in the plane.
Two planesDistinct parallel planes have no common points; otherwise they intersect in a line or coincide. They cannot have exactly one common point.

Loci and useful deductions

Expression or conditionGeometrical interpretation
\(|\mathbf r-\mathbf a|=|\mathbf r-\mathbf b|\)In 3D: the perpendicular bisector plane of AB.
\(|\mathbf r-\mathbf a|=k,\quad k>0\)Sphere with centre A and radius k in 3D. In a plane through A: a circle.
\((\mathbf r-\mathbf a)\cdot(\mathbf r-\mathbf b)=0\)Sphere with diameter AB in 3D; in a specified plane containing AB: the circle with diameter AB.
\((\mathbf a+\mathbf b)\cdot(\mathbf a-\mathbf b)=0\)The sum and difference vectors are perpendicular; these are the diagonal directions of the parallelogram with adjacent sides \(\mathbf a,\mathbf b\).
\(\dfrac{\mathbf a}{|\mathbf a|}+\dfrac{\mathbf b}{|\mathbf b|}\)Direction of the internal angle bisector of angle AOB, for non-collinear rays OA and OB.

Exam check: apply every restriction on the moving point. A sphere intersected by a plane may give a circle, one point or no points. A line equation with a restricted parameter may describe only a segment or ray.

Answering precisely

Identify the quantity or object; check for a unit vector; rewrite position-vector differences as named displacements; name the actual line, plane, triangle or projection; include parameter restrictions. For example, \(|\mathbf b\times\mathbf c|\) is the area of the parallelogram with adjacent sides OB and OC, or twice the area of triangle OBC.

Similar questions are unavailable for this question.

Need help? Join our JC Math tuition classes.

Learn more