Tutorials › Calculus III › Orthogonal Vectors in Coordinate Geometry

Dot Products and Projections · Tutorial 57 of 999

Orthogonal Vectors in Coordinate Geometry

Learn to translate perpendicularity into coordinate equations and solve for unknown components or geometric dimensions.

Beginner 9 min read

What You'll Learn

  • Test perpendicularity by setting a dot product equal to zero.
  • Form displacement vectors from point coordinates before applying the test.
  • Solve for unknown vector components and check the result directly.
  • Relate the perpendicular diagonals of a rectangle to its side lengths.
  • Find a box dimension that makes two body diagonals perpendicular.
  • Avoid sign errors and assumptions based only on how lines look in a diagram.

Perpendicularity Becomes an Equation

In coordinate geometry, a picture can suggest that two directions meet at a right angle, but a calculation establishes whether they really do. The dot product turns perpendicularity into an equation involving vector components. When one or more components are unknown, that equation can determine their values.

This method applies both to vectors given directly by their components and to vectors built from points, rectangle sides, or box diagonals. The key is to identify the direction vectors correctly before forming their dot product.

Criterion: Two nonzero vectors \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular, or orthogonal, exactly when \(\mathbf{u}\cdot\mathbf{v}=0\). In coordinates, if \(\mathbf{u}=(u_1,u_2,u_3)\) and \(\mathbf{v}=(v_1,v_2,v_3)\), this means \(u_1v_1+u_2v_2+u_3v_3=0\).

The zero vector also has dot product zero with every vector, but it does not have a direction and is not called perpendicular to a vector in the geometric sense. In the examples below, the direction vectors are nonzero.

From Points to Direction Vectors

When a problem gives points rather than vectors, first subtract the coordinates of the starting point from those of the ending point. For points \(A=(a_1,a_2,a_3)\) and \(B=(b_1,b_2,b_3)\), the displacement from \(A\) to \(B\) is \(\overrightarrow{AB}=B-A=(b_1-a_1,b_2-a_2,b_3-a_3)\). The dot-product test applies to these displacements, not to the point coordinates treated as if they were directions.

For two vectors with unknown components, take their dot product component by component, set the result equal to zero, and solve. If the unknown is a length, also enforce the geometric condition that the length must be positive. A solution to the algebraic equation is not automatically a valid geometric dimension.

Worked Example: Find a Point Coordinate for Perpendicular Displacements

Let \(A=(1,2,-1)\), \(B=(4,0,2)\), and \(C=(x,1,3)\). Find \(x\) so that \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are perpendicular.

First calculate both displacements:

$$ \overrightarrow{AB}=B-A=(4-1,0-2,2-(-1))=(3,-2,3), $$
$$ \overrightarrow{AC}=C-A=(x-1,1-2,3-(-1))=(x-1,-1,4). $$

Set their dot product to zero and show each component product:

$$ \overrightarrow{AB}\cdot\overrightarrow{AC} =3(x-1)+(-2)(-1)+3(4) =3x-3+2+12 =3x+11=0. $$

Thus \(3x=-11\), so \(x=-\frac{11}{3}\). To check, \(x-1=-\frac{14}{3}\), and therefore

$$ (3,-2,3)\cdot\left(-\frac{14}{3},-1,4\right) =3\left(-\frac{14}{3}\right)+(-2)(-1)+3(4) =-14+2+12=0. $$

The resulting displacements are nonzero and their dot product is zero, as required.

Why Setting the Dot Product to Zero Works

The coordinate method follows directly from the dot-product criterion. For vectors with an unknown parameter \(t\), write each component product explicitly. The condition becomes an equation in \(t\); solving it identifies precisely the parameter values for which the dot product vanishes. By the criterion, those and only those values make the vectors perpendicular, provided both vectors have direction.

This also explains why checking only one component, or just noticing that the vectors have a component with opposite signs, is not enough. Every component product contributes to the total. Positive and negative contributions can cancel, and it is that complete sum that must equal zero.

Rectangle Sides and Diagonals

Let \(\mathbf{u}\) and \(\mathbf{v}\) be the two side vectors meeting at a corner of a rectangle. They are perpendicular by the definition of the rectangle, so \(\mathbf{u}\cdot\mathbf{v}=0\). The two diagonal directions can be represented by \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). For example, in a rectangle with vertices \(O\), \(O+\mathbf{u}\), \(O+\mathbf{v}\), and \(O+\mathbf{u}+\mathbf{v}\), these directions can be taken along its two diagonals.

Their dot product shows exactly when the diagonals are perpendicular. Here \(\|\mathbf{u}\|^2=\mathbf{u}\cdot\mathbf{u}\) denotes the squared length of \(\mathbf{u}\), and similarly for \(\mathbf{v}\).

$$ \begin{aligned} (\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}-\mathbf{v}) &=\mathbf{u}\cdot\mathbf{u}-\mathbf{u}\cdot\mathbf{v} +\mathbf{v}\cdot\mathbf{u}-\mathbf{v}\cdot\mathbf{v}\\ &=\|\mathbf{u}\|^2-\|\mathbf{v}\|^2. \end{aligned} $$

The middle terms cancel because the dot product is symmetric and the rectangle’s side vectors have dot product zero. Consequently, the diagonals are perpendicular exactly when the two side lengths are equal. A rectangle has perpendicular diagonals only if it is a square. This conclusion is about a right angle between the diagonal directions; the fact that the diagonals cross at their midpoint does not by itself make that crossing a right angle.

Worked Example: Find the Rectangle’s Width from Its Diagonals

A rectangle has side vectors \(\mathbf{u}=(5,0)\) and \(\mathbf{v}=(0,t)\), where \(t\) is its positive height. Find \(t\) if its diagonals are perpendicular.

Use the diagonal directions \(\mathbf{u}+\mathbf{v}=(5,t)\) and \(\mathbf{u}-\mathbf{v}=(5,-t)\). Their dot product must be zero:

$$ (5,t)\cdot(5,-t)=5(5)+t(-t)=25-t^2=0. $$

Thus \(t^2=25\), which gives \(t=5\) or \(t=-5\) algebraically. Since \(t\) is a positive side length, \(t=5\). Check the diagonal directions \((5,5)\) and \((5,-5)\):

$$ (5,5)\cdot(5,-5)=5(5)+5(-5)=25-25=0. $$

So the diagonals are perpendicular when the height is \(5\), equal to the other side length.

Body Diagonals in a Rectangular Box

A rectangular box has three mutually perpendicular edge directions. If its edge lengths are \(a\), \(b\), and \(c\), choose coordinate axes along those edges. Its body-diagonal vectors then have components whose entries are signed versions of those lengths. The signs depend on the direction of each diagonal: a component is negative when that diagonal points opposite to the chosen positive axis.

For two such diagonals, calculate the dot product using those signs. The result is a sum of signed products of edge lengths. Setting it to zero gives the condition for perpendicular diagonals. Do not replace signed components with lengths alone: losing a sign can change the equation and therefore the answer.

Worked Example: Find a Box Dimension for Perpendicular Body Diagonals

A rectangular box has edge lengths \(2\), \(3\), and \(h\), with \(h>0\). Two body-diagonal directions are \(\mathbf{d}_1=(2,3,h)\) and \(\mathbf{d}_2=(2,-3,h)\). Find \(h\) so that these diagonals are perpendicular.

Their dot product is

$$ \mathbf{d}_1\cdot\mathbf{d}_2 =2(2)+3(-3)+h(h) =4-9+h^2 =h^2-5. $$

Perpendicularity requires \(h^2-5=0\), so \(h^2=5\). The algebraic roots are \(h=\sqrt{5}\) and \(h=-\sqrt{5}\), but only the positive value is a valid edge length. Hence \(h=\sqrt{5}\). Substituting this value back gives

$$ (2,3,\sqrt{5})\cdot(2,-3,\sqrt{5}) =4-9+5=0. $$

The diagonal directions are therefore perpendicular for a box whose third edge length is \(\sqrt{5}\).

A Common Mistake: Assuming Crossing Means Perpendicular

Diagonals of a rectangle cross, but that does not mean they meet at a right angle. For a rectangle with side vectors \((5,0)\) and \((0,3)\), take its two diagonal directions to be \((5,3)\) and \((5,-3)\). A mistaken conclusion might be: “They intersect, so they are perpendicular.” The direct test contradicts it:

$$ (5,3)\cdot(5,-3)=5(5)+3(-3)=25-9=16\ne 0. $$

The correction is to distinguish “the lines intersect” from “the lines meet at a right angle.” Their intersection says nothing by itself about the angle. Only a zero dot product of their direction vectors establishes perpendicularity. In this example the diagonals are not perpendicular, consistent with the unequal side lengths \(5\) and \(3\).

A Reliable Calculation Process

For coordinate-geometry problems of this kind, use the following sequence. It keeps the geometric interpretation and the algebra connected.

1
Identify the directions.
Write the vectors to be tested. If they come from points, form each displacement by subtracting starting coordinates from ending coordinates.
2
Keep the signs.
Record negative components when a vector points opposite to a coordinate direction; do not replace components by lengths.
3
Set up the condition.
Multiply corresponding components and add them. Set the entire dot product equal to zero.
4
Solve and check.
Solve for the unknown, retain only values allowed by the geometry, and substitute the result back into the dot product.

Why This Method Is Useful

Perpendicularity questions often appear in coordinate geometry as a mixture of algebra and shape information. The dot-product equation translates the right-angle condition into a manageable calculation without requiring a diagram to be drawn to scale. It can determine a missing point coordinate, a rectangle’s side length, or a box dimension. The final substitution is important: it confirms both that the algebra was solved correctly and that the proposed value actually satisfies the geometric condition.

Check Your Understanding

For each question, identify the relevant direction vectors before applying the perpendicularity criterion.

  1. Find \(x\) so that \((x,2,-1)\) and \((3,-4,5)\) are perpendicular. Check your answer by substitution.
  2. A rectangle has side vectors \((6,0)\) and \((0,t)\), with \(t>0\). What value of \(t\) makes its diagonals perpendicular?
  3. For the points \(P=(0,1,2)\), \(Q=(2,3,1)\), and \(R=(r,0,4)\), write the dot-product equation that requires \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) to be perpendicular.
  4. Why must a negative algebraic solution for a box edge length be rejected?
  5. Do intersecting rectangle diagonals have to be perpendicular? State the coordinate test that decides.