Construct oriented cross products, normals and exact areas

Build on the spatial-vector foundation

7A established three-dimensional components, scalar products, independent directions and resolutes. Cross products add an oriented normal. Keep that vector operation separate from the scalar dot product; both use the same orthonormal frame

Use three orthogonal unit directions

In an orthonormal Cartesian frame, the three basis vectors point along the coordinate axes. The coordinates of a vector are signed scalar components; each scalar times its basis vector is a vector component. The origin changes position vectors, while displacements between fixed points stay unchanged

Choose a right-handed orientation

For a right-handed orthonormal basis, i×j=k, j×k=i and k×i=j. Reversing either order gives the negative vector; crossing a basis vector with itself gives zero. For nonparallel vectors, the right-hand rule chooses the direction of a×b perpendicular to both, with magnitude |a||b|sinθ

In a right-handed basis, what is j×i?

Use the component determinant carefully

Expand the formal determinant along its basis-vector row. The middle term has a minus sign. This is a compact way to organise the three signed 2×2 minors; the result is a vector, not a scalar determinant

Work a cross product with all three components

For a=(1,2,0), b=(0,1,2), the cross product is (4,-2,1). Check a·(a×b)=4-4=0 and b·(a×b)=-2+2=0. The middle component is 0(0)-1(2)=-2.

For a=(1,2,0) and b=(0,1,2), find the middle component of a×b.

Move scalars outside and preserve order

The cross product is bilinear: (λa)×(μb)=λμ(a×b), and a×(b+c)=a×b+a×c. It is anticommutative: b×a=−a×b. Thus (-2b)×(3a)=6(a×b). A zero scalar or a repeated vector gives zero

If a×b=(4,-2,1), find the first component of (-2b)×(3a)

A cross product is not generally a unit normal

For a=(1,2,0), b=(0,1,2), n=(4,-2,1) has magnitude √21. The unit normal in that orientation is (4,-2,1)/√21. The opposite unit normal is its negative. A question asking for the ordered cross product does not permit arbitrary rescaling

Find the squared magnitude of n=(4,-2,1)

For n=(4,-2,1), find the coefficient of √21 in the first component of n/|n|

Magnitude gives area

For adjacent side vectors a,b, parallelogram area is |a×b| and triangle area is |a×b|/2. The norm is nonnegative; reversing order changes orientation but not area. The side angle lies in [0,π], so the sine in the area formula is nonnegative

What is the area of one triangle cut by this diagonal?

For a=(1,2,0), b=(0,1,2), enter the coefficient of √21 in the triangle area

Zero area and undefined normalisation are different conclusions

If a×b=0, the parallelogram area is zero. For nonzero a,b, the vectors are parallel or antiparallel. If either vector is zero, it has no direction, although the algebraic product is defined. A zero cross product cannot be divided by its norm to create a unit normal

Use two edges with a common starting point

For A=(1,-1,2), B=(3,0,2), C=(0,1,3), form AB=(2,1,0) and AC=(-1,2,1). Their ordered product is (1,-2,5). It is nonzero, so the points are noncollinear. The normal is perpendicular to both edges

For A=(1,-1,2), B=(3,0,2) and C=(0,1,3), find the third component of AB×AC

Point order determines normal orientation

Changing AB×AC to AC×AB negates the normal. A simultaneous translation of all three points leaves the edge vectors and their product unchanged. If two points coincide or the three are collinear, a unique plane is not determined. Plane equations are developed in the next section

Solve a parallel-vector parameter condition

For a=(1,2,-1), b=(3,t,-3), parallelism forces the multiplier to be 3 from the first coordinate, hence t=6. The cross product is (t-6,0,t-6), which vanishes at the same value. Check every coordinate rather than dividing by a possibly zero one

For a=(1,2,-1) and b=(3,t,-3), find t so that the vectors are parallel

Area conditions can leave two orientations

For a=(1,2,0), b=(1,2+t,3), the product is (6,-3,t), so squared area is 45+t². If the parallelogram area is 7, then t²=4 and t=−2 or 2. Both have the same area, but opposite signs for the third normal component

For a=(1,2,0) and b=(1,2+t,3), the parallelogram area is 7. Give the smaller value of t.

Do not assume associativity

For a=b=i,c=j, i×(i×j)=i×k=−j, but (i×i)×j=0. Therefore parentheses cannot be moved freely. A dot product is also a different operation: a·b is a scalar, while a×b is a vector

Derive the cross-product rules and area identity, ungraded

Starting with the right-handed basis products and bilinearity, derive the three component formulae. Prove that the result is perpendicular to both inputs and establish |a×b|²=|a|²|b|²-(a·b)². Deduce the parallelogram and triangle area formulas, including degenerate cases

Model: expand, cancel and use the included angle

Expand by bilinearity to obtain the three components below. In each dot product, terms cancel in pairs. Square and add the components, then collect terms to obtain the norm identity. For nonzero vectors, use the included-angle formula and take the nonnegative root. This gives base times perpendicular height. A zero or dependent side gives zero area; a triangle has half the parallelogram area

Prove the rhombus diagonal theorem, ungraded

Let adjacent edge vectors of a nondegenerate rhombus be u and v. Prove that its diagonals are perpendicular. State where the equal-side and nondegeneracy assumptions enter, and distinguish this from claiming that every parallelogram has perpendicular diagonals

Model: take the dot product of the diagonal vectors

The diagonal directions are u+v and u-v. Their dot product is |u|²-|v|² because the mixed terms cancel. In a rhombus the adjacent side lengths agree, so this is zero. Nondegeneracy makes both diagonal vectors nonzero; therefore the zero dot product gives a right angle. A general parallelogram need not have equal adjacent lengths, so the conclusion does not follow for every parallelogram

Prove the medians of a triangle are concurrent, ungraded

For noncollinear vertices with position vectors a,b,c, find a single point on all three medians and prove the vertex-to-midpoint ratio in which it divides each. Do not infer concurrency from an approximate drawing

Model: the common median point

Set G at the average of the three vertex position vectors. Each midpoint calculation below places G two-thirds of the way from its vertex to the opposite side, so all three medians contain G and each vertex-to-midpoint ratio is 2:1. Noncollinearity ensures these median segments are nonzero

Prove the angle in a semicircle theorem, ungraded

Let AB be a diameter of a circle and P another point on the circle, distinct from A and B. Use vectors to prove APB is a right angle. Explain why the endpoint exclusions matter

Model: choose the centre as origin

Write A=-u, B=u and P=p, with |p|=|u|>0. The two vectors from P are -u-p and u-p. Their dot product is |p|²-|u|²=0, since the mixed terms cancel. The condition P≠A,B makes both vectors nonzero, so the angle at P is 90°. A zero vector at an endpoint would not define that angle

Construct and verify an oriented normal diagram, ungraded

For A=(1,-1,2), B=(3,0,2), C=(0,1,3), calculate AB×AC, both corresponding unit normals and the triangle area. Draw and label an appropriate spatial diagram, state its viewing limitations and explain the effect of swapping B and C.

Model: check components, orientation and magnitude separately

AB=(2,1,0), AC=(-1,2,1) and AB×AC=(1,-2,5). Its dot product with either edge is zero. Its norm is √30, so the two unit normals are ±(1,-2,5)/√30 and the triangle area is √30/2. Swapping B and C reverses the ordered normal but preserves the area and the two-element set of unit normals. A flat view need not preserve all spatial angles; the algebra supplies the checks. The learner diagram is ungraded

Assessment boundary

Practice checks stated cross-product components, oriented normals, exact areas and bounded validity conclusions. The complete rhombus, median and semicircle arguments are written work with models; a correct numeric component or selected statement does not establish independent geometric proof