FP1 5.2 - Scalar triple product and volumes

FP1 5.2 - Scalar triple product and volumes

Combine a vector product with a scalar product to measure volume (FP1 5.2). You will calculate the signed scalar triple product, distinguish it from geometric volume and find volumes of parallelepipeds and tetrahedra, including degenerate cases.

Three vectors produce a signed number

Could a solid have a negative volume? Its geometric volume cannot, but an ordered set of three edge vectors can have a negative scalar triple product. The sign records orientation; the magnitude measures the space occupied.

The scalar triple product is a.(b×c)\mathbf a.(\mathbf b\times\mathbf c): first form the vector product inside the brackets, then take its scalar product with the remaining vector. The final answer is a scalar.

You already know

The vector product is a directed normal whose magnitude is an area. A scalar product measures the component of one vector along another, multiplied by the other's magnitude. A three by three determinant can be positive, negative or zero.

Expanding the cross product and then taking the dot product gives exactly the determinant with the three vectors as rows, in the same order.

Scalar triple product

a.(b×c)=∣a1a2a3b1b2b3c1c2c3∣=b.(c×a)=c.(a×b)\mathbf a.(\mathbf b\times\mathbf c)= \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} =\mathbf b.(\mathbf c\times\mathbf a) =\mathbf c.(\mathbf a\times\mathbf b)

A cyclic move preserves the value. Interchanging any two vectors reverses its sign; this sign-change rule follows from the determinant or the vector product.

In the formula booklet.

The booklet prints the determinant and cyclic identities. The sign change on interchanging two vectors is their learnt consequence. Cyclic order is a three-step rotation, such as (a,b,c)(\mathbf a,\mathbf b,\mathbf c) to (b,c,a)(\mathbf b,\mathbf c,\mathbf a); it is different from swapping just two entries.

Worked example 1: calculate a scalar triple product

Question. For a=(1,2,0)T\mathbf a=(1,2,0)^{\mathrm T}, b=(0,1,2)T\mathbf b=(0,1,2)^{\mathrm T}, c=(2,0,1)T\mathbf c=(2,0,1)^{\mathrm T}, find a.(b×c)\mathbf a.(\mathbf b\times\mathbf c).

Plan. Calculate the bracketed vector first, then dot it with a\mathbf a.

b×c=(1(1)−2(0)2(2)−0(1)0(0)−1(2))=(1,4,−2)Tvector producta.(b×c)=1(1)+2(4)+0(−2)=9scalar product\begin{aligned} \mathbf b\times\mathbf c &=\begin{pmatrix}1(1)-2(0)\\2(2)-0(1)\\0(0)-1(2)\end{pmatrix} =(1,4,-2)^{\mathrm T} && \text{vector product}\\ \mathbf a.(\mathbf b\times\mathbf c) &=1(1)+2(4)+0(-2)=9 && \text{scalar product} \end{aligned}

Why this step? The brackets select the cross product first. The intermediate answer is a vector, while the final answer is one number.

Answer. a.(b×c)=9\mathbf a.(\mathbf b\times\mathbf c)=9.

Check. Expanding the row determinant gives 1(1−0)−2(0−4)=1+8=91(1-0)-2(0-4)=1+8=9.

Change only the order of b\mathbf b and c\mathbf c, keeping all three vectors unchanged.

Worked example 2: swapping two inputs

Question. For the same vectors, find a.(c×b)\mathbf a.(\mathbf c\times\mathbf b).

Plan. Reverse the cross-product order and retain the resulting sign when taking the scalar product.

c×b=(−1,−4,2)Tnegative of b×ca.(c×b)=1(−1)+2(−4)+0(2)=−9dot with the same a\begin{aligned} \mathbf c\times\mathbf b&=(-1,-4,2)^{\mathrm T} && \text{negative of }\mathbf b\times\mathbf c\\ \mathbf a.(\mathbf c\times\mathbf b) &=1(-1)+2(-4)+0(2)=-9 && \text{dot with the same }\mathbf a \end{aligned}

Why this step? A swap changes orientation; taking an absolute value here would discard information requested by the scalar triple product.

Answer. The scalar triple product is −9-9.

Check. Swapping the last two determinant rows changes its sign. A cyclic expression such as b.(c×a)\mathbf b.(\mathbf c\times\mathbf a) instead remains 99.

The sign belongs to the ordered vectors. When the question asks for a geometric volume, its final answer will use the magnitude.

Watch out

Wrong: “all reorderings give the same triple product”. Right: cyclic reorderings preserve it; swapping two vectors negates it. Keep brackets around the cross product so the two operations are clear.

Keep the vector order and change one component in the final vector.

Base area times perpendicular height

A parallelepiped is the three-dimensional counterpart of a parallelogram: three edge vectors from one vertex generate its eight vertices. Its volume is base area multiplied by perpendicular height, even when the side edges slope.

Take u,v\mathbf u,\mathbf v as adjacent base edges, with n=u×v≠0\mathbf n=\mathbf u\times\mathbf v\ne\mathbf0. The base area is ∣n∣|\mathbf n|. The third edge w\mathbf w has signed component w.n/∣n∣\mathbf w.\mathbf n/|\mathbf n| along the directed unit normal, so the perpendicular height is ∣w.n∣/∣n∣|\mathbf w.\mathbf n|/|\mathbf n|. Multiplying cancels the normal's magnitude.

Volume of a parallelepiped

For edge vectors u,v,w\mathbf u,\mathbf v,\mathbf w from the same vertex,

V=∣w.(u×v)∣V=\big|\mathbf w.(\mathbf u\times\mathbf v)\big|

If u×v≠0\mathbf u\times\mathbf v\ne\mathbf0, the perpendicular height above that base is

h=∣w.(u×v)∣∣u×v∣h=\frac{|\mathbf w.(\mathbf u\times\mathbf v)|}{|\mathbf u\times\mathbf v|}

The volume formula also gives zero for dependent edges, but the displayed height formula requires a non-degenerate base. The booklet supplies the scalar triple product, not these geometric volume and height formulae.

Not in the formula booklet: learn it.

Do not substitute the length of the sloping edge w\mathbf w for the perpendicular height. Only its component normal to the base contributes.

Worked example 3: an oblique solid

Question. A parallelepiped has adjacent edge vectors

u=(2,0,0)T,v=(1,3,0)T,w=(1,1,4)T\mathbf u=(2,0,0)^{\mathrm T},\quad \mathbf v=(1,3,0)^{\mathrm T},\quad \mathbf w=(1,1,4)^{\mathrm T}

Find its volume and the perpendicular height above the base spanned by u,v\mathbf u,\mathbf v.

Plan. Find the directed base-area vector, dot it with the third edge, then distinguish signed projection from height.

u×v=(0,0,6)Tbase-area vectorw.(u×v)=24signed triple productV=∣24∣=24geometric volumeh=246=4divide by base area\begin{aligned} \mathbf u\times\mathbf v&=(0,0,6)^{\mathrm T} && \text{base-area vector}\\ \mathbf w.(\mathbf u\times\mathbf v)&=24 && \text{signed triple product}\\ V&=|24|=24 && \text{geometric volume}\\ h&=\frac{24}{6}=4 && \text{divide by base area} \end{aligned}

Why this step? The third edge has length 18\sqrt{18}, but its perpendicular height above the horizontal base is 44, not 18\sqrt{18}.

Answer. The volume is 2424, and the height is 44.

Check. The base has area 2(3)=62(3)=6; multiplying by height 44 gives volume 2424.

Change only the final component of w\mathbf w from 44 to −4-4. The third edge now reaches the other side of the same base plane.

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