The question: Consider the triangle delta OPQ as shown below. Let p = Op rightar

Question

The question:

Consider the triangle delta OPQ as shown below. Let p = Op rightarrow and q = OQ rightarrow . Write down an expression in terms of p and q for a vector which points directly out of the page and has magnitude equal to the area of delta OPQ. (Hint: consider a cross product, and be careful about using the right hand rule.) Consider the tetrahedron OPQR as shown below. Let p = Op rightarrow, q = OQ rightarrow , and r = OR rightarrow . Write down expressions for the vectors PQ rightarrow , PR rightarrow , QR rightarrow in terms of p, q, r. Let a be a vector with length equal to the area of delta OPQ which points out perpendicularly from the face OPQ. Let vectors b, c, d be defined analogously with respect to the faces OPR, ORQ, PQR (in that order). Write down expressions for a. b, c, d in terms of p. q. r. Then show that a + b + c + d = 0. Now suppose that POQ = POR = ROQ = 90 degree . Let A,B,C,D denote the areas of the faces OPQ. OPR, ORQ, PQR (in that order). Show that A2 + B2 + C2 = D2. This is another kind of 3D version of Pythagoras' Theorem. (Hint: One way is to consider the dot product of a + b + c + d with itself).

Explanation / Answer

a) 1/2*(pxq) (since 1/2pqsin(O) gives area)

b) i)PQ = q-p

PR = r-p

QR = r-q

ii)a = 1/2(pxq) b =1/2( rxp) c = 1/2(qxr) d = 1/2(QPxQR) = 1/2((p-q)x(r-q)) = 1/2(pxr-pxq-qxr)

clearly a+b+c+d = 0 vector (rxp = -pxr)

iii)Ia+b+c+dI^2 =0

(a+b+c+d).(a+b+c+d)=0

hence a^2+b^2+c^2+d^2+2(a.b+a.c+a.d+b.c+b.d+c.d)=0

Verify that a.b, a.c ,b.c = 0

a^2+b^2+c^2+d^2+2(0+0+a.d+0+b.d+c.d)

so d.(a+b+c) = d.(-d) = -d^2

hence a^2+b^2+c^2+d^2+2(-d^2)=0

so a^2+b^2+c^2=d^2

now a^2 = A^2 and others too

A^2+B^2+C^2=D^2

for a.b,b.c,c.a try to visualize where they would lie in 3D space with the angles as described; it's very easy to see they would be perpendicular; moreover a.a = IaI^2 & A=Ia

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