# 8: Let P = (1, 2, 3, 4), Q = (2,0,1,3), R-(4,0,-1,2) be three points in R. The
ID: 3116312 • Letter: #
Question
# 8: Let P = (1, 2, 3, 4), Q = (2,0,1,3), R-(4,0,-1,2) be three points in R. They form a triangle. a) Find the lengths of the sides PQ, PR, QR, that is, |lPQlI, 1lPRI, QRII b) Find the angle Op at the corner P between the sides PQ, PR. c) Compute the quantity How does it compare to the quantity IQRI? Hint: For vectors-(a, , a2 , . . . , an), w = (bi, by, . . . , Un) we may form the dot product The length of the vector u is just Ilt1-… and for u, u, non-zero the angle between u and u, is given by cos@) =Explanation / Answer
a).The vector PR is u1=(3,-2,-4,-2)T, the vector PQ is u2=(1,-2,-2,-1)T and the vector QR is u3=(2,0,-2,-1)T. Then ||PQ|| = ||u2||=[ 12+ (-2)2+(-2)2 +(-1)2] = 10, ||PR|| = ||u1||=|[ 32+ (-2)2+(-4)2 +(-2)2] = 33, and ||QR|| =||u3|| =[ 22+ (0)2+(-2)2 +(-1)2] = 9 =3.
b). cos p = u2.u1/||u2||||u1|| = (3+4+8+2)/( 10)( 33) = 17/330 so that p = cos-1(17/330) = cos-1 (0.9358192) = 20.640 (approximately).
c). ||PQ||2 +||PR||2 -2cos(p ) ||PQ||||PR||= 10+33 –(2*17/330)( 10)( 33) = 43-34 = 9 =||QR||2 .
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