2019-11-23T13:50:57-05:00
A parallelogram is formed in R3 by the vectors = (3, 2, –3) and = (4, 1, 5). The point P = (0, 2, 3). a. Determine the location of the vertices. b. Determine the vectors representing the diagonals. c. Determine the length of the diagonals.
1
2019-11-25T12:52:50-0500
Solution:a) Let PMNK - a parallelogram.
a → = P K → \overrightarrow {a}=\overrightarrow {PK} a = P K
P K → ( x ; y − 2 ; z − 3 ) \overrightarrow {PK} (x; y-2; z-3) P K ( x ; y − 2 ; z − 3 )
x = 3 ; y − 2 = 2 ; y = 4 ; z − 3 = − 3 ; z = 0. K ( 3 ; 4 ; 0 ) x=3; y-2=2; y=4; z-3=-3; z=0. K (3;4;0) x = 3 ; y − 2 = 2 ; y = 4 ; z − 3 = − 3 ; z = 0. K ( 3 ; 4 ; 0 )
b → = P M → \overrightarrow {b}= \overrightarrow {PM} b = PM
P M → ( x ; y − 2 ; z − 3 ) ; \overrightarrow {PM} (x; y-2; z-3); PM ( x ; y − 2 ; z − 3 ) ;
x = 4 ; y − 2 = 1 ; y = 3 ; z − 3 = 5 ; z = 8. M ( 4 ; 3 ; 8 ) x=4; y-2=1; y=3; z-3=5; z=8. M(4;3;8) x = 4 ; y − 2 = 1 ; y = 3 ; z − 3 = 5 ; z = 8. M ( 4 ; 3 ; 8 )
a → = M N → \overrightarrow {a} = \overrightarrow {MN} a = MN
M N → ( x − 4 ; y − 3 ; z − 8 ) ; \overrightarrow {MN} (x-4; y-3; z-8); MN ( x − 4 ; y − 3 ; z − 8 ) ;
x − 4 = 3 ; x = 7 ; y − 3 = 2 ; y = 5 ; z − 8 = − 3 ; z = 5. N ( 7 ; 5 ; 5 ) x-4=3; x=7; y-3=2; y=5; z-8=-3; z=5. N(7;5;5) x − 4 = 3 ; x = 7 ; y − 3 = 2 ; y = 5 ; z − 8 = − 3 ; z = 5. N ( 7 ; 5 ; 5 )
b)
P N → ( 7 − 0 ; 5 − 2 ; 5 − 3 ) ; P N → ( 7 ; 3 ; 2 ) \overrightarrow {PN} (7-0;5-2;5-3); \overrightarrow {PN} (7;3;2) PN ( 7 − 0 ; 5 − 2 ; 5 − 3 ) ; PN ( 7 ; 3 ; 2 )
M K → ( 3 − 4 ; 4 − 3 ; 0 − 8 ) ; M K → ( − 1 ; 1 ; − 8 ) \overrightarrow {MK} (3-4;4-3;0-8); \overrightarrow {MK} (-1;1;-8) M K ( 3 − 4 ; 4 − 3 ; 0 − 8 ) ; M K ( − 1 ; 1 ; − 8 )
c)
∣ P N → ∣ = 49 + 9 + 4 = 62 \vert \overrightarrow {PN} \vert= \sqrt {49+9+4} = \sqrt {62} ∣ PN ∣ = 49 + 9 + 4 = 62
∣ M K → ∣ = 1 + 1 + 64 = 66 \vert \overrightarrow {MK} \vert = \sqrt {1+1+64} = \sqrt {66} ∣ M K ∣ = 1 + 1 + 64 = 66
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