Question #99309
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
Expert's answer
2019-11-25T12:52:50-0500

Solution:a) Let PMNK - a parallelogram.


a=PK\overrightarrow {a}=\overrightarrow {PK}

PK(x;y2;z3)\overrightarrow {PK} (x; y-2; z-3)


x=3;y2=2;y=4;z3=3;z=0.K(3;4;0)x=3; y-2=2; y=4; z-3=-3; z=0. K (3;4;0)



b=PM\overrightarrow {b}= \overrightarrow {PM}


PM(x;y2;z3);\overrightarrow {PM} (x; y-2; z-3);




x=4;y2=1;y=3;z3=5;z=8.M(4;3;8)x=4; y-2=1; y=3; z-3=5; z=8. M(4;3;8)


a=MN\overrightarrow {a} = \overrightarrow {MN}



MN(x4;y3;z8);\overrightarrow {MN} (x-4; y-3; z-8);


x4=3;x=7;y3=2;y=5;z8=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)



PN(70;52;53);PN(7;3;2)\overrightarrow {PN} (7-0;5-2;5-3); \overrightarrow {PN} (7;3;2)


MK(34;43;08);MK(1;1;8)\overrightarrow {MK} (3-4;4-3;0-8); \overrightarrow {MK} (-1;1;-8)



c)


PN=49+9+4=62\vert \overrightarrow {PN} \vert= \sqrt {49+9+4} = \sqrt {62}

MK=1+1+64=66\vert \overrightarrow {MK} \vert = \sqrt {1+1+64} = \sqrt {66}




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