Question #99172

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.

Expert's answer

Solution:a) Let PMNK - a parallelogram.

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

PK→(x;y−2;z−3)\overrightarrow {PK} (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)

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


PM→(x;y−2;z−3)\overrightarrow {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)


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


MN→(x−4;y−3;z−8)\overrightarrow {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)

b)


PN→(7−0;5−2;5−3)=PN→(7;3;2)\overrightarrow {PN} (7-0;5-2;5-3)=\overrightarrow {PN} (7;3;2)


MK→(3−4;4−3;0−8)=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}


LATEST TUTORIALS
APPROVED BY CLIENTS