Area of parallelogram determined by two vectors Example-2

1. Find AreaParallelogram(A,B)
A=(5,6,1) $A=(5,6,1)$ ,B=(0,2,3) $B=(0,2,3)$

Solution:
Here

→A=(5,6,1),→B=(0,2,3),→C=(3,4,5) $vec A=(5,6,1),vec B=(0,2,3),vec C=(3,4,5)$

Area

=|→A×→B| $=|vec A xx vec B|$

1. Calculate cross product

→A×→B $vec A xx vec B$

=|ijkA1A2A3B1B2B3| $=|[i,j,k],[A_1,A_2,A_3],[B_1,B_2,B_3]|$

=|ijk561023| $=|[i,j,k],[5,6,1],[0,2,3]|$

=i(6×3-1×2)-j(5×3-1×0)+k(5×2-6×0) $=i(6xx3-1xx2)-j(5xx3-1xx0)+k(5xx2-6xx0)$

=i(18-2)-j(15-0)+k(10-0) $=i(18-2)-j(15-0)+k(10-0)$

=16i-15j+10k $=16i-15j+10k$

=(16,-15,10) $=(16,-15,10)$

2. Calculate magnitude

|→→A×→B| $|vec vec A xx vec B|$

=√162+(-15)2+102 $=sqrt(16^2 + (-15)^2 + 10^2)$

=√256+225+100 $=sqrt(256 + 225 + 100)$

=√581 $=sqrt(581)$

=24.1039 $=24.1039$

3. Calculate triangle area
Area

=24.1039 $= 24.1039$

2. Find AREAPARALLELOGRAM(B,C)
A=(5,6,1) $A=(5,6,1)$ ,B=(0,2,3) $B=(0,2,3)$ ,C=(3,4,5) $C=(3,4,5)$

Solution:
Here

→A=(5,6,1),→B=(0,2,3),→C=(3,4,5) $vec A=(5,6,1),vec B=(0,2,3),vec C=(3,4,5)$

Area

=|→B×→C| $=|vec B xx vec C|$

1. Calculate cross product

→B×→C $vec B xx vec C$

=|ijkB1B2B3C1C2C3| $=|[i,j,k],[B_1,B_2,B_3],[C_1,C_2,C_3]|$

=|ijk023345| $=|[i,j,k],[0,2,3],[3,4,5]|$

=i(2×5-3×4)-j(0×5-3×3)+k(0×4-2×3) $=i(2xx5-3xx4)-j(0xx5-3xx3)+k(0xx4-2xx3)$

=i(10-12)-j(0-9)+k(0-6) $=i(10-12)-j(0-9)+k(0-6)$

=-2i+9j-6k $=-2i+9j-6k$

=(-2,9,-6) $=(-2,9,-6)$

2. Calculate magnitude

|→→B×→C| $|vec vec B xx vec C|$

=√(-2)2+92+(-6)2 $=sqrt((-2)^2 + 9^2 + (-6)^2)$

=√4+81+36 $=sqrt(4 + 81 + 36)$

=√121 $=sqrt(121)$

=11 $=11$

3. Calculate triangle area
Area

=11 $= 11$

This material is intended as a summary. Use your textbook for detail explanation.
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