Volume of the parallelepiped having vertices at O≡(0,0,0),A≡(2,-2,1),B≡(5,-4,4) and C≡(1,-2,4) is
5 cu. units
10 cu. units
15 cu. units
20 cu. units
Explanation for the correct option
Given: O≡(0,0,0),A≡(2,-2,1),B≡(5,-4,4) and C≡(1,-2,4)
OA(0,0,0),(2,-2,1)=(2-0)i+(-2-0)j+(1-0)k=2i-2j+k
OB(0,0,0),(5,-4,4)=5-0i+(-4-0)j+(4-0)k=5i-4j+4k
OC(0,0,0),(1,-2,4)=(1-0)i+(-2-0)j+(4-0)k=i-2j+4k
Therefore volume of a parallelepiped=l×b×h
=OA×OB×OC=(2i-2j+k)×(5i-4j+4k)×(i-2j+4k)=2-215-441-24
We know that a1a2a3b1b2b3c1c2c3=a1(b2c3-b3c2)-a2(b1c3-b3c1)+a3(b1c2-b2c1)
⇒2-215-441-24=2-4×4-(4)×(-2)-(-2)5×4-4×1+(1)5×(-2)-(-4)×1=2(-16+8)+2(20-4)+(-10+4)=2(-8)+2(16)-6=-16+32-6=10cuunits
Hence, option B is correct.