Need steps for 3rd and 4th
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Cross Product:
As, vectors are in it's component form, we can use “cross product” directly, to multiply it's sides to obtain area of parallelogram.
AxB=(3i+2j+0k)x(2i+oj-2k).
AxB= -4i+6j-4k
Magnitude of AxB= √(-4^2+6^2+-4^2)=√68 =8.24unit square. (approx)
This, the area of the parallelogram is 8.24unit square.
As, vectors are in it's component form, we can use “cross product” directly, to multiply it's sides to obtain area of parallelogram.
AxB=(3i+2j+0k)x(2i+oj-2k).
AxB= -4i+6j-4k
Magnitude of AxB= √(-4^2+6^2+-4^2)=√68 =8.24unit square. (approx)
This, the area of the parallelogram is 8.24unit square.
NihaGiha:
But the answer is √224
so there must be some calculation mistake
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(iii) →A×B=A.B.sin@ (@=angle bw A&B=0)
=A.B.0=0
(iv)→area=cross product of adjacent sides
=A×B
=-2i+6j-4k
So it's magnitude=√(4+36+16)=√64
=8
=A.B.0=0
(iv)→area=cross product of adjacent sides
=A×B
=-2i+6j-4k
So it's magnitude=√(4+36+16)=√64
=8
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