How do you prove by the vector method that the parallelograms on the same base, and between the same parallels, are equal in area .HERE I HAVE A DOUBT WHY WE TAKE AB× AD ,IF I CAN TAKE AB XBC IS THE ANSWER COME...
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Answer:
Step-by-step explanation:
Let AB
=a
andAD
=b
Area of parallelogram ABCD =a
×b
→1
(Cross product, Since Area= baseXheight)
=a
(b
sinθ)
Now in parallelogram ABB'A
AB=a
and A′D=ma
(let)
(∵ A'D is parallel to AB)
Consider triangle ADA'
By triangular law of vectors
⇒AA′=ma
+b
∴ Vector area of ABB'A' =a
×(ma
+b
)=(a
×ma
)+(a
×b
)
(∵a
×(ma
)=0, both are parallel, ∴sin0=0)
=0+(a
×b
)
=a
×b
→2
∵(1)=(2),
Hence proved
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