Given :

→

a

×

→

b

=

→

c

×

→

d

and

→

a

×

→

c

=

→

b

×

→

d

To show

→

a

−

→

d

is parallel to

→

b

−

→

c

i.e (

→

a

−

→

d

) × (

→

b

−

→

c

) = 0
Consider (

→

a

−

→

d

) × (

→

b

−

→

c

) =

→

a

×(

→

b

−

→

c

)−

→

d

×(

→

b

−

→

c

)
=

→

a

×

→

b

-

→

a

×

→

c

-

→

d

×

→

b

+

→

d

×

→

c

→

c

×

→

d

-

→

b

×

→

d

-

→

d

×

→

b

+

→

d

×

→

c

[Since

→

a

×

→

b

=

→

c

×

→

d

and

→

a

×

→

c

=

→

b

×

→

d

]

→

c

×

→

d

-

→

b

×

→

d

+

→

b

×

→

d

-

→

c

×

→

d

[Since

→

d

×

→

c

= −

→

c

×

→

d

and

→

d

×

→

b

= -

→

b

×

→

d

]
= 0
Therefore

→

a

−

→

d

is parallel to

→

b

−

→

c