TS EAMCET 2017 Code A Solved Paper

The vector c is parallel to a×b
a×b,=|

∧ i	∧ j	∧ k
2	1	−3
1	3	2

|
=11

∧

i

−7

∧

j

+5

∧

k

=11

∧

i

−7

∧

j

+5

∧

k

The magnitude is calculated as
d=‌‌

a×b

|a×b|

=‌‌

1‌li−7j+5k

√195

Thus, vector c is calculated as
c=|c|

∧

d

=‌‌

2

√195

(

∧

11i

−7

∧

j

+5

∧

k

)
The volume of the parallelepiped is calculated as
[

a

b

c

] =|

2	1	−3
1	3	2
‌‌ 22 √195	‌‌ −14 √195	‌‌ 10 √195

|
=‌‌

2

√195

[2(29)−1(−17)−3(−40)]
=‌‌

2

√195

×195
=2√195