AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Consider given data,
m=√3× unit vectors of [(

^

i

+

^

j

)×(

^

j

−

^

k

)]
This implies,
(

^

i

+

^

j

)×(

^

j

−

^

k

)=[

^ i	^ j	^ k
1	1	0
0	1	−1

]
=

^

i

(−1−0)−

^

j

(−1−0)+

^

k

(1)
=−

^

i

+

^

j

+

^

k

So,
m=√3×

− ^ i + ^ j + ^ k

√1+1+1

=√3×

− ^ i + ^ j + ^ k

√3

−

^

i

+

^

j

+

^

k

And,
n=2√6× unit vectors of [(2

^

i

−

^

j

)×(

^

j

+2

^

k

)]
This implies,
(2

^

i

−

^

j

)×(

^

j

+2

^

k

)=[

^ i	^ j	^ k
0	−1	0
0	1	2

]
=

^

i

(−2−0)−

^

j

(4−0)+

^

k

(2+0)
=−2

^

i

−4

^

j

+2

^

k

So,
n=2√6×

−2 ^ i −4 ^ j +2 ^ k

√4+16+4

=2√6×

−2 ^ i −4 ^ j +2 ^ k

2√6

= −2

^

i

−4

^

j

+2

^

k

Now, the required area is,

1

2

(m×n)=

1

2

(−

^

i

+

^

j

+

^

k

) ×(−2

^

i

−4

^

j

+2

^

k

)
=

1

2

|

^ i	^ j	^ k
−1	1	1
−2	−4	2

|
=

1

2

×|

^

i

(2+4)−

^

j

(−2+2)+

^

k

(4+2)|
=

1

2

×|6

^

i

+6

^

k

|
=

1

2

√62+62
Further simplify the above,

1

2

(m×n)=

1

2

×6×√2
=3√2