JEE Main 29-Jan-2024 Shift 2 Solved Paper

Unit vector

^

u

=x

^

i

+y

^

j

+z

^

k

‌

→

p

1=‌

1

√2

^

i

+‌

1

√2

^

k

,

→

p

2=‌

1

√2

^

j

+‌

1

√2

^

k

‌

→

p

3=‌

1

√2

^

i

+‌

1

√2

^

j

Now angle between

^

u

and

→

p

1=‌

π

2

^

u

⋅

→

p

1=0⇒‌

x

√2

+‌

z

√2

=0
⇒x+z=0 ..............(i)
Angle between

^

u

and

→

p

2=‌

π

3

^

u

⋅

→

p

2=|

^

u

|⋅|

→

p

2|‌cos‌

π

3

⇒‌

y

√2

+‌

z

√2

=‌

1

2

⇒y+z=‌

1

√2

.............(ii)
Angle between

^

u

and

→

p

3=‌

2π

3

‌

^

u

⋅

→

p

3=|

^

u

|⋅|

→

p

3|‌cos‌

2π

3

‌⇒‌

x

√2

+‌

4

√2

=‌

−1

2

⇒x+y=‌

−1

√2

.............(iii)
from equation (i), (ii) and (iii) we get
x=‌

−1

√2

‌‌y=0‌‌z=‌

1

√2

‌‌ Thus ‌

^

u

−

→

v

=‌

−1

√2

^

i

+‌

1

√2

^

k

−‌

1

√2

^

i

−‌

1

√2

^

j

−‌

1

√2

^

k

‌

^

u

−

→

v

=‌

−2

√2

^

i

−‌

1

√2

^

j

‌∴|

^

u

−

→

v

|2=(√‌

4

2

+‌

1

2

)2 =‌

5

2