UPSC NDA I 2021 Mathematics Solved Paper

Concept:
The cross product of two vectors

→

a

and

→

b

is given by

→

a

×

→

b

=|a|⋅|b|sin‌θ‌

∧

1

‌‌ and |

→

a

×

→

b

|=|a|⋅|b|sin‌θ
The scalar product of two vectors

→

a

and

→

b

is given by

→

a

⋅

→

b

=|a|×|b|cos‌θ
If

→

a

is a unit vector then |a|=1
Calculations:
Statement 1: The cross product of two unit vectors is always a unit vector.
Let

→

a

and

→

b

are two unit vectors.
i.e ∣a]=1 and |

→

b

|=1
As we knowthat, the cross product of two vectors

→

a

and

→

b

is given by

→

a

×

→

b

=∣a]⋅|b|sin‌θ‌

∧

f

‌‌ and |

→

a

×

→

b

|=|a|
⇒|

→

a

×

→

b

|=|a|⋅|b|sin‌θ=sin‌θ
The range of sin‌θ is [−1,1]
So, it is notnecessarily true that the cross product of two unit vectors is always a unit vector.
Hence, statement 1 is false.
Statement 2: The dot product of two unit vectors is always unity.
Let

→

a

and

→

b

are two unit vectors.
i.e |a|=1 and ∣b∣=1
As we know that, the scalar product of two vectors

→

a

and

→

b

is given by

→

a

⋅

→

b

=|

→

a

|×|b|cos‌θ
⇒|

→

a

⋅

→

b

|=cos‌θ
The range of cos‌θ is [−1,1] So, it is not necessarily true that the dot product of two unit vectors is always a unit vector.
Hence, statement 2 is false.
Statement 3: The magnitude of sum of two unit vectors is always greater than the magnitude of their difference. Let

→

a

=

∧

i

and

→

b

=

∧

j

As we can see that, the vectors

→

a

and

→

b

are two unit vectors
⇒|

∧

i

+

∧

j

|=√2‌ and ‌|

∧

i

−

∧

j

|=√2
⇒|

→

a

+

→

b

|=|

→

a

−

→

b

|
So, statement 3 is also false.
Hence, the correct option is 4 .