COMEDK 2023 Shift 1 Paper

For the vectors

→

a

and

→

b

which are unit vectors, the expression given is

→

a

−√2

→

b

. We are looking for the angle θ between

→

a

and

→

b

such that

→

a

−√2

→

b

is a unit vector.
First, let's calculate the magnitude of the vector

→

a

−√2b. If this is a unit vector, its magnitude must be 1 . We use the property that the square of the magnitude of the sum (or difference) of two vectors

→

u

and

→

v

can be found using:
Here, |

→

a

|=1 and |

→

b

|=1 because they are unit vectors. Therefore, the expression modifies to:
The equation simplifies to:
We want this magnitude squared to be 1 , hence:
3−2√2‌cos(θ)=1
Solving for cos(θ) :
‌2√2‌cos(θ)=2
‌cos(θ)=‌

2

2√2

=‌

1

√2

The value of cos(θ)=‌

1

√2

corresponds to an angle θ=‌

π

4

.
Thus, the correct answer is Option D:

π

4