UPSC NDA Math Model Paper 4

Given: |

→

a

|=3,|

→

b

|=4 and |

→

a

×

→

b

|=6
As we know,

→

a

×

→

b

=|

→

a

|×|

→

b

|×sin‌θ×

^

n

|

→

a

×

→

b

|=|

→

a

|×|

→

b

|×|sin‌θ|×|

^

n

| =|

→

a

|×|+

→

b

|×sin‌θ (∵ Magnitude of a unit vector is one)
⇒sin‌θ=

| → a × → b |

| → a |×| → b |

=

6

12

=

1

2

∴θ=30°
Now,
cos‌θ=cos‌30=

√3

2

As we know that, If

→

a

and

→

b

are two vectors, then the scalar product between the given by:

→

a

.

→

b

=|

→

a

|×|

→

b

|×cos‌θ
∴

→

a

.

→

b

=3×4×

√3

2

=6√3