Given two vectors are

^

i

−2

^

j

+

^

k

and 3

^

i

+2

^

j

+

^

k

⇒ sin θ =

| → a × → b |

| → a || → b |

... (i)
To find

→

a

×

→

b

→

a

×

→

b

= |

^ i	^ j	^ k
1	−2	3
3	−2	1

|
⇒

→

a

×

→

b

= [- 2 - (- 6)]

^

i

- (1 - 9)

^

j

+ [- 2 - (- 6)]

^

k

⇒

→

a

×

→

b

= 4

^

i

+8

^

j

+4

^

k

⇒ |

→

a

×

→

b

| = √42+82+42 = √96 = 4 √6 ... (i)
|

→

a

| = √12+(−2)2+32 = √14 ... (ii) and |

→

b

| = √32+(−2)2+32 = √14 ... (iii)
Since sin θ =

| → a × → b |

| → a || → b |

⇒ sin θ =

4√6

√14√14

... From (i), (ii) and (iii)
⇒ sin θ =

2√6

7