UPSC NDA Math Model Paper 9

Given, triangle whose vertices are at A = (3, -1, 2) =3

→

i

−

→

j

+2

→

k

B=(1,−1,−3)=

→

i

−

→

j

−3

→

k

and C=(4,−3,1)=4

→

i

−3

→

j

+

→

k

Let A, B , and C be the vertices of the △ABC , then the area of that triangle =

1

2

×|

→

AB

×

→

AC

|

→

AB

=(

→

i

−

→

j

−3

→

k

)−(3

→

i

−

→

j

+2

→

k

)
⇒

→

AB

=−2

→

i

+0

→

k

−5

→

k

→

AC

=(4

→

i

−3

→

j

+

→

k

)−(3

→

i

−

→

j

+2

→

k

)
⇒

→

AC

=

→

i

−2

→

j

−

→

k

Now,

→

AB

×

→

AC

=|

→ i	→ j	→ k
−2	0	−5
1	−2	−1

|
⇒

→

AB

×

→

AC

=

→

i

(−10)−

→

j

(2+5)+

→

k

(4−0)
⇒

→

AB

×

→

AC

=−10

→

i

−7

→

j

+4

→

k

⇒|

→

AB

×

→

AC

|=√(−10)2+(−7)2+(4)2
⇒|

→

AB

×

→

AC

|=√165
The area of that triangle =

1

2

×|

→

AB

×

→

AC

|
⇒ The area of that triangle =

1

2

×√165
Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is

√165

2