Three Dimensional Geometry Part 1

Refer diagram, the normal vector be n and it is perpendicular to both AB and AC
AB×AC=n Now, A(1,2,3),B(2,3,1) and C(2,4,2)
Then, AB=(2−1)

∧

i

+(3−2)

∧

j

+(1−3)

∧

k

‌‌=

∧

i

+

∧

j

−2

∧

k

‌‌−1)‌

∧

i

+(4−2)

∧

j

+(2−3)

∧

k

AC‌‌=(2−1)

∧

i

+(4
‌‌=

∧

i

+2

∧

j

−

∧

k

Now, AB×AC=|

∧ i	∧ j	∧ k
1	1	−2
1	2	−1

|
‌‌=

∧

i

(−1+4)−

∧

j

(−1+2)+

∧

k

(2−1)
‌‌=3

∧

i

−

∧

j

+

∧

k

n‌‌=3

∧

i

−

∧

j

+

∧

k

Let P be any point on normal vector and O be origin. Then refer the diagram, projection of OP on plane have length OM .‌ OP ‌‌‌=2

∧

i

−

∧

j

+

∧

k

‌ and ‌n=3

∧

i

−

∧

j

+

∧

k

‌ OP. ‌n‌‌=|OP||n|cos‌θ
(6+1+1)‌‌=√4+1+1(√9+1+1)‌cos‌θ
8‌‌=√6√11‌cos‌θ⇒cos‌θ=‌

8

√66

‌ Again, ‌sin‌θ‌‌=‌

|OM|

|OP|

,‌ gives ‌|OM|=sin‌θ⋅|OP|
⇒‌‌|OM|‌‌=√1−cos2θ|OP|
‌=√1−‌

64

66

√4+1+1‌‌‌ (use ‌cos‌θ=8√66)
‌‌=√‌

2

66

⋅√6
∴‌‌|OM|=√‌

2

11