Three Dimensional Geometry Part 3

If the lines ‌

x−1

2

=‌

2−y

−3

=‌

z−3

α

and ‌

x−4

5

=‌

y−1

2

=‌

z

β

intersect Point of first line (1,2,3) and point on second line (4,1,0).
Vector joining both points is −3

^

i

+

^

j

+3k
Now vector along second line is 2

^

i

+3

^

j

+αk
Also vector along second line is 5

^

i

+2

^

j

+βk
Now these three vectors must be coplanar
⇒|

2	3	α
5	2	β
−3	1	3

|
‌⇒2(6−β)−3(15+3β)+α(11)=0
‌⇒α−β=3
‌‌ Now ‌α=3+β
‌‌ Given expression ‌8(3+β)⋅β=8(β2+3β)
‌=8(β2+3β+‌

9

4

−‌

9

4

)=8(β+‌

3

2

)2−18
So magnitude of minimum value =18