Three Dimensional Geometry Part 4

‌L1:‌

x−2

1

=‌

y

−1

=‌

z−1

1

=λ
‌L2:‌

x+1

1 2

=‌

y−1

1 2

=‌

z+1

1

=µ
dr of line MN will be
<3+λ−‌

µ

2

,−1−λ−‌

µ

2

,2+λ−µ>& it will be proportional to <3,1,2>
∴‌

3+λ− µ 2

3

=‌

−1−λ− µ 2

1

=‌

2+λ−µ

2

⇒λ=−‌

4

3&

µ=−‌

2

3

∴ Coordinate of M will be <(‌

2

3

,‌

4

3

,−‌

1

3

) and equation of required line will be.
‌

x− 2 3

3

=‌

y− 4 3

1

=‌

z+ 1 3

2

=k
So any point on this line will be
‌(‌

2

3

+3k,‌

4

3

+k,−‌

1

3

+2k)
‌∵‌

2

3

+3k=−‌

1

3

⇒k=−‌

1

3

∴ Point lie on the line for
k=−‌

1

3

‌ is ‌(−‌

1

3

,1,−1)