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Show that the following two lines are coplanar: (x−a+d)/(α−δ)= (y−a)/α=(z−a−d)/(α+δ) and (x−b+c)/(β−γ)=(y−b)/β=(z−b−c)/(β+γ) - Mathematics

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Show that the following two lines are coplanar:

`(x−a+d)/(α−δ)= (y−a)/α=(z−a−d)/(α+δ) and (x−b+c)/(β−γ)=(y−b)/β=(z−b−c)/(β+γ)`

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Solution

We know that the lines 

`(x−x_1)/l_1=(y−y_1)/m_1=(z−z_1)/n_1 and (x−x_2)/l_2=(y−y_2)/m_2=(z−z_2)/n_2` are coplanar, if

`|[x_2-x_1,y_2-y_1,z_2-z_1],[1_1,m_1,n_1],[l_2,m_2,n_2]|=0`

Now, the equations of the given lines are

`(x−a+d)/(α−δ)= (y−a)/α=(z−a−d)/(α+δ) and (x−b+c)/(β−γ)=(y−b)/β=(z−b−c)/(β+γ)`

These can be rewritten as

`therefore |[(b-c)-(a-d),b-a,(b+c)-(a+d)],[alpha-beta,alpha,alpha+delta],[beta-gama,beta,beta+gama]|`

`=|[2(b-a),(b-a),(b+c)-(a+d)],[2alpha,alpha,alpha+delta],[2beta,beta,beta+gama]|  [`

`=2|[(b-a),b-a,(b+c)-(a+d)],[alpha,alpha,alpha+delta],[beta,beta,beta+gama]| `

=0  [C1 and C2 are identical]

Hence, the given lines are coplanar.

Concept: Shortest Distance Between Two Lines
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