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Question: The line \(\frac{x–1}{2}\) = \(\frac{y + 1}{3}\) = \(\frac{2–z}{1}\) and y – x = 0 = x – z + l are ...

The line x–12\frac{x–1}{2} = y+13\frac{y + 1}{3} = 2–z1\frac{2–z}{1}

and y – x = 0 = x – z + l are coplanar for :

A

l = 0

B

l = – 2

C

l = – 5

D

l = 2

Answer

l = – 5

Explanation

Solution

Shortest Distance = ∣(a→2–a→1).(b→1×b→2)∣b→1×b→2∣∣\left| \frac{\left( {\overrightarrow{a}}_{2}–{\overrightarrow{a}}_{1} \right).\left( {\overrightarrow{b}}_{1} \times {\overrightarrow{b}}_{2} \right)}{\left| {\overrightarrow{b}}_{1} \times {\overrightarrow{b}}_{2} \right|} \right|