Lines $r = a_{1} + t b_{1}$ and $r = a_{2} + s b_{2}$ lie on a plane if

a_{1} \times a_{2} = 0

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b_{1} \times b_{2} = 0

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(a_{2} - a_{1}) . (b_{1} \times b_{2}) = 0

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Solution

The correct option is D $(a_{2} - a_{1}) . (b_{1} \times b_{2}) = 0$
Since the lines are co-planar
Therefore shortest distance between them is zero
$\Rightarrow \frac{(a_{2} - a_{1}) . (b_{1} \times b_{2})}{| b_{1} \times b_{2} |} = 0$
Therefore, $(a_{2} - a_{1}) . (b_{1} \times b_{2})$

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Lines r=a1+tb1 and r=a2+sb2 lie on a plane if

The correct option is D (a2−a1).(b1×b2)=0Since the lines are co-planar Therefore shortest distance between them is zero⇒(a2−a1).(b1×b2)|b1×b2|=0Therefore, (a2−a1).(b1×b2)

Lines $r = a_{1} + t b_{1}$ and $r = a_{2} + s b_{2}$ lie on a plane if

The correct option is D $(a_{2} - a_{1}) . (b_{1} \times b_{2}) = 0$
Since the lines are co-planar
Therefore shortest distance between them is zero
$\Rightarrow \frac{(a_{2} - a_{1}) . (b_{1} \times b_{2})}{| b_{1} \times b_{2} |} = 0$
Therefore, $(a_{2} - a_{1}) . (b_{1} \times b_{2})$