Question

Find an equation for the plane containing the two (parallel) lines v_1=(0,1,-2)+t(2,3,-1) and v_2=(2,-1,0)+t(2,3,-1).

Vectors
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asked 2021-06-10
Find an equation for the plane containing the two (parallel) lines
\(v_1=(0,1,-2)+t(2,3,-1)\)
and \(v_2=(2,-1,0)+t(2,3,-1).\)

Answers (1)

2021-06-11
The given two lines are,
\(v_1=(0,1,-2)+t(2,3,-1)\) and
\(v_2=(2,-1,0)+t(2,3,-1)\)
Let A=(0,1,-2)
B=(2,-1,0)
Thus the plane containing these two lines will contain A and B and hence the vector AB=(2,-2,2)
Normal to the plane shall be the cross product of AB and the direction ratio of the lines.
Thus \(\vec{n}=\vec{AB}\times\text{direction ratio of the line}\)
\(\vec{n}=(2,-2,2)\times(2,3,-1)\)
\(\vec{n}=\begin{bmatrix}i&j&k\\2&-2&2\\2&3&-1\end{bmatrix}=-4i+6j+10k\)
Thus the equation of the plane can be written as,
\([(x-0)i+(y-1)j+(z+2)\cdot\vec{n}=0\)
\(-4(x-0)+6(y-1)+10(z+2)=0\)
\(-4x+6y+10z+14=0\)
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