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to represent a plane, you need `normal vector` and a `point`
cross the direction vectors of given lines for `normal vector`
yes cross product gives you the normal
are you dont with question 1 ?
*done
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looks good!
two vectors are perpendicular iff the dot product is 0
direction vector of first line : <k, 2, k-1> direction vector of second line : <-2, -1, 1>
take their dot product, set it equal to 0 and solve k
change the given lines to parametric form
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x-1/k = y-2/2 = z+1/k-1 and x+3/-2 = z/1, y=-1
\[\large \rm \dfrac{x-1}{\color{red}{k}} = \dfrac{y-2}{\color{Red}{2}} = \dfrac{z+1}{\color{red}{k-1}}\]
in symmetric form of line, the bottom stuff givs u the direction vector
k=3 is correct!
np:)
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