what am i doing wrong?

[earless]

Do these two lines intersectx/3)=(y/2)=(z/2) and (x/5)=(y/3)=(z-4/2)??

I found the parallel vector to each line, v1=3i+2j+2k and v2=5i+3j+2k and did the cross product (v1)X(v2)=-2i+4j-k, since the cross product wasnt 0 then i the vectors must intesect and since they are parallel to the given lines they to must intersect.But the book says they dont intersect.what am i doing wrong?

 

[pka]

\(\displaystyle l_1 :\left\{ {\begin{array}{l}
{x = 3t} \\
{y = 2t} \\
{z = 2t} \\
\end{array}} \right.\quad \& \quad l_2 :\left\{ {\begin{array}{l}
{x = 5s} \\
{y = 3s} \\
{z = 4 + 2s} \\
\end{array}} \right.\)

When written in parametric form using different parameters, it is clean that they can not intersect.

 

[pka]

\(\displaystyle l_1 :\left\{ {\begin{array}{l}
{x = 3t} \\
{y = 2t} \\
{z = 2t} \\
\end{array}} \right.\quad \& \quad l_2 :\left\{ {\begin{array}{l}
{x = 5s} \\
{y = 3s} \\
{z = 4 + 2s} \\
\end{array}} \right.\)

When written in parametric form using different parameters, it is clean that they can not intersect.

 

[Deleted member 4993]

Do these two lines intersectx/3)=(y/2)=(z/2) and (x/5)=(y/3)=(z-4/2)??

I found the parallel vector to each line, v1=3i+2j+2k and v2=5i+3j+2k and did the cross product (v1)X(v2)=-2i+4j-k,

since the cross product wasnt 0 then i the vectors must intesect

Not necessarily true all the time.

Those two lines could be (and are) skew lines - lying in different planes.

and since they are parallel to the given lines they to must intersect.But the book says they dont intersect.what am i doing wrong?

If I were to do this problem - I would convert the equations into parametric forms.