Determine even if it is the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, uncover the suggest of intersection.\(L1:\fracx1=\fracy-1-1=\fracz-23\)\(L2:\fracx-22=\fracy-3-2=\fracz7\)
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The direction vectors space the number in the denominators of the symmetric equations.\(L_1:\)\(L_2:\)One vector is not a many of the other. 2 times 1 is 2, however 2 time 3 isn"t 7. The lines space not parallel.Parametric type of \(L_1\)\(\fracx1=t\Rightarrow x=t\)\(\fracy-1-1=t\Rightarrow y=1-t\)\(\fracz-23=t\Rightarrow z=2+3t\)Parametric kind of \(L_2\)\(\fracx-22=s\Rightarrow x=2+2s\)\(\fracy-3-2=s\Rightarrow y=3-2s\)\(\fracz7=s\Rightarrow z=7s\)See if the lines intersect by setup the component parts equal to every other and seeing if there is a solution.\(t=2+2s\)\(1-t=3-2s\)\(2+3t=7s\)Substitute for t\(1-(2+2s)=3-2s\)\(1-2-2s=3-2s\)\(-1=3\)This is a contradiction, so there is no solution. The lines carry out not crossing so they space skew.

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Step 1

Rationalization. Geometrically speak (and together taught in the publication as a corollary) , 2 lines \(l_1\) and \(l_2 \)are parallel if and also only if the cross assets of your directional vectors a and b is the zero vector \(=\vec0\). That is:

\(a \times b =\vec0\)

Two lines space intersecting if over there exists a point (x,y,z) that is common in your domain . Lastly , 2 lines room skew if they space neither parallel or intersecting. We are provided two set of symmetric equations:



Two lines need to be among the three (parallel, intersecting, or skew)

Step 2

Acquiring the directional vectors. ‘he directional vectors of a parametric equation correspond to the denominators of every line. Thus, we get:

\(\fracx1=\fracy-1-1=\fracz-23 \Rightarrow v_1=\)

\(\fracx-22=\fracy-3-2=\fracz7 \Rightarrow v_2=\)

Step 3

Checking if the vectors space parallel. Us can examine if the vectors space parallel through the cross product or by ratio and also proportion. The book suggests using the cross product an approach corollary, so let’s do simply that:

\(v_1 \times v_2 = \beginvmatrixi & j &k \\1 & -1&-3\\2&-2&7 \endvmatrix\)

\(=\beginvmatrix-1 & -3 \\-2 & 7 \endvmatrixi-\beginvmatrix1 & -3 \\2 & 7 \endvmatrixj+\beginvmatrix1 & -1 \\2 & -2 \endvmatrixk\)

\(=<(-1)(7)-(-3)(-2)>i-<(1)(7)-(-3)(2)>j \)\(<(1)(-2)-(-1)(2)>k\)\(=-13i-13j+0k \neq 0\)

Therefore, the lines space not parallel.

Step 4

Checkingifthevectorsareintersecting:rationalization. A common means of law this is by equating x and also yand fixing for parameters s and t.. However, the offered equations are still in symmetric form. Let's rewrite these symmetric equations into parametric form by equating to parametrs s and also t ,respectively:

\(L_1: t=\fracx1=\fracy-1-1=\fracz-23 \Rightarrow \begincasesx=t\\y=-t+1\\z=3t+2\endcases\)

\(L_2: s=\fracx-22=\fracy-3-2=\fracz7 \Rightarrow \begincasesx=2s+2 \\ y=-2s+3 \\ z=7s \endcases\)

Substituting s and also tto the parametric equations for zmust yield an equality. We space checking the consistency of the straight system. Doing that, we get:


\(t=2s+2 \ \ \ \ (1)\)

\(-t+1=-2s+3 \ \ \ \ (2)\)

\(3t+2=7s \ \ \ \ \ \ (3)\)

Step 5

Checking if the vectors are intersecting: solving. Let’s settle for s and £ using equations (1) and also (2). Law that, we get:\(\begincasest&=2s+2&&t-2s&=2(4)\\&& \Rightarrow&&\\-t+1&=-2s+3&&-t+2s&=2(5)\endcases \)

Equating (4) and (5) by the removed of t, we get:

\(\begincasest-2s = 2 (4) \\ -t+2s=2(5) \endcases \Rightarrow t-t-2s+2s=2+2\)

There is a contradiction since \(0 \neq 4\). Therefore, this is one inconsistent system as there is no solution to s and also t. Therefore, the lines room not intersecting.

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Step 6

Conclusion.There space no worths of s and tsuch the both equations space equal because of the inconsistency the the system. Therefore, the equipment are also not intersecting. Therefore, the lines are skew.