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Determine whether the lines \( l_1 \) and \( l_2 \) are parallel, skew, or intersecting. If they intersect, find the point of intersection.

\( l_1 : x \)

Answer :

There is no point of intersection of L1 and L2. Given, L1: x-1/2=y+1/3=z+1/4 and L2: 2x-3/5=y+1/6=3z-5/6.To determine whether the lines L1 and L2 are parallel, skew or intersecting, we will use the following steps:

Step 1: We will convert the equations of the given lines in vector form.

Step 2: We will calculate the cross product of the direction vectors of the given lines.

Step 3: If the cross product of the direction vectors of the given lines is non-zero, then the given lines L1 and L2 are skew lines. If the cross product of the direction vectors of the given lines is zero, then the given lines L1 and L2 are parallel lines.

If L1 and L2 are not parallel and do not have a unique point of intersection, then the given lines L1 and L2 are skew lines.

L1: x-1/2=y+1/3=z+1/4

=> r = (1/2)i - (1/3)j - (1/4)k + ai + bj + ckL2: 2x-3/5=y+1/6=3z-5/6

=> r = (3/10)i - (5/6)j + (1/2)k + (2/5)a + (1/6)b + (3/6)c.

The point of intersection vector of L1 = i + 3j + 4kThe direction vector of L2 = 2i - 6j + 9kCross product of direction vectors of L1 and L2 = i(-3×9-4×(-6))+j(-4×2-1×9)+k(1×(-6)-3×(-6))= -23i + 46j + 12kSince the cross product of the direction vectors of L1 and L2 is non-zero, hence L1 and L2 are skew lines. Since they are not intersecting, there is no point of intersection.

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