35. (a) Represent the equations of lines l1 and l2 in vector form and check whether they are intersecting or not.
l1:−3x+3=1y−1=5z−5 l2:−1x+1=−22−y=5z−5
(lines, vectors, intersection, 3D geometry, symmetric equations)
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Step-by-Step Solution
Step 1: Convert Line l1 to Vector Form
The standard symmetric form of a line is ax−x1=by−y1=cz−z1. From this, we can extract a point (x1,y1,z1) on the line and its direction vector b1=ai^+bj^+ck^. The vector form of a line is r=a1+λb1, where a1 is the position vector of a point on the line and b1 is the direction vector.
Step 2: Extract Point and Direction Vector for l1
For l1:−3x−(−3)=1y−1=5z−5, we identify (x1,y1,z1)=(−3,1,5) and direction ratios (a,b,c)=(−3,1,5). Therefore, the position vector of a point on l1 is a1=−3i^+j^+5k^ and its direction vector is b1=−3i^+j^+5k^. Notice that the direction vector and the position vector of the point given by the symmetric equation coincide in this particular problem because the point components are −3,1,5 and the direction ratios are also −3,1,5. This implies that the point (−3,1,5) is on the line l1 and the line passes through the origin if a1 was (0,0,0).
Step 3: Convert Line l2 to Symmetric Form
The given equation for l2 has (2−y) in the numerator, which is not in the standard form (y−y1). We rewrite −22−y as −2−(y−2), which simplifies to 2y−2. This puts l2 in proper symmetric form.
Step 4: Extract Point and Direction Vector for l2
From the symmetric form l2:−1x−(−1)=2y−2=5z−5, we identify a point (x1,y1,z1)=(−1,2,5) and direction ratios (a,b,c)=(−1,2,5). So, the position vector of a point on l2 is a2=−i^+2j^+5k^ and the direction vector is b2=−i^+2j^+5k^.
Step 5: Check for Intersection of Lines
Two lines intersect if and only if the shortest distance between them is zero. This condition is met when the scalar triple product of (a2−a1), b1, and b2 is zero. First, we compute the vector joining the points on the lines, (a2−a1). Then, we compute the cross product of the direction vectors, b1×b2. Finally, we find their dot product.
Step 6: Calculate a2−a1
We subtract the components of a1 from a2 to find the vector connecting a point on l1 to a point on l2. This vector is important for determining if the lines are coplanar.
Step 7: Calculate b1×b2
We compute the cross product of the direction vectors b1 and b2. This vector is normal to the plane containing the two lines (if they are coplanar). If the lines are parallel, their cross product would be zero. Since this is not zero, the lines are not parallel.
Step 8: Calculate Scalar Triple Product
We calculate the dot product of the vector (a2−a1) and the cross product (b1×b2). Since the scalar triple product is 0, the lines are coplanar. If two non-parallel lines are coplanar, they must intersect.