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
(vector form, lines, intersection, parametric equations, direction vectors)
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Step-by-Step Solution
Step 1: Convert Line 1 to Vector Form
A line in Cartesian form ax−x1=by−y1=cz−z1 passes through the point (x1,y1,z1) and has direction ratios (a, b, c). The vector form of this line is r=(x1i^+y1j^+z1k^)+λ(ai^+bj^+ck^). For l1, comparing it with the standard form, we identify the point and direction vector.
Step 2: Identify Point and Direction Vector for l1
From the given equation for line l1:−3x−(−3)=1y−1=5z−5, we can see that it passes through the point (−3,1,5) and has direction ratios (−3,1,5). So, the position vector a1 is −3i^+j^+5k^, and the direction vector b1 is −3i^+j^+5k^.
Step 3: Convert Line 2 to Vector Form
Similarly, we convert the Cartesian equation of line l2 into its vector form. First, we need to adjust the y term in the equation to match the standard form by−y1.
Step 4: Adjust l2 Cartesian Form
The given equation for line l2 is −1x+1=−22−y=5z−5. To get it into the standard form (y−y1)/b, we rewrite −22−y as −2−(y−2)=2y−2. Now the equation is in the correct format.
Step 5: Identify Point and Direction Vector for l2
From the adjusted equation for line l2:−1x−(−1)=2y−2=5z−5, we identify that it passes through the point (−1,2,5) and has direction ratios (−1,2,5). So, the position vector a2 is −1i^+2j^+5k^, and the direction vector b2 is −1i^+2j^+5k^.
Step 6: Check for Intersection: Condition for Intersecting Lines
Two lines intersect if the shortest distance between them is zero. This condition is met if the scalar triple product of the difference in position vectors and the cross product of their direction vectors is zero. We calculate a2−a1, b1×b2, and then their dot product.
Step 7: Calculate a2−a1
We subtract the position vector of the first line from the position vector of the second line to get the vector connecting a point on l1 to a point on l2. This results in a vector 2i^+j^.
Step 8: Calculate b1×b2
Next, we compute the cross product of the direction vectors b1 and b2. This vector is perpendicular to both b1 and b2, and its magnitude is related to the area of the parallelogram formed by the vectors.
Step 9: Calculate (a2−a1)⋅(b1×b2)
Finally, we compute the dot product of the vector (a2−a1) and the cross product (b1×b2). Since the result is 0, the lines intersect.
Step 10: Conclusion about Intersection
The scalar triple product being zero indicates that the vectors a2−a1, b1, and b2 are coplanar. When this condition is met, and the lines are not parallel (which they are not, as b1 is not a scalar multiple of b2), the lines intersect.