If the square of the shortest distance between the lines 1x−2=2y−1=3z+3 and 2x+1=4y+3=5z+5 is nm, where m, n are coprime numbers, then m+n is equal to:
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Step-by-Step Solution
Step 1: Identify points and direction vectors
First, we identify the position vectors of points on each line and their respective direction vectors. For the first line, a point on the line is A1(2,1,−3) and its direction vector is b1=i^+2j^+3k^. For the second line, a point on the line is A2(−1,−3,−5) and its direction vector is b2=2i^+4j^+5k^.
Step 2: Calculate vector connecting points
Next, we calculate the vector connecting a point on the first line to a point on the second line, which is a2−a1. This vector is obtained by subtracting the coordinates of A1 from A2.
Step 3: Calculate cross product of direction vectors
We then compute the cross product of the direction vectors b1 and b2. This cross product gives a vector that is perpendicular to both direction vectors, which is essential for finding the shortest distance.
Step 4: Calculate shortest distance
The shortest distance d between two skew lines is given by the formula involving the scalar triple product. We substitute the calculated vectors into the formula and simplify to find the distance.
Step 5: Calculate the square of the shortest distance and m+n
The problem asks for the square of the shortest distance. We square the calculated distance to get 54. Here, m=4 and n=5. Since 4 and 5 are coprime, we find m+n=4+5=9.