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標題:
幾何兩直線間的距離問題..
發問:
求兩直綫(((((({x+y-z=1}{2x+z=3})))))和x=y=z-1間的距離....要有過程.... 更新: maximal_ideal_space你個答案錯咗.........
i+j-k is a normal vector for the plane x+y-z=1. 2i+k is a normal vector for the plane 2x+z=3 The line (((((({x+y-z=1}{2x+z=3}))))) which is the intersection of the two planes, must be perpendicular to both normal vectors, so its direction is along the vector (i+j-k)x(2i+k)=i-3j-2k. On the other hand, the direction of the line x=y=z-1 is along the vector i+j+k. Hence, a common normal for the two lines is (i-3j-2k)x(i+j+k)=-i-3j+4k. Choose two points on each line respectively. For example, you can put x=0 in the equations to get A=(0,4,3) on the line (((((({x+y-z=1}{2x+z=3}))))) B=(0,0,-1) on the line x=y=z-1. Then, AB=-4j-4k. The distant between the two lines is | AB．(unit vector along common normal) | (the dot denotes the dot product) =|(-4j-4k)．(-i-3j+4k)/sqrt(26)| =4/sqrt(26)
其他解答:
幾何兩直線間的距離問題..
發問:
求兩直綫(((((({x+y-z=1}{2x+z=3})))))和x=y=z-1間的距離....要有過程.... 更新: maximal_ideal_space你個答案錯咗.........
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最佳解答:i+j-k is a normal vector for the plane x+y-z=1. 2i+k is a normal vector for the plane 2x+z=3 The line (((((({x+y-z=1}{2x+z=3}))))) which is the intersection of the two planes, must be perpendicular to both normal vectors, so its direction is along the vector (i+j-k)x(2i+k)=i-3j-2k. On the other hand, the direction of the line x=y=z-1 is along the vector i+j+k. Hence, a common normal for the two lines is (i-3j-2k)x(i+j+k)=-i-3j+4k. Choose two points on each line respectively. For example, you can put x=0 in the equations to get A=(0,4,3) on the line (((((({x+y-z=1}{2x+z=3}))))) B=(0,0,-1) on the line x=y=z-1. Then, AB=-4j-4k. The distant between the two lines is | AB．(unit vector along common normal) | (the dot denotes the dot product) =|(-4j-4k)．(-i-3j+4k)/sqrt(26)| =4/sqrt(26)
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