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The shortest distance between the line x + y + 2z – 3 = 2x + 3y + 4z – 4 = 0 and the z-axis is : |
1 unit 2 units 3 units 4 units |
2 units |
We have, $x+y+2 z-3=0, x+2z-2+\frac{3}{2} y=0$ Solving these equations, we get Y = –2 Thus required shortest distance is 2 units. |
