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Let : P_(1):3y+z+1=0 and P_(2):2x-y+3z-7=0 and the equation of line AB is (x-1)/(2)=(y-3)/(-1)=(z-4)/(3) in 3D space. Shortest distance between the line of intersection of planes P_(1) and P_(2) and the line AB is equal to |
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Answer» `(7)/(sqrt(10))" UNITS"` |
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