Evaluate ab + c for a = 2, b = 3, and c = 4. 9 10 27
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Answer:
Step-by-step explanation:
Let the I line be PQ and II line be RS
Direction ratios of PQ = (9,0,0) and direction ratios of RS = (0, -4, -12)
Distance between skew lines
=(a1-a2).(b1xb2)/|b1xb2|
where a1 and a2 are points
Here a1-a2 = (-3,2,-4) taking P and Q as points.
b1 = (9,0,0) and b2 = (0,-4,-12)
Shortest Distance = (a2-a1).(b1xb2)/|b1xb2|
De
Numerator = -9(-24-16)=360
Denominator vector = i(0)-j(-108)+k(-36)
Modulus of vector =\sqrt{108^2+36^2} =36\sqrt{10}
Hence Shortest distance = \sqrt{10}