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Find the shortest distance from (6,-4,4) to the line joining (2,1,2) and (3,-1,4).
Hi Find the vector equation of the line joining pt (2,1,2) & (3,-1,4) using r(t) = a + t (b - a ). hence eq wud be r =(2,1,2) +t(1, -2,2). Suppose the position vector of P from the origin O is OP = p = (6,-4,-4). The shortest distance of P from L is given by the length of the perpendicular from P to L. Suppose this perpendicular meets L at H. Then we want to find the length of PH. Using the dot product Since H lies on L we can say that OH = h = (2,1,2) +t(1, -2,2). = (2+t , 1-2t , 2+2t) for some value of t which we need to find. Also, vector PH = -p + h = -(6 , -4, -4) + (2+t , 1-2t , 2+2t) = (t-4, 5-2t, 6+2t) But PH is perpendicular to the direction vector (1, -2,2) of line L. So the dot product of vector PH and (4, 1, -2) is zero. So t-4, 5-2t, 6+2t).(1, -2,2) = 0 so solve for t, and find pt h, then find the distance between the two points.
Hi
Find the vector equation of the line joining pt (2,1,2) & (3,-1,4) using r(t) = a + t (b - a ). hence eq wud be r =(2,1,2) +t(1, -2,2).
Suppose the position vector of P from the origin O is OP = p = (6,-4,-4). The shortest distance of P from L is given by the length of the perpendicular from P to L. Suppose this perpendicular meets L at H. Then we want to find the length of PH.
Using the dot product
Since H lies on L we can say that OH = h = (2,1,2) +t(1, -2,2). = (2+t , 1-2t , 2+2t) for some value of t which we need to find. Also, vector PH = -p + h = -(6 , -4, -4) + (2+t , 1-2t , 2+2t) = (t-4, 5-2t, 6+2t)
But PH is perpendicular to the direction vector (1, -2,2) of line L. So the dot product of vector PH and (4, 1, -2) is zero. So t-4, 5-2t, 6+2t).(1, -2,2) = 0 so solve for t, and find pt h, then find the distance between the two points.
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