Question
Mathematics Question on Vector Algebra
The ratio in which the plane r.(i^ -2j^ + 3k^ ) =17 divides the line joining the points -2i^+4j^+7k^ and 3i^-5j^+8k^ is
5 : 3
4 : 5
3 : 10
10 : 3
3 : 10
Solution
First we need to find the point of intersection between the line and the plane.The direction vector of the line AB is given by the difference between the coordinates of the two points:
d = B - A = ( 3i^-5j^+8) - (-2i^+4j^+7k^) = 5i^ - 9j^ + k^
Now by substituting the parametric form of the line equation into the plane equation:r = A + t.d
Substituting the values: r = (-2i^+4j^+7k^) + t x (5i^ - 9j^ + k^)
Substituting this into the plane equation:
((-2 + 5t)i^ + (4 - 9t)j^ + (7 + t)k^) . (i^ - 2j^ + 3k^) = 17
(-2 + 5t)i^2 + (4 - 9t)(-2) + (7 + t)3 = 17 (5t - 2) + (-8 + 18t) + (21 + 3t) = 17 26t + 11 = 17 26t = 6 t = 6/26 t = 3/13.
Now, we substitute this value of t back into the parametric equation of the line to find the point of intersection:
r = A + t.d.r = (-2i^+4j^+7k^) +133 x (5i^ - 9j^ + k^) r = (-2i^+4j^+7) +\frac {15}{13}$$\hat i - \frac {27}{13}$$\hat j + \frac {3}{13}$$\hat k
r = \frac {15}{13}$$\hat i + (4 - \frac {27}{13})$$\hat j + (7 + \frac {3}{13})$$\hat k
r =\frac {15}{13}$$\hat i + \frac {35}{13}$$\hat j +\frac {100}{13}$$\hat k
Therefore, the point of intersection is:
P = \frac {15}{13}$$\hat i + \frac {35}{13}$$\hat j + \frac {100}{13}$$\hat k
Now, find the ratios by calculating the distances between the points:
AP = (1315−(−2))2+(1335−4)2+(13100−7)2
BP = (3−1315)2+(−5−1335)2+(8−13100)2
Calculating the distances:
AP = 169225+1691089+1698100 = 16910414 = 1310414
BP = 169468+16925+169392 = 169885 = 13885
Now express the ratio AP:BP
AP:BP = 1310414 : 13885=10414 : 885
Simplifying by rationalizing the denominators:
AP:BP = 88510414 x 885885= 885×88510414×885 =8859207690
Therefore, the ratio in which the plane divides the line joining the points is (C) 3:10