NCERT Solutions
Class 12 Maths
Vector Algebra

Ex.Misc. Q.8
Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.
The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).
AB = (5 - 1)i + (0 + 2)j + (-2 + 8)k = 4i + 2j + 6k
BC = (11 - 5)i + (3 - 0)j + (7 + 2)k = 6i + 3j + 9k
AC = (11 - 1)i + (3 + 2)j + (7 + 8)k = 10i + 5j + 15k
|AB| = √{(4)2 + (2)2 + (6)2} = √(16 + 4 + 36) = √56 = 2√14
|BC| = √{62 + 32 + 92} = √(36 + 9 + 81) = √126 = 3√14
|AC| = √{(10)2 + 52 + 152} = √(100 + 25 + 225) = √350 = 5√14
Now, |AC| = |AB| + |BC|
Thus, the given points A, B and C are collinear.
Now, let the point B divide AC in the ratio k : 1
Then, we have OB =
=> 5i – 2k =
=> (5i – 2k)(k + 1) = k* (11i + 3j + 7k) + (i – 2j – 8k)
=> 5(k + 1)i – 2(k + 1)k = (11k + 1)i + (3k - 2)j + (7k - 8)k
On equating the corresponding components, we get
5(k + 1) = 11k + 1
=> 5k + 5 = 11k + 1
=> 6k = 4
=> k =
Hence, the point B divides AC in the ratio 2 : 3