Question

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.

Answer 1

The three given points A( 1,−2,−8 ),B( 5,0,−2 ) and C( 11,3,7 ) are collinear if,

| AB → |+| BC → |=| AC → |(1)

The vector AB → is,

AB → =( 5−1 ) i ^ +( 0−( −2 ) ) j ^ +( −2−( −8 ) ) k ^ AB → =4 i ^ +2 j ^ +6 k ^

The magnitude of AB → is,

| AB → |= 4 2 + 2 2 + 6 2 | AB → |= 56 | AB → |=2 14

The vector BC → is,

BC → =( 11−5 ) i ^ +( 3−( 0 ) ) j ^ +( 7−( −2 ) ) k ^ BC → =6 i ^ +3 j ^ +9 k ^

The magnitude of BC → is,

| BC → |= 6 2 + 3 2 + 9 2 | BC → |= 126 | BC → |=3 14

The vector AC → is,

AC → =( 11−1 ) i ^ +( 3−( −2 ) ) j ^ +( 7−( −8 ) ) k ^ AC → =10 i ^ +5 j ^ +15 k ^

The magnitude of AC → is,

| AC → |= 10 2 + 5 2 + 15 2 | AC → |= 350 | AC → |=5 14

Substitute the values of | AB → |, | BC → | and | AC → | in the LHS of equation (1).

| AB → |+| BC → |=2 14 +3 14 =5 14 =| AC → |

Therefore, A, B and C are collinear.

Consider that Bdivides ACin the ratio m:1. Then, the position vector will be, OB → = m. OC → +1. OA → m+1 5 i ^ +0 j ^ −2 k ^ = m.( 11 i ^ +3 j ^ +7 k ^ )+1.( 1 i ^ −2 j ^ −8 k ^ ) m+1 5 i ^ +0 j ^ −2 k ^ = 11m i ^ +3m j ^ +7m k ^ +1 i ^ −2 j ^ −8 k ^ m+1 5 i ^ +0 j ^ −2 k ^ = ( 11m+1 ) i ^ +( 3m−2 ) j ^ +( 7m−8 ) k ^ m+1

Simplifying further,

5 i ^ +0 j ^ −2 k ^ = ( 11m+1 ) i ^ m+1 + ( 3m−2 ) j ^ m+1 + ( 7m−8 ) k ^ m+1

Comparing both sides,

( 11m+1 ) m+1 =5 11m+1=5m+5 6m=4 m= 2 3

Thus, Bdivides AC in ratio 2:3.