Show that the points whose position vectors are `(5hat(i) + 6 hat(j) + 7 hat(k)), ( 7 hat(i) - 8hat(j) + 9 hat(k)) and (3 hat(i) + 20 hat(j) + 5 hat(k))` are collinear.

Explanation

Let the given points be A,B and C respectively. Then,
`vec(AB)= " (position vector of B )- (position vector of A) "`
` = (7 hat(i) - 8 hat(j) + 9 hat(k)) - ( 5 hat(i) + 6 hat(j) + 7 hat(k))= ( 2 hat(i) - 14 hat(j)+2 hat(k)).`
And, `vec(AC) = " ( position vector of C)- ( position vector of A) "`
`=(3 hat(i) + 20 hat(j) + 5 hat(k))-( 5 hat(i) + 6 hat(j) + 7 hat(k))`
` = (-2hat(i) + 14 hat(j) - 2 hat(k)).`
`:. (vec(AB) xx vec(AC))=|(hat(i),hat(j), hat(k)),(2,-14,2),(-2,14,-2)|`
`= 2 xx (-2). |(hat(i),hat(j), hat(k)),(1,-7,1),(1,-7,1)|=(-4) xx vec(0) =vec(0)`
`[:' R_(2) and R_(3)" are identical "]`.
So, `vec(AB) and vec(AC)` are parallel vectors, having a common end point , A.
Hence, the given points A,B and C are collinear.