Sponsor Area

Three Dimensional Geometry

Question
CBSEENMA12033275

Show that the points whose position vectors are given by 
2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top comma space space minus space 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space space 5 space straight i with hat on top space plus space 5 space straight k with hat on top are collinear. 

Solution
Given points have position vectors as
   2 straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top comma space space minus space 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space space 5 space straight i with hat on top space plus space 5 space straight k with hat on top.
∴ points are (2, 1, 3), (– 4, 3, – 1), (5, 0, 5).
The equations of straight lines through the points (2, 1, 3) and (– 4, 3, – 1) are
    fraction numerator straight x minus 2 over denominator negative 4 minus 2 end fraction space equals space fraction numerator straight y minus 1 over denominator 3 minus 1 end fraction space equals space fraction numerator straight z minus 3 over denominator negative 1 minus 3 end fraction space space space or space space fraction numerator straight x minus 2 over denominator negative 6 end fraction space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 3 over denominator negative 4 end fraction
or     fraction numerator straight x minus 2 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 1 over denominator negative 1 end fraction space equals space fraction numerator straight z minus 3 over denominator 2 end fraction
The points (5, 0, 5) will lie on it
 if fraction numerator 5 minus 2 over denominator 3 end fraction space equals space fraction numerator 0 minus 1 over denominator negative 1 end fraction space equals space fraction numerator 5 minus 3 over denominator 2 end fraction
i.e.. if 1 = 1 = 1, which is true.
∴   the points (2, 1, 3), (– 4, 3, – 1), (5, 0, 5) are collinear
∴ points with position vectors 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top comma space space minus 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space 5 space straight i with hat on top space plus space 5 space straight k with hat on top are collinear.
Another Method:
Let straight a with rightwards arrow on top space equals space 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 straight k with hat on top comma space space straight b with rightwards arrow on top space equals space minus 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space 5 space straight i with hat on top space plus space 5 space straight k with hat on top.
The equation of line through two points with positions vectors straight a with rightwards arrow on top space and space straight b with rightwards arrow on top is straight r with rightwards arrow on top space equals space straight a with rightwards arrow on top plus straight lambda open parentheses straight b with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses
or space space straight r with rightwards arrow on top space equals space open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space plus space straight lambda space open parentheses negative 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top space minus space 2 space straight i with hat on top space minus space straight j with hat on top space minus space space 3 space straight k with hat on top close parentheses
or space space straight r with rightwards arrow on top space equals space open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space plus space straight lambda space open parentheses negative 6 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top close parentheses
Now the point straight c with rightwards arrow on top space equals space 5 space straight i with hat on top space plus space 5 space straight k with hat on top will lie on it
 if 5 space straight i with hat on top space plus space 5 space straight k with hat on top space equals space open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space plus space straight lambda space open parentheses negative 6 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top close parentheses
i.e.,  if 3 straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top space equals space minus 6 space straight lambda space straight i with hat on top space plus space 2 space straight lambda space straight j with hat on top space plus space 4 space straight lambda space straight k with hat on top
i.e., if 3 equals negative 6 straight lambda comma space space minus 1 space equals space 2 straight lambda comma space space space 2 space equals space minus space 4 space straight lambda
i.e., if  straight lambda space equals space minus 1 half
therefore  for straight lambda space equals space minus 1 half comma space space space given space points space are space collinear. space

Some More Questions From Three Dimensional Geometry Chapter