Page 77 - Mathematics Class - XII

6. Area of parallelogram OAPC = (OA) (CL) = (OA) (LN + NC)
= (OA) (BM + NC)
= (OA) (BM) + (OA) (NC)

= Area of parallelogram OAQB + Area of parallelogram BQPC
 
= ab ac
So, from step 3
      
a c
a b c a b

      
c
Direction of each of these vectors a b c , a b and a are perpendicular to the same plane.
      
Hence, a b c a b a c

OBSERVATION

1. a = OA = OA = OM + ML + LA

2. CL = CN + NL
  

3. a b c = Area of parallelogram OAPC = (OA) (CL) = __________________ sq. units ... (i)
 
a × b = Area of parallelogram OAQB

= (OA) (BM) = ________ × ________ = ________ ... (ii)
 
a × c = Area of parallelogram BQPC
= (OA) (CN) = ________ × ________ = ________ ... (iii)

from (i), (ii) and (iii)

Area of parallelogram OAPC = Area of parallelogram OAQB + Area of parallelogram BQPC
      
So, a b c a b a c
      
4. a b c , a b and a c are perpendicular to the same plane.

      

a b c a b a c (Condition of co-planarity)

CONCLUSION
        

From the above activity it is verified that for any three vectors ab, and c we have a b c a b a c

APPLICATION
This activity is useful to understand distributive property of vectors.

Knowledge Booster


Regardless of the value of λ, the vector λ a is always collinear to



the vector a . For that matter, two vectors a and b are collinear


if and only if there exists a non-zero scalar λ such that b = λ a .
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