1. Show that a × (b × c), b × (c × a) and c × (a × b) are coplanar.
Proof: a × (b × c) = (a . c) b - (a . b)c     → (1)
             b × (c × a) = (b . a) c - (b . c)a     → (2)
             c × (a × b) = (c . b) a - (c . a)b     → (3)
                              (1) + (2) + (3)
             a × (b × c) + b × (c × a) + c × (a × b) = 0
       a × (b × c), b × (c × a), c × (a × b) are coplanar
 

2. Find the value of   i × (j × k) + j × (k × i) + k × (i × j)
Sol.  Given  i × (j × k) + j × (k × i) + k × (i × j)
                        = i × i + j × j + k × k
                        = 0 + 0 + 0
                        = 0
                                           (or)
 

 = (i.k)j - (i.j)k + (j.i)k - (j.k)i + (k.j)i - (k.i)j
                     = (0)j - (0)k + (0)k - (0)i + (0)i - (0)j
                     = 0
 

3. If  = i - 2j + 3k,  = 2i + j - k and  = i + j + 2k then find the value of  (a × b) × c

4. Show that  [a × b b × c c × a] = [abc]²
Sol. L.H.S     [a × b b × c c × a]
                      = (a × b) . ((b × c) × (c × a))
                Let b × c = p
                      = (a × b) . (p × (c × a))
                      = (a × b) ((p . a) c - (p . c) a)
                      = (a × b) . ((b × c . a) c - (b × c . c) a)
                      = (a × b) . ([abc] c - [cbc] a)
                      = (a × b) . ([abc] c - 0)
                      = (a × b) . c [abc]
                      = [abc] [abc]
                      = [abc]²
                      = R.H.S

5. If = 2i + 3j + 4k,  = i + j - k,  = i - j + k then verify that a × (b × c) is perpendicular to  .
Sol. Given      = 2i + 3j + 4k
                        = i + j - k
                        = i - j + k

              
           

6. If i, j, k are orthogonal unit vector kind and for any vector

, show that
   

7. Show that d . [a × {b × (c × d)}] = (b . d) [acd]
Sol:   L.H.S d . [ a × {b × (c × d)}]
               = d . [ a × {(b . d) c - (b . c) d}]              ( By using Conceptual Theorem (1) )
               = d . [(a × c)(b . d) - (a × d)(b . c)]
               = d . (a × c)(b . d) - d (a × d) (b . c)
               = [ d a c ] (b . d) - [d a d] (b . c)
               = [ a c d ] (b . d) - 0
                = (b . d) [a c d]
 

8. Show that (a × b) . (c × d) = (a . c) (b . d) - (a . d) (b . c) = 

            = (a . c) (b . d) - (a . d) (b . c) --→ (2)
     From (1) and (2) The given statement is true.

9. Show that (a × b) × (c × d) = [a b d] c - [a b c] d = [a c d] b - [b c d] a

 

10. Show that (b × c) . (a × d) + (c × a) . (b × d) + (a × b) . (c × d) = 0

11. If   = i + 2j - k,   = 3i - 4k,   = -i + j and = 2i - j + 3k then find
      (i) (a × b) . (c × d)      (ii) (a × b) × (c × d)
Sol: Given   = i + 2j - k
                        = 3i - 4k
                        = -i + j
                     

  = 2i - j + 3k
     

(ii) (a × b) × (c × d)  = - 35 (-i + j) - 9 (2i - j + 3k)
                                              = 35i - 35j - 18i + 9j - 27k
                                              = 17i - 26j - 27k

Writer: Sayyad Anwar