text: Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.3 Textbook Questions and Answers, Notes.
  
  ## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3
  
  Question 1.
  If $$\overline { a }$$ = $$\hat { i }$$ – 2$$\hat { j }$$ + 3$$\hat { k }$$, $$\overline { b }$$ = 2$$\hat { i }$$ + $$\hat { j }$$ – 2$$\hat { k }$$, $$\overline { c }$$ = 3$$\hat { i }$$ + 2$$\hat { j }$$ + $$\hat { k }$$
  Find (i) ($$\overline { a }$$ × $$\overline { b }$$ ) × $$\overline { c }$$
  (ii) $$\overline { a }$$ × ($$\overline { b }$$ × $$\overline { c }$$)
  Solution:
  (i) $$\overline { a }$$ × $$\overline { b }$$ = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -2 \end{array}\right|$$
  = $$\hat { i }$$(4 – 3) – $$\hat { j }$$(-2 – 6) + $$\hat { k }$$(1 + 4)
  = $$\hat { i }$$ + 8$$\hat { j }$$ + 5$$\hat { k }$$
  $$\overline { a }$$ × $$\overline { b }$$ × $$\overline { c }$$ = $$\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 8 & 5 \\ 3 & 2 & 1 \end{array}\right|$$
  = $$\hat { i }$$(8 – 10) – $$\hat { j }$$(1 – 15) + $$\hat { k }$$(2 – 24)
  = -2$$\hat { i }$$ + 14$$\hat { j }$$ – 22$$\hat { k }$$
  
  (ii) $$\overline { b }$$ × $$\overline { c }$$ = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{array}\right|$$
  = $$\hat { i }$$(1 + 4) – $$\hat { j }$$(2 + 6) + $$\hat { k }$$(4 – 3)
  = 5$$\hat { i }$$ – 8$$\hat { j }$$ + $$\hat { k }$$
  $$\overline { a }$$ × $$\overline { b }$$ × $$\overline { c }$$ = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 5 & -8 & 1 \end{array}\right|$$
  = $$\hat { i }$$(-2 + 24) – $$\hat { j }$$(1 – 15) + $$\hat { k }$$(-8 + 10)
  = 22$$\hat { i }$$ + 14$$\hat { j }$$ + 2$$\hat { k }$$
  
  Question 2.
  For any vector a, prove that
  $$\hat { i }$$($$\overline { a }$$ × $$\hat { i }$$) + $$\hat { j }$$ × ($$\overline { a }$$ × $$\hat { j }$$) + $$\hat { k }$$ × ($$\overline { a }$$ × $$\hat { k }$$) = 2$$\overline { a }$$.
  Solution:
  Let $$\overline { a }$$ = a1$$\hat { i }$$ + a2$$\hat { j }$$ + a3$$\hat { k }$$
  $$\overline { a }$$ × $$\hat { i }$$ = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 0 & 0 \end{array}\right|$$
  = $$\hat { i }$$(0) – $$\hat { j }$$(-a3) + $$\hat { k }$$(0 – a2)
  $$\hat { i }$$ × ($$\overline { a }$$ × $$\hat { i }$$) = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & a_{3} & -a_{2} \end{array}\right|$$
  = $$\hat { i }$$(0) – $$\hat { j }$$(-a2) + $$\hat { k }$$(a3)
  = a2$$\hat { j }$$ + a3$$\hat { k }$$
  Similarly $$\hat { j }$$ × ($$\overline { a }$$ × $$\hat { j }$$) = a1$$\hat { i }$$ + a3$$\hat { k }$$
  $$\hat { k }$$ × ($$\overline { a }$$ × $$\hat { k }$$) = a1$$\hat { i }$$ + a2$$\hat { j }$$
  $$\hat { i }$$ × ($$\overline { a }$$ × $$\hat { i }$$) + $$\hat { j }$$ × ($$\overline { a }$$ × $$\hat { j }$$) + $$\hat { k }$$ × ($$\overline { a }$$ × $$\hat { k }$$)
  = 2a1$$\hat { i }$$ + 2a2$$\hat { j }$$ + 2a3$$\hat { k }$$
  = 2(a1$$\hat { i }$$ + a2$$\hat { j }$$ + a3$$\hat { k }$$)= 2 $$\overline { a }$$
  
  Question 3.
  Prove that [$$\overline { a }$$ – $$\overline { b }$$, $$\overline { b }$$ – $$\overline { c }$$, c – a] = 0.
  Solution:
  
  Hence proved.
  
  Question 4.
  If $$\overline { a }$$ = 2$$\hat { i }$$ + 3$$\hat { j }$$ – $$\hat { k }$$, $$\overline { b }$$ = 3$$\hat { i }$$ + 5$$\hat { j }$$ + 2$$\hat { k }$$, $$\overline { c }$$ = –$$\hat { i }$$ – 2$$\hat { j }$$ + 3$$\hat { k }$$
  (i) ($$\overline { a }$$ × $$\overline { b }$$) × $$\overline { c }$$ = ($$\overline { a }$$. $$\overline { c }$$)$$\overline { b }$$ – ($$\overline { b }$$.$$\overline { c }$$)$$\overline { a }$$
  (ii) $$\overline { a }$$ ($$\overline { b }$$ × $$\overline { c }$$) = ($$\overline { a }$$. $$\overline { c }$$)$$\overline { b }$$ – ($$\overline { a }$$.$$\overline { b }$$)$$\overline { c }$$
  Solution:
  
  Question 5.
  $$\overline { a }$$ = 2$$\hat { i }$$ + 3$$\hat { j }$$ – $$\hat { k }$$, $$\overline { b }$$ = –$$\hat { i }$$ + 2$$\hat { j }$$ – 4$$\hat { k }$$, $$\overline { c }$$ = $$\hat { i }$$ + $$\hat { j }$$ + $$\hat { k }$$ then find the value of ($$\overline { a }$$ × $$\overline { b }$$) – ($$\overline { a }$$ × $$\overline { c }$$)
  Solution:
  
  Question 6.
  If $$\overline { a }$$, $$\overline { b }$$, $$\overline { c }$$ and $$\overline { d }$$ are coplanar vectors, show that ($$\overline { a }$$ × $$\overline { b }$$) × ($$\overline { c }$$ × $$\overline { d }$$) = $$\overline { 0 }$$
  Solution:
  
  Question 7.
  If $$\overline { a }$$ = $$\hat { i }$$ + 2$$\hat { j }$$ + 3$$\hat { k }$$, $$\overline { b }$$ = 2$$\hat { i }$$ – $$\hat { j }$$ + $$\hat { k }$$, $$\overline { c }$$ = 3$$\hat { i }$$ + 2$$\hat { j }$$ + $$\hat { k }$$ and $$\overline { a }$$ × ($$\overline { b }$$ × $$\overline { c }$$) = l$$\overline { a }$$ + m$$\overline { b }$$ + n$$\overline { c }$$, find the values of l, m, n.
  Solution:
  
  m = 3 + 4 + 3; n = -(2 – 2 + 3)
  m = 10; n = -3
  l = 0; m = 10; n = -3
  
  Question 8.
  If $$\hat { a }$$, $$\hat { b }$$, $$\hat { c }$$ are three unit vectors such that $$\hat { b }$$ and $$\hat { c }$$ are non-parallel and $$\hat { a }$$ × ($$\hat { b }$$ × $$\hat { c }$$) = $$\frac { 1 }{ 2 }$$ $$\hat { b }$$ find the angle between $$\hat { a }$$ and $$\hat { c }$$.
  Solution: