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:
explanation: The document is a set of textbook questions and answers from a Tamil Nadu State Board 12th-grade mathematics guide, specifically from the chapter on Applications of Vector Algebra. It contains multiple questions and their corresponding solutions, all of which are related to vector algebra operations such as cross products and dot products.
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distilabel_metadata.content: 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:
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a set of textbook questions and answers from a Tamil Nadu State Board 12th-grade mathematics guide, specifically from the chapter on Applications of Vector Algebra. It contains multiple questions and their corresponding solutions, all of which are related to vector algebra operations such as cross products and dot products.
  
  Label: multiple-question-and-answer