Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.3

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) \( (\overline { a } \times \overline { b } ) \times \overline { c } \) (ii) \( \overline { a } \times (\overline { b } \times \overline { c }) \)
Answer:
(i) To find \( (\overline { a } \times \overline { b } ) \times \overline { c } \):
First, calculate the cross product of \( \overline { a } \) and \( \overline { b } \):
\( \overline { a } \times \overline { b } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -2 \end{array}\right| \)
\( = \hat { i }((-2)(-2) - (3)(1)) - \hat { j }((1)(-2) - (3)(2)) + \hat { k }((1)(1) - (-2)(2)) \)
\( = \hat { i }(4 - 3) - \hat { j }(-2 - 6) + \hat { k }(1 + 4) \)
\( = 1\hat { i } + 8\hat { j } + 5\hat { k } \)
Next, calculate the cross product of \( (\overline { a } \times \overline { b }) \) with \( \overline { c } \):
\( (\overline { a } \times \overline { b }) \times \overline { c } = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 8 & 5 \\ 3 & 2 & 1 \end{array}\right| \)
\( = \hat { i }((8)(1) - (5)(2)) - \hat { j }((1)(1) - (5)(3)) + \hat { k }((1)(2) - (8)(3)) \)
\( = \hat { i }(8 - 10) - \hat { j }(1 - 15) + \hat { k }(2 - 24) \)
\( = -2\hat { i } + 14\hat { j } - 22\hat { k } \)
(ii) To find \( \overline { a } \times (\overline { b } \times \overline { c }) \):
First, calculate the cross product of \( \overline { b } \) and \( \overline { c } \):
\( \overline { b } \times \overline { c } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{array}\right| \)
\( = \hat { i }((1)(1) - (-2)(2)) - \hat { j }((2)(1) - (-2)(3)) + \hat { k }((2)(2) - (1)(3)) \)
\( = \hat { i }(1 + 4) - \hat { j }(2 + 6) + \hat { k }(4 - 3) \)
\( = 5\hat { i } - 8\hat { j } + 1\hat { k } \)
Next, calculate the cross product of \( \overline { a } \) with \( (\overline { b } \times \overline { c }) \):
\( \overline { a } \times (\overline { b } \times \overline { c }) = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 5 & -8 & 1 \end{array}\right| \)
\( = \hat { i }((-2)(1) - (3)(-8)) - \hat { j }((1)(1) - (3)(5)) + \hat { k }((1)(-8) - (-2)(5)) \)
\( = \hat { i }(-2 + 24) - \hat { j }(1 - 15) + \hat { k }(-8 + 10) \)
\( = 22\hat { i } + 14\hat { j } + 2\hat { k } \)
In simple words: We find the vector products step-by-step. For part (i), we first cross product 'a' and 'b', then cross that result with 'c'. For part (ii), we first cross product 'b' and 'c', then cross 'a' with that result. Vector cross products are important in physics for calculating torque or magnetic forces.

🎯 Exam Tip: Remember to calculate determinants carefully for each cross product. Pay close attention to the order of operations in triple products, as \( (\overline{a} \times \overline{b}) \times \overline{c} \) is generally not equal to \( \overline{a} \times (\overline{b} \times \overline{c}) \).

Question 2. For any vector, prove that \( \hat { i } \times (\overline { a } \times \hat { i }) + \hat { j } \times (\overline { a } \times \hat { j }) + \hat { k } \times (\overline { a } \times \hat { k }) = 2\overline { a } \).
Answer:
Let \( \overline { a } = a_{1}\hat { i } + a_{2}\hat { j } + a_{3}\hat { k } \).
First, find \( \overline { a } \times \hat { i } \):
\( \overline { a } \times \hat { i } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 0 & 0 \end{array}\right| \)
\( = \hat { i }(a_{2}(0) - a_{3}(0)) - \hat { j }(a_{1}(0) - a_{3}(1)) + \hat { k }(a_{1}(0) - a_{2}(1)) \)
\( = \hat { i }(0) - \hat { j }(-a_{3}) + \hat { k }(-a_{2}) \)
\( = a_{3}\hat { j } - a_{2}\hat { k } \)
Next, find \( \hat { i } \times (\overline { a } \times \hat { i }) \):
\( \hat { i } \times (\overline { a } \times \hat { i }) = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & a_{3} & -a_{2} \end{array}\right| \)
\( = \hat { i }(0 - 0) - \hat { j }((1)(-a_{2}) - (0)(0)) + \hat { k }((1)(a_{3}) - (0)(0)) \)
\( = \hat { i }(0) - \hat { j }(-a_{2}) + \hat { k }(a_{3}) \)
\( = a_{2}\hat { j } + a_{3}\hat { k } \)
Similarly, for the other terms:
\( \hat { j } \times (\overline { a } \times \hat { j }) = a_{1}\hat { i } + a_{3}\hat { k } \)
\( \hat { k } \times (\overline { a } \times \hat { k }) = a_{1}\hat { i } + a_{2}\hat { j } \)
Now, sum all three terms:
\( \hat { i } \times (\overline { a } \times \hat { i }) + \hat { j } \times (\overline { a } \times \hat { j }) + \hat { k } \times (\overline { a } \times \hat { k }) \)
\( = (a_{2}\hat { j } + a_{3}\hat { k }) + (a_{1}\hat { i } + a_{3}\hat { k }) + (a_{1}\hat { i } + a_{2}\hat { j }) \)
\( = 2a_{1}\hat { i } + 2a_{2}\hat { j } + 2a_{3}\hat { k } \)
\( = 2(a_{1}\hat { i } + a_{2}\hat { j } + a_{3}\hat { k }) \)
\( = 2\overline { a } \)
Hence proved.
In simple words: We break down the main vector 'a' into its parts along the x, y, and z axes. Then, we apply the cross product operations with the unit vectors 'i', 'j', and 'k'. When we add up the results, it simply gives us two times the original vector 'a'. This identity shows a fascinating relationship between a vector and its cross products with unit vectors along the coordinate axes.

🎯 Exam Tip: This identity is a common result that can be directly used in problems. Remember the vector triple product identity \( \overline{A} \times (\overline{B} \times \overline{C}) = (\overline{A} \cdot \overline{C})\overline{B} - (\overline{A} \cdot \overline{B})\overline{C} \) as an alternative way to prove this.

Question 3. Prove that \( [\overline { a } - \overline { b }, \overline { b } - \overline { c }, \overline { c } - \overline { a }] = 0 \).
Answer:
We need to prove \( [\overline { a } - \overline { b }, \overline { b } - \overline { c }, \overline { c } - \overline { a }] = 0 \).
Using the definition of the scalar triple product, \( [\overline{u}, \overline{v}, \overline{w}] = \overline{u} \cdot (\overline{v} \times \overline{w}) \):
\( [\overline { a } - \overline { b }, \overline { b } - \overline { c }, \overline { c } - \overline { a }] = (\overline { a } - \overline { b }) \cdot ((\overline { b } - \overline { c }) \times (\overline { c } - \overline { a })) \)
First, expand the cross product part: \( (\overline { b } - \overline { c }) \times (\overline { c } - \overline { a }) \)
\( = (\overline { b } \times \overline { c }) - (\overline { b } \times \overline { a }) - (\overline { c } \times \overline { c }) + (\overline { c } \times \overline { a }) \)
Since the cross product of a vector with itself is zero, \( \overline{c} \times \overline{c} = \overline{0} \):
\( = (\overline { b } \times \overline { c }) - (\overline { b } \times \overline { a }) + (\overline { c } \times \overline { a }) \)
We know that \( -(\overline { b } \times \overline { a }) = (\overline { a } \times \overline { b }) \):
\( = (\overline { b } \times \overline { c }) + (\overline { a } \times \overline { b }) + (\overline { c } \times \overline { a }) \)
Now, substitute this back into the scalar triple product:
\( (\overline { a } - \overline { b }) \cdot [(\overline { b } \times \overline { c }) + (\overline { a } \times \overline { b }) + (\overline { c } \times \overline { a })] \)
Distribute the dot product:
\( = \overline { a } \cdot (\overline { b } \times \overline { c }) + \overline { a } \cdot (\overline { a } \times \overline { b }) + \overline { a } \cdot (\overline { c } \times \overline { a }) - \overline { b } \cdot (\overline { b } \times \overline { c }) - \overline { b } \cdot (\overline { a } \times \overline { b }) - \overline { b } \cdot (\overline { c } \times \overline { a }) \)
Recall that \( \overline{x} \cdot (\overline{x} \times \overline{y}) = 0 \) (since \( \overline{x} \times \overline{y} \) is perpendicular to \( \overline{x} \)) and \( [\overline{x}, \overline{y}, \overline{z}] = \overline{x} \cdot (\overline{y} \times \overline{z}) \):
\( = [\overline { a }, \overline { b }, \overline { c }] + 0 + 0 - 0 - 0 - [\overline { b }, \overline { c }, \overline { a }] \)
Since \( [\overline { b }, \overline { c }, \overline { a }] = [\overline { a }, \overline { b }, \overline { c }] \) (cyclic permutation property of scalar triple product):
\( = [\overline { a }, \overline { b }, \overline { c }] - [\overline { a }, \overline { b }, \overline { c }] \)
\( = 0 \)
Hence proved.
In simple words: This problem asks us to show that if we have three vectors and create new vectors by subtracting them cyclically, the scalar triple product of these new vectors will always be zero. This happens because the three new vectors are coplanar. This property highlights that if three vectors are coplanar, their scalar triple product is zero, reflecting that the volume formed by them is zero.

🎯 Exam Tip: Remember the properties of the scalar triple product, especially that \( \overline{X} \cdot (\overline{Y} \times \overline{Z}) \) is zero if any two vectors are identical or if all three vectors are coplanar. Cyclic permutation \( [\overline{a}, \overline{b}, \overline{c}] = [\overline{b}, \overline{c}, \overline{a}] = [\overline{c}, \overline{a}, \overline{b}] \) is also key.

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) \( (\overline { a } \times \overline { b }) \times \overline { c } = (\overline { a } \cdot \overline { c })\overline { b } – (\overline { b } \cdot \overline { c })\overline { a } \)
(ii) \( \overline { a } (\overline { b } \times \overline { c }) = (\overline { a } \cdot \overline { c })\overline { b } – (\overline { a } \cdot \overline { b })\overline { c } \)
Answer:
(i) To prove \( (\overline { a } \times \overline { b }) \times \overline { c } = (\overline { a } \cdot \overline { c })\overline { b } – (\overline { b } \cdot \overline { c })\overline { a } \):
Calculate LHS: \( (\overline { a } \times \overline { b }) \times \overline { c } \)
First, find \( \overline { a } \times \overline { b } \):
\( \overline { a } \times \overline { b } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 3 & 5 & 2 \end{array}\right| \)
\( = \hat { i }( (3)(2) - (-1)(5) ) - \hat { j }( (2)(2) - (-1)(3) ) + \hat { k }( (2)(5) - (3)(3) ) \)
\( = \hat { i }(6 + 5) - \hat { j }(4 + 3) + \hat { k }(10 - 9) \)
\( = 11\hat { i } - 7\hat { j } + 1\hat { k } \)
Next, find \( (\overline { a } \times \overline { b }) \times \overline { c } \):
\( (\overline { a } \times \overline { b }) \times \overline { c } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 11 & -7 & 1 \\ -1 & -2 & 3 \end{array}\right| \)
\( = \hat { i }( (-7)(3) - (1)(-2) ) - \hat { j }( (11)(3) - (1)(-1) ) + \hat { k }( (11)(-2) - (-7)(-1) ) \)
\( = \hat { i }(-21 + 2) - \hat { j }(33 + 1) + \hat { k }(-22 - 7) \)
\( = -19\hat { i } - 34\hat { j } - 29\hat { k } \) ... (1)
Calculate RHS: \( (\overline { a } \cdot \overline { c })\overline { b } – (\overline { b } \cdot \overline { c })\overline { a } \)
First, find \( \overline { a } \cdot \overline { c } \):
\( \overline { a } \cdot \overline { c } = (2\hat { i } + 3\hat { j } - \hat { k }) \cdot (-\hat { i } - 2\hat { j } + 3\hat { k }) \)
\( = (2)(-1) + (3)(-2) + (-1)(3) = -2 - 6 - 3 = -11 \)
Then, find \( (\overline { a } \cdot \overline { c })\overline { b } \):
\( (\overline { a } \cdot \overline { c })\overline { b } = -11(3\hat { i } + 5\hat { j } + 2\hat { k }) = -33\hat { i } - 55\hat { j } - 22\hat { k } \)
Next, find \( \overline { b } \cdot \overline { c } \):
\( \overline { b } \cdot \overline { c } = (3\hat { i } + 5\hat { j } + 2\hat { k }) \cdot (-\hat { i } - 2\hat { j } + 3\hat { k }) \)
\( = (3)(-1) + (5)(-2) + (2)(3) = -3 - 10 + 6 = -7 \)
Then, find \( (\overline { b } \cdot \overline { c })\overline { a } \):
\( (\overline { b } \cdot \overline { c })\overline { a } = -7(2\hat { i } + 3\hat { j } - \hat { k }) = -14\hat { i } - 21\hat { j } + 7\hat { k } \)
Finally, calculate \( (\overline { a } \cdot \overline { c })\overline { b } – (\overline { b } \cdot \overline { c })\overline { a } \):
\( = (-33\hat { i } - 55\hat { j } - 22\hat { k }) - (-14\hat { i } - 21\hat { j } + 7\hat { k }) \)
\( = -33\hat { i } - 55\hat { j } - 22\hat { k } + 14\hat { i } + 21\hat { j } - 7\hat { k } \)
\( = -19\hat { i } - 34\hat { j } - 29\hat { k } \) ... (2)
From (1) and (2), LHS = RHS. Hence proved.

(ii) To prove \( \overline { a } \times (\overline { b } \times \overline { c }) = (\overline { a } \cdot \overline { c })\overline { b } – (\overline { a } \cdot \overline { b })\overline { c } \) (interpreting the question as vector triple product, as per solution):
Calculate LHS: \( \overline { a } \times (\overline { b } \times \overline { c }) \)
First, find \( \overline { b } \times \overline { c } \):
\( \overline { b } \times \overline { c } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ -1 & -2 & 3 \end{array}\right| \)
\( = \hat { i }( (5)(3) - (2)(-2) ) - \hat { j }( (3)(3) - (2)(-1) ) + \hat { k }( (3)(-2) - (5)(-1) ) \)
\( = \hat { i }(15 + 4) - \hat { j }(9 + 2) + \hat { k }(-6 + 5) \)
\( = 19\hat { i } - 11\hat { j } - 1\hat { k } \)
Next, find \( \overline { a } \times (\overline { b } \times \overline { c }) \):
\( \overline { a } \times (\overline { b } \times \overline { c }) = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 19 & -11 & -1 \end{array}\right| \)
\( = \hat { i }( (3)(-1) - (-1)(-11) ) - \hat { j }( (2)(-1) - (-1)(19) ) + \hat { k }( (2)(-11) - (3)(19) ) \)
\( = \hat { i }(-3 - 11) - \hat { j }(-2 + 19) + \hat { k }(-22 - 57) \)
\( = -14\hat { i } - 17\hat { j } - 79\hat { k } \) ... (1)
Calculate RHS: \( (\overline { a } \cdot \overline { c })\overline { b } – (\overline { a } \cdot \overline { b })\overline { c } \)
From part (i), we know \( \overline { a } \cdot \overline { c } = -11 \).
So, \( (\overline { a } \cdot \overline { c })\overline { b } = -11(3\hat { i } + 5\hat { j } + 2\hat { k }) = -33\hat { i } - 55\hat { j } - 22\hat { k } \)
Next, find \( \overline { a } \cdot \overline { b } \):
\( \overline { a } \cdot \overline { b } = (2\hat { i } + 3\hat { j } - \hat { k }) \cdot (3\hat { i } + 5\hat { j } + 2\hat { k }) \)
\( = (2)(3) + (3)(5) + (-1)(2) = 6 + 15 - 2 = 19 \)
Then, find \( (\overline { a } \cdot \overline { b })\overline { c } \):
\( (\overline { a } \cdot \overline { b })\overline { c } = 19(-\hat { i } - 2\hat { j } + 3\hat { k }) = -19\hat { i } - 38\hat { j } + 57\hat { k } \)
Finally, calculate \( (\overline { a } \cdot \overline { c })\overline { b } – (\overline { a } \cdot \overline { b })\overline { c } \):
\( = (-33\hat { i } - 55\hat { j } - 22\hat { k }) - (-19\hat { i } - 38\hat { j } + 57\hat { k }) \)
\( = -33\hat { i } - 55\hat { j } - 22\hat { k } + 19\hat { i } + 38\hat { j } - 57\hat { k } \)
\( = -14\hat { i } - 17\hat { j } - 79\hat { k } \) ... (2)
From (1) and (2), LHS = RHS. Hence proved.
In simple words: This problem asks us to verify a special rule for how three vectors multiply. For both parts, we calculate the left side (LHS) and the right side (RHS) of the given equations separately. If both sides give the same vector, then the rule is proven. This problem demonstrates the important vector triple product identity, which simplifies complex vector operations.

🎯 Exam Tip: The vector triple product identity \( \overline{A} \times (\overline{B} \times \overline{C}) = (\overline{A} \cdot \overline{C})\overline{B} - (\overline{A} \cdot \overline{B})\overline{C} \) is fundamental in vector algebra. Memorize this identity and practice its application for maximum marks.

Question 5. If \( \overline { a } = 2\hat { i } + 3\hat { j } – \hat { k }, \overline { b } = −\hat { i } + 2\hat { j } – 4\hat { k }, \overline { c } = \hat { i } + \hat { j } + \hat { k } \) then find the value of \( (\overline { a } \times \overline { b }) – (\overline { a } \times \overline { c }) \)
Answer:
First, calculate \( \overline { a } \times \overline { b } \):
\( \overline { a } \times \overline { b } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{array}\right| \)
\( = \hat { i }( (3)(-4) - (-1)(2) ) - \hat { j }( (2)(-4) - (-1)(-1) ) + \hat { k }( (2)(2) - (3)(-1) ) \)
\( = \hat { i }(-12 + 2) - \hat { j }(-8 - 1) + \hat { k }(4 + 3) \)
\( = -10\hat { i } + 9\hat { j } + 7\hat { k } \)
Next, calculate \( \overline { a } \times \overline { c } \):
\( \overline { a } \times \overline { c } = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{array}\right| \)
\( = \hat { i }( (3)(1) - (-1)(1) ) - \hat { j }( (2)(1) - (-1)(1) ) + \hat { k }( (2)(1) - (3)(1) ) \)
\( = \hat { i }(3 + 1) - \hat { j }(2 + 1) + \hat { k }(2 - 3) \)
\( = 4\hat { i } - 3\hat { j } - 1\hat { k } \)
Finally, find the value of \( (\overline { a } \times \overline { b }) – (\overline { a } \times \overline { c }) \):
\( = (-10\hat { i } + 9\hat { j } + 7\hat { k }) - (4\hat { i } - 3\hat { j } - \hat { k }) \)
\( = -10\hat { i } + 9\hat { j } + 7\hat { k } - 4\hat { i } + 3\hat { j } + \hat { k } \)
\( = (-10 - 4)\hat { i } + (9 + 3)\hat { j } + (7 + 1)\hat { k } \)
\( = -14\hat { i } + 12\hat { j } + 8\hat { k } \)
In simple words: We first find the cross product of vector 'a' with vector 'b', and then the cross product of vector 'a' with vector 'c'. After finding these two new vectors, we subtract the second one from the first one. This calculation demonstrates how the distributive property works with vector cross products when a common vector is involved.

🎯 Exam Tip: When performing vector subtraction, remember to change the signs of all components of the vector being subtracted. Double-check your determinant calculations to avoid errors.

Question 6. If \( \overline { a }, \overline { b }, \overline { c } \) and \( \overline { d } \) are coplanar vectors, show that \( (\overline { a } \times \overline { b }) \times (\overline { c } \times \overline { d }) = \overline { 0 } \).
Answer:
Given that the vectors \( \overline { a }, \overline { b }, \overline { c }, \) and \( \overline { d } \) are coplanar.
We know that the vector \( \overline { a } \times \overline { b } \) is perpendicular to the plane containing \( \overline { a } \) and \( \overline { b } \).
Similarly, the vector \( \overline { c } \times \overline { d } \) is perpendicular to the plane containing \( \overline { c } \) and \( \overline { d } \).
Since all four vectors \( \overline { a }, \overline { b }, \overline { c }, \overline { d } \) lie in the same plane, both \( \overline { a } \times \overline { b } \) and \( \overline { c } \times \overline { d } \) must be perpendicular to this common plane.
This means that \( \overline { a } \times \overline { b } \) and \( \overline { c } \times \overline { d } \) are parallel vectors.
The cross product of two parallel vectors is always the zero vector.
Therefore, \( (\overline { a } \times \overline { b }) \times (\overline { c } \times \overline { d }) = \overline { 0 } \).
Hence proved.
In simple words: When four vectors are in the same flat surface (coplanar), the cross product of the first two vectors and the cross product of the last two vectors will both be straight up or straight down from that surface. This means these two resulting vectors are parallel to each other. When you cross product two parallel vectors, the answer is always zero. This property is a powerful consequence of vector perpendicularity and parallelism within a single plane.

🎯 Exam Tip: The key idea here is that a vector resulting from a cross product is always perpendicular to the plane formed by the original two vectors. If multiple vectors are coplanar, any cross product formed from pairs of these vectors will be parallel to each other.

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 \( \overline { a } \times (\overline { b } \times \overline { c }) = l\overline { a } + m\overline { b } + n\overline { c } \), find the values of \( l, m, n \).
Answer:
We know the vector triple product identity:
\( \overline { a } \times (\overline { b } \times \overline { c }) = (\overline { a } \cdot \overline { c })\overline { b } – (\overline { a } \cdot \overline { b })\overline { c } \)
Given that \( \overline { a } \times (\overline { b } \times \overline { c }) = l\overline { a } + m\overline { b } + n\overline { c } \).
Comparing the two expressions, we can deduce:
\( l = 0 \)
\( m = \overline { a } \cdot \overline { c } \)
\( n = -(\overline { a } \cdot \overline { b }) \)
Now, calculate the dot products:
\( \overline { a } \cdot \overline { c } = (\hat { i } + 2\hat { j } + 3\hat { k }) \cdot (3\hat { i } + 2\hat { j } + \hat { k }) \)
\( = (1)(3) + (2)(2) + (3)(1) = 3 + 4 + 3 = 10 \)
So, \( m = 10 \).
Next, calculate \( \overline { a } \cdot \overline { b } \):
\( \overline { a } \cdot \overline { b } = (\hat { i } + 2\hat { j } + 3\hat { k }) \cdot (2\hat { i } – \hat { j } + \hat { k }) \)
\( = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3 \)
So, \( n = -(3) = -3 \).
Therefore, the values are \( l = 0, m = 10, \) and \( n = -3 \).
In simple words: This problem uses a special rule for multiplying three vectors to simplify the left side of the equation. Once simplified, we compare it to the right side of the equation to find the unknown numbers 'l', 'm', and 'n'. This problem highlights how the vector triple product identity simplifies complex expressions into a linear combination of vectors.

🎯 Exam Tip: Always remember the vector triple product identity. This identity is extremely useful for expressing a vector triple product as a linear combination of the other two vectors involved, making it easier to solve for coefficients like \( l, m, n \).

Question 8. If \( \hat { a }, \hat { b }, \hat { c } \) are three unit vectors such that \( \hat { b } \) and \( \hat { c } \) are non-parallel and \( \hat { a } \times (\hat { b } \times \hat { c }) = \frac { 1 }{ 2 } \hat { b } \), find the angle between \( \hat { a } \) and \( \hat { c } \).
Answer:
Given that \( \hat { a }, \hat { b }, \hat { c } \) are unit vectors, which means their magnitudes are 1:
\( |\hat{a}| = |\hat{b}| = |\hat{c}| = 1 \).
We know the vector triple product identity:
\( \hat { a } \times (\hat { b } \times \hat { c }) = (\hat { a } \cdot \hat { c })\hat { b } – (\hat { a } \cdot \hat { b })\hat { c } \)
We are given \( \hat { a } \times (\hat { b } \times \hat { c }) = \frac { 1 }{ 2 } \hat { b } \).
Equating the two expressions:
\( (\hat { a } \cdot \hat { c })\hat { b } – (\hat { a } \cdot \hat { b })\hat { c } = \frac { 1 }{ 2 } \hat { b } \)
Rearrange the terms to form a linear combination of \( \hat { b } \) and \( \hat { c } \):
\( (\hat { a } \cdot \hat { c } - \frac{1}{2})\hat { b } – (\hat { a } \cdot \hat { b })\hat { c } = \overline{0} \)
Since \( \hat { b } \) and \( \hat { c } \) are non-parallel, they are linearly independent. For the linear combination to be a zero vector, their coefficients must both be zero.
Therefore, we have two equations:
1. \( \hat { a } \cdot \hat { c } - \frac{1}{2} = 0 \)
\( \hat { a } \cdot \hat { c } = \frac{1}{2} \)
2. \( -(\hat { a } \cdot \hat { b }) = 0 \)
\( \hat { a } \cdot \hat { b } = 0 \)
We need to find the angle between \( \hat { a } \) and \( \hat { c } \). Let this angle be \( \theta \).
Using the dot product formula:
\( \hat { a } \cdot \hat { c } = |\hat{a}||\hat{c}|\cos\theta \)
Substitute the known values:
\( \frac{1}{2} = (1)(1)\cos\theta \)
\( \cos\theta = \frac{1}{2} \)
This means the angle \( \theta \) is \( \frac{\pi}{3} \) radians or \( 60^\circ \).
In simple words: We use a special rule to change the complicated multiplication of three vectors into a simpler form. Then, we compare it with what the question gives us. Since two of the vectors are not parallel, we can easily find the dot product of 'a' and 'c'. From that dot product, we can calculate the angle between vectors 'a' and 'c'. Understanding the properties of unit vectors and their products is crucial for solving problems in geometry and physics.

🎯 Exam Tip: The condition that two vectors are "non-parallel" is crucial, as it implies linear independence, allowing you to equate their coefficients to zero in a vector equation. Remember the dot product definition for finding angles.