Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.2

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 \( \overline{a} \cdot (\overline{b} \times \overline{c}) \).
Answer: To find \( \overline{a} \cdot (\overline{b} \times \overline{c}) \), we calculate the scalar triple product, which can be represented as a determinant of the coefficients of the vectors. This value shows the volume of the parallelepiped formed by the three vectors.
\[ \overline{a} \cdot (\overline{b} \times \overline{c}) = [ \overline{a}, \overline{b}, \overline{c} ] = \left|\begin{array}{ccc} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{array}\right| \] We expand the determinant:
\[ = 1((1)(1) - (-2)(2)) - (-2)((2)(1) - (-2)(3)) + 3((2)(2) - (1)(3)) \] \[ = 1(1 + 4) + 2(2 + 6) + 3(4 – 3) \] \[ = 5 + 16 + 3 \] \[ = 24 \] The scalar triple product is 24.
In simple words: We put the numbers from each vector into a grid and solve it like a puzzle. The final number, 24, is how we find the specific product of these three vectors.

🎯 Exam Tip: Remember that \( \overline{a} \cdot (\overline{b} \times \overline{c}) \) is the scalar triple product, which can be found using a determinant of the vector components. Ensure you correctly apply the signs for each term in the determinant expansion.

Question 2. Find the volume of the parallelepiped whose coterminous edges are represented by the vectors \( -6\hat{i} + 14\hat{j} + 10\hat{k} \), \( 14\hat{i} – 10\hat{j} – 6\hat{k} \) and \( 2\hat{i} + 4\hat{j} – 2\hat{k} \).
Answer: The volume of a parallelepiped formed by three coterminous vectors \( \overline{a}, \overline{b}, \overline{c} \) is given by the absolute value of their scalar triple product, \( |[\overline{a}, \overline{b}, \overline{c}]| \). The scalar triple product is calculated as the determinant of a matrix formed by the vector components.
Let \( \overline{a} = -6\hat{i} + 14\hat{j} + 10\hat{k} \)
Let \( \overline{b} = 14\hat{i} – 10\hat{j} – 6\hat{k} \)
Let \( \overline{c} = 2\hat{i} + 4\hat{j} – 2\hat{k} \)
Volume of the parallelepiped \( = [ \overline{a}, \overline{b}, \overline{c} ] \)
\[ = \left|\begin{array}{ccc} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{array}\right| \] Now, we expand the determinant:
\[ = -6((-10)(-2) - (-6)(4)) - 14((14)(-2) - (-6)(2)) + 10((14)(4) - (-10)(2)) \] \[ = -6(20 + 24) - 14(-28 + 12) + 10(56 + 20) \] \[ = -6(44) - 14(-16) + 10(76) \] \[ = -264 + 224 + 760 \] \[ = 720 \] So, the volume of the parallelepiped is 720 cubic units. The volume is always a positive value.
In simple words: We arrange the numbers from the vectors into a special grid and solve it. The answer we get, 720, tells us how much space the 3D shape (parallelepiped) takes up.

🎯 Exam Tip: Always remember that volume is a scalar quantity and must be positive. If your determinant calculation results in a negative value, take its absolute value for the final answer.

Question 3. The volume of the parallelepiped whose coterminous edges are \( 7\hat{i} + \lambda\hat{j} – 3\hat{k} \), \( \hat{i} + 2\hat{j} – \hat{k} \), \( -3\hat{i} + 7\hat{j} + 5\hat{k} \) is 90 cubic units. Find the value of \( \lambda \).
Answer: We are given the volume of the parallelepiped as 90 cubic units. We can set up the scalar triple product (determinant) of the three vectors and equate it to 90 to find the unknown value of \( \lambda \).
Let \( \overline{a} = 7\hat{i} + \lambda\hat{j} – 3\hat{k} \)
Let \( \overline{b} = \hat{i} + 2\hat{j} – \hat{k} \)
Let \( \overline{c} = -3\hat{i} + 7\hat{j} + 5\hat{k} \)
Volume of the parallelepiped \( = [\overline{a}, \overline{b}, \overline{c}] \)
\[ \left|\begin{array}{ccc} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{array}\right| = 90 \] Now, expand the determinant:
\[ 7((2)(5) - (-1)(7)) - \lambda((1)(5) - (-1)(-3)) - 3((1)(7) - (2)(-3)) = 90 \] \[ 7(10 + 7) – \lambda(5 – 3) – 3(7 + 6) = 90 \] \[ 7(17) – \lambda(2) – 3(13) = 90 \] \[ 119 – 2\lambda – 39 = 90 \] Next, combine the constant terms:
\[ 80 – 2\lambda = 90 \] Now, isolate \( \lambda \):
\[ -2\lambda = 90 - 80 \] \[ -2\lambda = 10 \]
\( \implies \) \[ \lambda = \frac{-10}{2} \]
\( \implies \) \[ \lambda = -5 \] The value of \( \lambda \) is -5.
In simple words: We know how much space the 3D shape takes up. We use a formula with a missing number, \( \lambda \). By solving the equation, we find that the missing number \( \lambda \) is -5.

🎯 Exam Tip: When solving for an unknown variable in a determinant, be very careful with the signs during expansion and algebraic manipulation. Double-check your calculations, especially when dealing with negative numbers.

Question 4. If \( \overline{a} \), \( \overline{b} \), \( \overline{c} \) are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of \( (\overline{a} + \overline{b}) \cdot (\overline{b} \times \overline{c}) + (\overline{b} + \overline{c}) \cdot (\overline{c} \times \overline{a}) + (\overline{c} + \overline{a}) \cdot (\overline{a} \times \overline{b}) \).
Answer: We are given that \( [\overline{a}, \overline{b}, \overline{c}] = \pm 4 \), since volume is a magnitude. We need to evaluate the expression by expanding each scalar triple product and using the properties that \( [\overline{x}, \overline{y}, \overline{z}] = [\overline{y}, \overline{z}, \overline{x}] = [\overline{z}, \overline{x}, \overline{y}] \) and if any two vectors in a scalar triple product are identical, the product is zero. Also, \( (\overline{x} + \overline{y}) \cdot (\overline{p} \times \overline{q}) = \overline{x} \cdot (\overline{p} \times \overline{q}) + \overline{y} \cdot (\overline{p} \times \overline{q}) \).
Let's expand each term:
1. \( (\overline{a} + \overline{b}) \cdot (\overline{b} \times \overline{c}) = \overline{a} \cdot (\overline{b} \times \overline{c}) + \overline{b} \cdot (\overline{b} \times \overline{c}) = [\overline{a}, \overline{b}, \overline{c}] + [\overline{b}, \overline{b}, \overline{c}] \) Since \( [\overline{b}, \overline{b}, \overline{c}] = 0 \) (two vectors are identical), this simplifies to \( [\overline{a}, \overline{b}, \overline{c}] \).
2. \( (\overline{b} + \overline{c}) \cdot (\overline{c} \times \overline{a}) = \overline{b} \cdot (\overline{c} \times \overline{a}) + \overline{c} \cdot (\overline{c} \times \overline{a}) = [\overline{b}, \overline{c}, \overline{a}] + [\overline{c}, \overline{c}, \overline{a}] \) Again, \( [\overline{c}, \overline{c}, \overline{a}] = 0 \). Using the cyclic property, \( [\overline{b}, \overline{c}, \overline{a}] = [\overline{a}, \overline{b}, \overline{c}] \). So this term simplifies to \( [\overline{a}, \overline{b}, \overline{c}] \).
3. \( (\overline{c} + \overline{a}) \cdot (\overline{a} \times \overline{b}) = \overline{c} \cdot (\overline{a} \times \overline{b}) + \overline{a} \cdot (\overline{a} \times \overline{b}) = [\overline{c}, \overline{a}, \overline{b}] + [\overline{a}, \overline{a}, \overline{b}] \) Here, \( [\overline{a}, \overline{a}, \overline{b}] = 0 \). Using the cyclic property, \( [\overline{c}, \overline{a}, \overline{b}] = [\overline{a}, \overline{b}, \overline{c}] \). So this term also simplifies to \( [\overline{a}, \overline{b}, \overline{c}] \).
Adding these three simplified terms together:
The total expression \( = [\overline{a}, \overline{b}, \overline{c}] + [\overline{a}, \overline{b}, \overline{c}] + [\overline{a}, \overline{b}, \overline{c}] \) \[ = 3[\overline{a}, \overline{b}, \overline{c}] \] Since the volume of the parallelepiped is 4 cubic units, we have \( [\overline{a}, \overline{b}, \overline{c}] = \pm 4 \). Therefore, the value of the expression is \( 3(\pm 4) = \pm 12 \).
In simple words: We have a special calculation with three vectors. Because of how these vector products work, many parts cancel out or repeat. In the end, we find that the whole complicated expression is just three times the volume of the shape made by the vectors, which is \( 3 \times (\pm 4) = \pm 12 \).

🎯 Exam Tip: Remember the key properties of the scalar triple product: cyclic permutation doesn't change the value, and if any two vectors are identical, the product is zero. These properties significantly simplify such expressions.

Question 5. Find the altitude of a parallelepiped determined by the vectors \( \overline{a} = -2\hat{i} + 5\hat{j} + 3\hat{k} \), \( \overline{b} = \hat{i} + 3\hat{j} – 2\hat{k} \) and \( \overline{c} = -3\hat{i} + \hat{j} + 4\hat{k} \) if the base is taken as the parallelogram determined by \( \overline{b} \) and \( \overline{c} \).
Answer: The altitude (height) of a parallelepiped is found by dividing its volume by the area of its base. Here, the base is formed by vectors \( \overline{b} \) and \( \overline{c} \), so its area is \( |\overline{b} \times \overline{c}| \). The volume of the parallelepiped is \( |[\overline{a}, \overline{b}, \overline{c}]| \).
First, let's find the volume \( V = |[\overline{a}, \overline{b}, \overline{c}]| \):
\[ [\overline{a}, \overline{b}, \overline{c}] = \left|\begin{array}{ccc} -2 & 5 & 3 \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{array}\right| \] Expand the determinant:
\[ = -2((3)(4) - (-2)(1)) - 5((1)(4) - (-2)(-3)) + 3((1)(1) - (3)(-3)) \] \[ = -2(12 + 2) - 5(4 – 6) + 3(1 + 9) \] \[ = -2(14) - 5(-2) + 3(10) \] \[ = -28 + 10 + 30 \] \[ = 12 \] So, the volume \( V = 12 \) cubic units.
Next, let's find the area of the base, which is \( |\overline{b} \times \overline{c}| \). We first calculate \( \overline{b} \times \overline{c} \):
\[ \overline{b} \times \overline{c} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{array}\right| \] Expand this determinant:
\[ = \hat{i}((3)(4) - (-2)(1)) - \hat{j}((1)(4) - (-2)(-3)) + \hat{k}((1)(1) - (3)(-3)) \] \[ = \hat{i}(12 + 2) - \hat{j}(4 – 6) + \hat{k}(1 + 9) \] \[ = 14\hat{i} + 2\hat{j} + 10\hat{k} \] Now, find the magnitude of \( \overline{b} \times \overline{c} \):
\[ |\overline{b} \times \overline{c}| = \sqrt{14^2 + 2^2 + 10^2} \] \[ = \sqrt{196 + 4 + 100} \] \[ = \sqrt{300} \] We can simplify \( \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3} \).
So, the Area \( = 10\sqrt{3} \) square units.
Finally, calculate the altitude \( h \):
\[ h = \frac{V}{\text{Area}} = \frac{12}{10\sqrt{3}} \] To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\[ h = \frac{12 \times \sqrt{3}}{10\sqrt{3} \times \sqrt{3}} = \frac{12\sqrt{3}}{10 \times 3} = \frac{12\sqrt{3}}{30} \] Simplify the fraction:
\[ h = \frac{2\sqrt{3}}{5} \] The altitude of the parallelepiped is \( \frac{2\sqrt{3}}{5} \) units. This calculation finds the perpendicular distance from the top face to the base.
In simple words: We first find the total volume of the 3D shape using all three vectors. Then we find the area of the bottom part using two of the vectors. To get the height, we just divide the total volume by the area of the bottom part.

🎯 Exam Tip: Remember that altitude is Volume / Base Area. Ensure you correctly identify which vectors form the base for calculating its area (magnitude of their cross product) and which are used for the volume (scalar triple product).

Question 6. Determine whether the three vectors \( 2\hat{i} + 3\hat{j} + \hat{k} \), \( \hat{i} – 2\hat{j} + 2\hat{k} \) and \( 3\hat{i} + \hat{j} + 3\hat{k} \) are coplanar.
Answer: Three vectors are coplanar if their scalar triple product is zero. We will calculate the scalar triple product of the given vectors.
Let \( \overline{a} = 2\hat{i} + 3\hat{j} + \hat{k} \)
Let \( \overline{b} = \hat{i} – 2\hat{j} + 2\hat{k} \)
Let \( \overline{c} = 3\hat{i} + \hat{j} + 3\hat{k} \)
If vectors are coplanar, then \( [\overline{a}, \overline{b}, \overline{c}] = 0 \).
We set up the determinant:
\[ [\overline{a}, \overline{b}, \overline{c}] = \left|\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -2 & 2 \\ 3 & 1 & 3 \end{array}\right| \] Now, expand the determinant:
\[ = 2((-2)(3) - (2)(1)) - 3((1)(3) - (2)(3)) + 1((1)(1) - (-2)(3)) \] \[ = 2(-6 – 2) - 3(3 – 6) + 1(1 + 6) \] \[ = 2(-8) - 3(-3) + 1(7) \] \[ = -16 + 9 + 7 \] \[ = 0 \] Since the scalar triple product is 0, the given vectors are coplanar, meaning they lie on the same plane.
In simple words: We calculate a special number for these three vectors. If this number is zero, it means all three vectors lie flat on the same surface, like drawing lines on a piece of paper. Since we got zero, they are coplanar.

🎯 Exam Tip: The condition for coplanarity of three vectors is that their scalar triple product is zero. A common mistake is forgetting to take the determinant, or making calculation errors during expansion.

Question 7. Let \( \overline{a} = \hat{i} + \hat{j} + \hat{k} \), \( \overline{b} = \hat{i} \) and \( \overline{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \). If \( c_1 = 1 \) and \( c_2 = 2 \), find \( c_3 \) such that \( \overline{a} \), \( \overline{b} \) and \( \overline{c} \) are coplanar.
Answer: For three vectors to be coplanar, their scalar triple product must be zero. We are given the vectors \( \overline{a} \), \( \overline{b} \), and \( \overline{c} \) with some known and unknown coefficients. We can set up the determinant and solve for \( c_3 \).
The vectors are:
\( \overline{a} = 1\hat{i} + 1\hat{j} + 1\hat{k} \)
\( \overline{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} \)
\( \overline{c} = 1\hat{i} + 2\hat{j} + c_3\hat{k} \) (since \( c_1 = 1 \) and \( c_2 = 2 \))
If \( \overline{a} \), \( \overline{b} \) and \( \overline{c} \) are coplanar, then \( [\overline{a}, \overline{b}, \overline{c}] = 0 \).
\[ \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_3 \end{array}\right| = 0 \] Now, expand the determinant, which is easiest along the second row:
\[ 1((1)(c_3) - (1)(2)) - 0(\text{any value}) + 0(\text{any value}) = 0 \] \[ 1(0 - (1)(c_3)) - 1(0 - (0)(1)) + 1((1)(2) - (0)(1)) = 0 \] Using expansion along the first row (as per the source):
\[ 1((0)(c_3) - (0)(2)) - 1((1)(c_3) - (0)(1)) + 1((1)(2) - (0)(1)) = 0 \] \[ 1(0) - 1(c_3) + 1(2) = 0 \] \[ 0 - c_3 + 2 = 0 \] \[ -c_3 + 2 = 0 \]
\( \implies \) \[ c_3 = 2 \] The value of \( c_3 \) is 2. This ensures all three vectors lie on the same plane.
In simple words: We have three vectors, but one number is missing. We know that if these vectors lie on the same flat surface, a special calculation involving their numbers must equal zero. By solving this calculation, we find the missing number, \( c_3 \), is 2.

🎯 Exam Tip: When dealing with determinants that have zeros, expand along the row or column containing the most zeros to simplify calculations. Always remember the coplanarity condition: scalar triple product must be zero.

Question 8. If \( \overline{a} = \hat{i} – \hat{k} \), \( \overline{b} = x\hat{i} + \hat{j} + (1-x)\hat{k} \) and \( \overline{c} = y\hat{i} + x\hat{j} + (1+x-y)\hat{k} \), show that \( [\overline{a}, \overline{b}, \overline{c}] \) is independent of x and y.
Answer: We need to calculate the scalar triple product \( [\overline{a}, \overline{b}, \overline{c}] \) and show that its value does not depend on the variables x and y. If the result is a constant, then it is independent of x and y.
The vectors are:
\( \overline{a} = 1\hat{i} + 0\hat{j} - 1\hat{k} \)
\( \overline{b} = x\hat{i} + 1\hat{j} + (1-x)\hat{k} \)
\( \overline{c} = y\hat{i} + x\hat{j} + (1+x-y)\hat{k} \)
We set up the determinant:
\[ [\overline{a}, \overline{b}, \overline{c}] = \left|\begin{array}{ccc} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{array}\right| \] Now, expand the determinant:
\[ = 1((1)(1+x-y) - (1-x)(x)) - 0(\text{any value}) - 1((x)(x) - (1)(y)) \] \[ = 1(1 + x - y - (x - x^2)) - 0 - 1(x^2 - y) \] \[ = 1 + x - y - x + x^2 - x^2 + y \] Combine like terms:
\[ = (1) + (x - x) + (-y + y) + (x^2 - x^2) \] \[ = 1 \] The scalar triple product simplifies to 1. Since the result is a constant number (1) and does not contain x or y, it means that \( [\overline{a}, \overline{b}, \overline{c}] \) is independent of x and y. This shows that no matter what values x and y take, the volume formed by these vectors remains the same.
In simple words: We calculated a special value using the numbers from three vectors that had letters 'x' and 'y' in them. After doing all the math, the 'x's and 'y's disappeared, and we were left with just the number 1. This means the result never changes, no matter what numbers 'x' and 'y' are.

🎯 Exam Tip: When proving independence from variables, calculate the expression carefully. If the variables cancel out and you're left with a constant, the proof is complete. Pay close attention to negative signs during expansion.

Question 9. If \( a\hat{i} + a\hat{j} + c\hat{k} \), \( \hat{i} + \hat{k} \) and \( c\hat{i} + c\hat{j} + b\hat{k} \) are coplanar, prove that c is the geometric mean of a and b.
Answer: For three vectors to be coplanar, their scalar triple product must be zero. We can use this condition to prove the relationship between a, b, and c.
Let \( \overline{u} = a\hat{i} + a\hat{j} + c\hat{k} \)
Let \( \overline{v} = 1\hat{i} + 0\hat{j} + 1\hat{k} \)
Let \( \overline{w} = c\hat{i} + c\hat{j} + b\hat{k} \)
Since the vectors are coplanar, their scalar triple product is 0:
\[ [\overline{u}, \overline{v}, \overline{w}] = \left|\begin{array}{ccc} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{array}\right| = 0 \] Now, expand the determinant:
\[ a((0)(b) - (1)(c)) - a((1)(b) - (1)(c)) + c((1)(c) - (0)(c)) = 0 \] \[ a(0 – c) – a(b - c) + c(c) = 0 \] \[ -ac – ab + ac + c^2 = 0 \] Combine the terms:
\[ (-ac + ac) - ab + c^2 = 0 \] \[ 0 - ab + c^2 = 0 \] \[ c^2 - ab = 0 \] \[ c^2 = ab \] This equation \( c^2 = ab \) means that c is the geometric mean of a and b. This relationship shows that when three specific vectors are on the same plane, their coefficients follow this pattern.
In simple words: If three vectors lie on the same flat surface, a specific math calculation (called the scalar triple product) using their numbers must be zero. When we do this calculation and solve the equation, we find that the square of 'c' is equal to 'a' multiplied by 'b'. This means 'c' is the geometric mean of 'a' and 'b'.

🎯 Exam Tip: Clearly state the coplanarity condition and set up the determinant correctly. Be meticulous with algebraic simplification, especially when dealing with multiple variables, to arrive at the desired geometric mean relationship.

Question 10. Let \( \overline{a} \), \( \overline{b} \) and \( \overline{c} \) be three non-zero vectors such that \( \overline{c} \) is a unit vector perpendicular to both \( \overline{a} \) and \( \overline{b} \). If the angle between \( \overline{a} \) and \( \overline{b} \) is \( \frac{\pi}{6} \), show that \( [\overline{a}, \overline{b}, \overline{c}]^2 = \frac{1}{4} |\overline{a}|^2 |\overline{b}|^2 \).
Answer: We need to use the properties of the scalar triple product and vector products to prove the given identity. The scalar triple product \( [\overline{a}, \overline{b}, \overline{c}] \) can be written as \( (\overline{a} \times \overline{b}) \cdot \overline{c} \).
We are given that:
1. \( \overline{c} \) is a unit vector, so \( |\overline{c}| = 1 \).
2. \( \overline{c} \) is perpendicular to both \( \overline{a} \) and \( \overline{b} \). This means \( \overline{c} \) is parallel to \( \overline{a} \times \overline{b} \) (the cross product is always perpendicular to both original vectors). So the angle between \( (\overline{a} \times \overline{b}) \) and \( \overline{c} \) is either 0 or \( \pi \).
3. The angle between \( \overline{a} \) and \( \overline{b} \) is \( \theta = \frac{\pi}{6} \).
First, let's consider \( [\overline{a}, \overline{b}, \overline{c}] \):
\[ [\overline{a}, \overline{b}, \overline{c}] = (\overline{a} \times \overline{b}) \cdot \overline{c} \] We know that for any two vectors \( \overline{P} \) and \( \overline{Q} \), \( \overline{P} \cdot \overline{Q} = |\overline{P}| |\overline{Q}| \cos \phi \), where \( \phi \) is the angle between them.
So, \( (\overline{a} \times \overline{b}) \cdot \overline{c} = |\overline{a} \times \overline{b}| |\overline{c}| \cos \phi \).
Since \( \overline{c} \) is perpendicular to both \( \overline{a} \) and \( \overline{b} \), it must be parallel to \( \overline{a} \times \overline{b} \). This means \( \phi = 0 \) or \( \phi = \pi \), so \( \cos \phi = \pm 1 \).
Therefore, \( [\overline{a}, \overline{b}, \overline{c}] = |\overline{a} \times \overline{b}| (1) (\pm 1) = \pm |\overline{a} \times \overline{b}| \).
Now, let's find \( [\overline{a}, \overline{b}, \overline{c}]^2 \):
\[ [\overline{a}, \overline{b}, \overline{c}]^2 = (\pm |\overline{a} \times \overline{b}|)^2 = |\overline{a} \times \overline{b}|^2 \] We also know that \( |\overline{a} \times \overline{b}| = |\overline{a}| |\overline{b}| \sin \theta \).
So, \( |\overline{a} \times \overline{b}|^2 = (|\overline{a}| |\overline{b}| \sin \theta)^2 = |\overline{a}|^2 |\overline{b}|^2 \sin^2 \theta \).
We are given that \( \theta = \frac{\pi}{6} \).
So, \( \sin \theta = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \).
Therefore, \( \sin^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
Substitute this back into the expression for \( [\overline{a}, \overline{b}, \overline{c}]^2 \):
\[ [\overline{a}, \overline{b}, \overline{c}]^2 = |\overline{a}|^2 |\overline{b}|^2 \left(\frac{1}{4}\right) \] \[ [\overline{a}, \overline{b}, \overline{c}]^2 = \frac{1}{4} |\overline{a}|^2 |\overline{b}|^2 \] Thus, the identity is proven. This shows a direct link between the scalar triple product and the magnitudes and angle of the component vectors.
In simple words: We are given three special vectors: one is a unit vector (length 1) that stands straight up from the other two. We also know the angle between the first two vectors. By using math rules for vector multiplication, we can show that the square of a certain product of these three vectors is always equal to one-fourth of the squared lengths of the first two vectors multiplied together.

🎯 Exam Tip: This question tests your understanding of the definitions and properties of scalar triple product, cross product, dot product, unit vectors, and angles between vectors. Clearly state each property used in your proof.