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JEE Main 2020
Vector Algebra
Vector Algebra
Medium

Question

If a.b=1,b.c=2\overrightarrow a \,.\,\overrightarrow b = 1,\,\overrightarrow b \,.\,\overrightarrow c = 2 and c.a=3\overrightarrow c \,.\,\overrightarrow a = 3, then the value of [a×(b×c),b×(c×a),c×(b×a)]\left[ {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right),\,\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right)} \right] is :

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Solution

Key Concepts and Formulas

  1. Vector Triple Product (VTP): For any three vectors A\overrightarrow A, B\overrightarrow B, and C\overrightarrow C, the vector triple product A×(B×C)\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) can be expanded using the formula: A×(B×C)=(AC)B(AB)C\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C This identity is often remembered by the mnemonic "BAC-CAB".

  2. Scalar Triple Product (STP): The scalar triple product of three vectors u\overrightarrow u, v\overrightarrow v, and w\overrightarrow w is denoted by [u,v,w][\overrightarrow u, \overrightarrow v, \overrightarrow w] or u(v×w)\overrightarrow u \cdot (\overrightarrow v \times \overrightarrow w).

    • Geometric Significance: The absolute value of the scalar triple product represents the volume of the parallelepiped formed by the three vectors.
    • Condition for Coplanarity: If three vectors are coplanar (i.e., they lie in the same plane), their scalar triple product is zero. This implies that the volume of the parallelepiped formed by them is zero.
    • Linear Dependence: Three vectors are coplanar if and only if they are linearly dependent. This means one vector can be expressed as a linear combination of the other two, or a non-trivial linear combination of the three vectors equals the zero vector (e.g., c1u+c2v+c3w=0c_1\overrightarrow u + c_2\overrightarrow v + c_3\overrightarrow w = \overrightarrow 0 for some non-zero scalars c1,c2,c3c_1, c_2, c_3).

Step-by-Step Solution

We are given the dot products:

  • ab=1\overrightarrow a \cdot \overrightarrow b = 1
  • bc=2\overrightarrow b \cdot \overrightarrow c = 2
  • ca=3\overrightarrow c \cdot \overrightarrow a = 3

We need to find the value of the scalar triple product [a×(b×c),b×(c×a),c×(b×a)]\left[ {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right),\,\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right)} \right]. Let's simplify each vector term first.

Step 1: Simplify the first vector term, u=a×(b×c)\overrightarrow u = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)

  • Why this step? To evaluate the scalar triple product of complex vector expressions, it's essential to simplify each individual vector term using the vector triple product formula. This reduces the complexity of the expression.
  • Applying the VTP formula A×(B×C)=(AC)B(AB)C\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C with A=a\overrightarrow A = \overrightarrow a, B=b\overrightarrow B = \overrightarrow b, and C=c\overrightarrow C = \overrightarrow c: u=a×(b×c)=(ac)b(ab)c\overrightarrow u = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = (\overrightarrow a \cdot \overrightarrow c)\overrightarrow b - (\overrightarrow a \cdot \overrightarrow b)\overrightarrow c
  • Substituting the given dot product values, ac=3\overrightarrow a \cdot \overrightarrow c = 3 and ab=1\overrightarrow a \cdot \overrightarrow b = 1: u=(3)b(1)c\overrightarrow u = (3)\overrightarrow b - (1)\overrightarrow c u=3bc\overrightarrow u = 3\overrightarrow b - \overrightarrow c

Step 2: Simplify the second vector term, v=b×(c×a)\overrightarrow v = \overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right)

  • Why this step? We continue the simplification process by applying the VTP formula to the second vector expression.
  • Applying the VTP formula with A=b\overrightarrow A = \overrightarrow b, B=c\overrightarrow B = \overrightarrow c, and C=a\overrightarrow C = \overrightarrow a: v=b×(c×a)=(ba)c(bc)a\overrightarrow v = \overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right) = (\overrightarrow b \cdot \overrightarrow a)\overrightarrow c - (\overrightarrow b \cdot \overrightarrow c)\overrightarrow a
  • Using the commutative property of the dot product (ba=ab=1\overrightarrow b \cdot \overrightarrow a = \overrightarrow a \cdot \overrightarrow b = 1) and the given value bc=2\overrightarrow b \cdot \overrightarrow c = 2: v=(1)c(2)a\overrightarrow v = (1)\overrightarrow c - (2)\overrightarrow a v=c2a\overrightarrow v = \overrightarrow c - 2\overrightarrow a

Step 3: Simplify the third vector term, w=c×(b×a)\overrightarrow w = \overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right)

  • Why this step? We complete the simplification of all three vector terms that form the scalar triple product.
  • Applying the VTP formula with A=c\overrightarrow A = \overrightarrow c, B=b\overrightarrow B = \overrightarrow b, and C=a\overrightarrow C = \overrightarrow a: w=c×(b×a)=(ca)b(cb)a\overrightarrow w = \overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right) = (\overrightarrow c \cdot \overrightarrow a)\overrightarrow b - (\overrightarrow c \cdot \overrightarrow b)\overrightarrow a
  • Substituting the given dot product values, ca=3\overrightarrow c \cdot \overrightarrow a = 3 and cb=bc=2\overrightarrow c \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow c = 2: w=(3)b(2)a\overrightarrow w = (3)\overrightarrow b - (2)\overrightarrow a w=3b2a\overrightarrow w = 3\overrightarrow b - 2\overrightarrow a

Step 4: Analyze the relationship between the simplified vectors u\overrightarrow u, v\overrightarrow v, and w\overrightarrow w

  • Why this step? The scalar triple product of three vectors is zero if they are linearly dependent. We will now check if such a dependency exists among the simplified vectors u\overrightarrow u, v\overrightarrow v, and w\overrightarrow w.
  • We have the simplified expressions:
    • u=3bc\overrightarrow u = 3\overrightarrow b - \overrightarrow c
    • v=c2a\overrightarrow v = \overrightarrow c - 2\overrightarrow a
    • w=3b2a\overrightarrow w = 3\overrightarrow b - 2\overrightarrow a
  • Let's try to express one vector as a linear combination of the others, or check if their sum or difference leads to a simpler relationship. Consider the sum of u\overrightarrow u and v\overrightarrow v: u+v=(3bc)+(c2a)\overrightarrow u + \overrightarrow v = (3\overrightarrow b - \overrightarrow c) + (\overrightarrow c - 2\overrightarrow a) u+v=3bc+c2a\overrightarrow u + \overrightarrow v = 3\overrightarrow b - \overrightarrow c + \overrightarrow c - 2\overrightarrow a u+v=3b2a\overrightarrow u + \overrightarrow v = 3\overrightarrow b - 2\overrightarrow a
  • Comparing this result with the expression for w\overrightarrow w, we observe that: u+v=w\overrightarrow u + \overrightarrow v = \overrightarrow w
  • This equation can be rearranged to show linear dependence: u+vw=0\overrightarrow u + \overrightarrow v - \overrightarrow w = \overrightarrow 0

Step 5: Conclude using the coplanarity condition

  • Why this step? The relationship u+vw=0\overrightarrow u + \overrightarrow v - \overrightarrow w = \overrightarrow 0 demonstrates that the vectors u\overrightarrow u, v\overrightarrow v, and w\overrightarrow w are linearly dependent.
  • As established by the properties of the scalar triple product, three linearly dependent vectors are always coplanar.
  • The scalar triple product of any three coplanar vectors is zero.
  • Therefore, the value of the scalar triple product of u\overrightarrow u, v\overrightarrow v, and w\overrightarrow w is 00. [u,v,w]=[a×(b×c),b×(c×a),c×(b×a)]=0\left[ {\overrightarrow u, \overrightarrow v, \overrightarrow w} \right] = \left[ {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right),\,\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right)} \right] = 0

The final answer is 0\boxed{0}.

Common Mistakes & Tips

  • VTP Formula Accuracy: Ensure the vector triple product formula A×(B×C)=(AC)B(AB)C\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C is recalled and applied correctly, paying close attention to the order of vectors and the signs.
  • Dot Product Commutativity: Always remember that XY=YX\overrightarrow X \cdot \overrightarrow Y = \overrightarrow Y \cdot \overrightarrow X. This is frequently used when substituting given values, e.g., using ab\overrightarrow a \cdot \overrightarrow b for ba\overrightarrow b \cdot \overrightarrow a.
  • Recognizing Linear Dependence: The key to solving this problem efficiently is to identify the linear relationship between the simplified vectors. If the vectors are linearly dependent, their scalar triple product is zero, saving further calculations.

Summary

The problem requires evaluating a scalar triple product of three complex vector terms. The strategy involves simplifying each term using the vector triple product identity. After applying the identity and substituting the given dot products, the three vector terms are found to be u=3bc\overrightarrow u = 3\overrightarrow b - \overrightarrow c, v=c2a\overrightarrow v = \overrightarrow c - 2\overrightarrow a, and w=3b2a\overrightarrow w = 3\overrightarrow b - 2\overrightarrow a. By observing that u+v=w\overrightarrow u + \overrightarrow v = \overrightarrow w, or u+vw=0\overrightarrow u + \overrightarrow v - \overrightarrow w = \overrightarrow 0, we establish that the three vectors are linearly dependent. Linearly dependent vectors are coplanar, and the scalar triple product of coplanar vectors is always zero.

The final answer is 0\boxed{0}.

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