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JEE Main 2021
Vector Algebra
Vector Algebra
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Question

If a→=αi^+βj^+3k^\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat k, b→=−βi^−αj^−k^\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k and c→=i^−2j^−k^\overrightarrow c = \widehat i - 2\widehat j - \widehat k such that a→ . b→=1\overrightarrow a \,.\,\overrightarrow b = 1 and b→ . c→=−3\overrightarrow b \,.\,\overrightarrow c = - 3, then 13((a→×b→) . c→){1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right) is equal to _____________.

Answer: 1

Solution

Key Concepts and Formulas

  1. Dot Product of Vectors: For vectors u→=uxi^+uyj^+uzk^\overrightarrow u = u_x \widehat i + u_y \widehat j + u_z \widehat k and v→=vxi^+vyj^+vzk^\overrightarrow v = v_x \widehat i + v_y \widehat j + v_z \widehat k, the dot product is u→ . v→=uxvx+uyvy+uzvz\overrightarrow u \,.\,\overrightarrow v = u_x v_x + u_y v_y + u_z v_z.
  2. Scalar Triple Product: For vectors a→\overrightarrow a, b→\overrightarrow b, and c→\overrightarrow c, the scalar triple product (a→×b→) . c→(\overrightarrow a \times \overrightarrow b) \,.\,\overrightarrow c can be computed as the determinant of the matrix formed by their components: (a→×b→) . c→=∣axayazbxbybzcxcycz∣(\overrightarrow a \times \overrightarrow b) \,.\,\overrightarrow c = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix}

Step-by-Step Solution

Step 1: Use the given dot product condition a→ . b→=1\overrightarrow a \,.\,\overrightarrow b = 1 to find a relationship between α\alpha and β\beta.

  • Reasoning: The dot product of two vectors yields a scalar. By applying the dot product formula to the given vectors a→\overrightarrow a and b→\overrightarrow b, we can form an equation involving α\alpha and β\beta.
  • Given vectors: a→=αi^+βj^+3k^\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat k b→=−βi^−αj^−k^\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k
  • Calculating the dot product: a→ . b→=(α)(−β)+(β)(−α)+(3)(−1)\overrightarrow a \,.\,\overrightarrow b = (\alpha)(-\beta) + (\beta)(-\alpha) + (3)(-1) =−αβ−αβ−3= -\alpha\beta - \alpha\beta - 3 =−2αβ−3= -2\alpha\beta - 3
  • Using the given condition a→ . b→=1\overrightarrow a \,.\,\overrightarrow b = 1: −2αβ−3=1-2\alpha\beta - 3 = 1 −2αβ=4-2\alpha\beta = 4 αβ=−2(Equation 1)\alpha\beta = -2 \quad \text{(Equation 1)}

Step 2: Use the given dot product condition b→ . c→=−3\overrightarrow b \,.\,\overrightarrow c = - 3 to find another relationship between α\alpha and β\beta.

  • Reasoning: Similar to Step 1, we use the second dot product condition to derive a second equation involving α\alpha and β\beta. This will allow us to solve for the specific values of α\alpha and β\beta.
  • Given vectors: b→=−βi^−αj^−k^\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k c→=i^−2j^−k^\overrightarrow c = \widehat i - 2\widehat j - \widehat k
  • Calculating the dot product: b→ . c→=(−β)(1)+(−α)(−2)+(−1)(−1)\overrightarrow b \,.\,\overrightarrow c = (-\beta)(1) + (-\alpha)(-2) + (-1)(-1) =−β+2α+1= -\beta + 2\alpha + 1
  • Using the given condition b→ . c→=−3\overrightarrow b \,.\,\overrightarrow c = -3: 2α−β+1=−32\alpha - \beta + 1 = -3 2α−β=−4(Equation 2)2\alpha - \beta = -4 \quad \text{(Equation 2)}

Step 3: Solve the system of equations to find the values of α\alpha and β\beta.

  • Reasoning: We now have a system of two algebraic equations with two unknowns. Solving this system will give us the numerical values for α\alpha and β\beta.
  • The system of equations is:
    1. αβ=−2\alpha\beta = -2
    2. 2α−β=−42\alpha - \beta = -4
  • From Equation 2, we can express β\beta in terms of α\alpha: β=2α+4\beta = 2\alpha + 4
  • Substitute this expression for β\beta into Equation 1: α(2α+4)=−2\alpha(2\alpha + 4) = -2 2α2+4α=−22\alpha^2 + 4\alpha = -2 2α2+4α+2=02\alpha^2 + 4\alpha + 2 = 0
  • Divide the equation by 2: α2+2α+1=0\alpha^2 + 2\alpha + 1 = 0
  • Factor the quadratic equation: (α+1)2=0(\alpha + 1)^2 = 0
  • This gives us α=−1\alpha = -1.
  • Now substitute α=−1\alpha = -1 back into the expression for β\beta: β=2(−1)+4\beta = 2(-1) + 4 β=−2+4\beta = -2 + 4 β=2\beta = 2
  • So, α=−1\alpha = -1 and β=2\beta = 2.

Step 4: Determine the component forms of the vectors a→\overrightarrow a and b→\overrightarrow b using the found values of α\alpha and β\beta.

  • Reasoning: With the values of α\alpha and β\beta determined, we can now write the explicit component forms of vectors a→\overrightarrow a and b→\overrightarrow b. Vector c→\overrightarrow c is already given.
  • Substituting α=−1\alpha = -1 and β=2\beta = 2: a→=(−1)i^+(2)j^+3k^=−i^+2j^+3k^\overrightarrow a = (-1)\widehat i + (2)\widehat j + 3\widehat k = - \widehat i + 2\widehat j + 3\widehat k b→=−(2)i^−(−1)j^−k^=−2i^+j^−k^\overrightarrow b = -(2)\widehat i - (-1)\widehat j - \widehat k = - 2\widehat i + \widehat j - \widehat k
  • Vector c→\overrightarrow c: c→=i^−2j^−k^\overrightarrow c = \widehat i - 2\widehat j - \widehat k

Step 5: Calculate the scalar triple product (a→×b→) . c→\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c.

  • Reasoning: The scalar triple product can be efficiently calculated using the determinant of a matrix formed by the components of the three vectors. The order of the vectors in the determinant must match their order in the scalar triple product.
  • The scalar triple product is given by the determinant: (a→×b→) . c→=∣axayazbxbybzcxcycz∣=∣−123−21−11−2−1∣(\overrightarrow a \times \overrightarrow b) \,.\,\overrightarrow c = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} = \begin{vmatrix} -1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix}
  • Expanding the determinant: =−1∣1−1−2−1∣−2∣−2−11−1∣+3∣−211−2∣= -1 \begin{vmatrix} 1 & -1 \\ -2 & -1 \end{vmatrix} - 2 \begin{vmatrix} -2 & -1 \\ 1 & -1 \end{vmatrix} + 3 \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} =−1((1)(−1)−(−1)(−2))−2((−2)(−1)−(−1)(1))+3((−2)(−2)−(1)(1))= -1 ((1)(-1) - (-1)(-2)) - 2 ((-2)(-1) - (-1)(1)) + 3 ((-2)(-2) - (1)(1)) =−1(−1−2)−2(2+1)+3(4−1)= -1 (-1 - 2) - 2 (2 + 1) + 3 (4 - 1) =−1(−3)−2(3)+3(3)= -1 (-3) - 2 (3) + 3 (3) =3−6+9= 3 - 6 + 9 =6= 6
  • Therefore, (a→×b→) . c→=6\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c = 6.

Step 6: Compute the final required value.

  • Reasoning: The problem asks for 13\frac{1}{3} of the scalar triple product. We substitute the calculated value from Step 5.
  • The required value is: 13((a→×b→) . c→)=13(6){1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right) = {1 \over 3}(6) =2= 2

Common Mistakes & Tips

  1. Algebraic Errors: Be meticulous with algebraic manipulations, especially when solving the system of equations for α\alpha and β\beta. Sign errors are particularly common.
  2. Determinant Calculation: Double-check the expansion of the determinant for the scalar triple product. Ensure the signs and multiplications are correct.
  3. Order of Vectors: The order of vectors in the scalar triple product (a→×b→) . c→(\overrightarrow a \times \overrightarrow b) \,.\,\overrightarrow c is crucial for the determinant calculation. If the order is changed, the sign of the result might change (though for this problem, the specific order is given).

Summary The problem required us to first find the unknown scalar components α\alpha and β\beta of vectors a→\overrightarrow a and b→\overrightarrow b by utilizing the given dot product conditions. This led to a system of algebraic equations which was solved to yield α=−1\alpha = -1 and β=2\beta = 2. With the complete vector components determined, we then calculated the scalar triple product (a→×b→) . c→(\overrightarrow a \times \overrightarrow b) \,.\,\overrightarrow c using the determinant method. Finally, we took one-third of this scalar triple product to arrive at the final answer.

The final answer is 2\boxed{2}.

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