Problem 22 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(a, b\) and \(c\) be distinct non- negative numbers. If the vectors \(a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}\) and \(\hat{c} \hat{i}+c \hat{j}+b \hat{k}\) lie in a plane, then \(c\) is (a) the Geometric Mean of \(a\) and \(b\) (b) the Arithmetic Mean of \(a\) and \(b\) (c) equal to zero (d) the Harmonic Mean of \(a\) and \(b\)

Step-by-Step Solution

Answer

(a) The Geometric Mean of \(a\) and \(b\).

Step 1: Understand the Condition for Coplanarity

Three vectors are coplanar if their scalar triple product is zero. If the vectors \( \mathbf{v_1} = a \hat{i} + a \hat{j} + c \hat{k} \), \( \mathbf{v_2} = \hat{i} + \hat{k} \), and \( \mathbf{v_3} = c \hat{i} + c \hat{j} + b \hat{k} \) lie in a plane, then the condition \( (\mathbf{v_1} \times \mathbf{v_2}) \cdot \mathbf{v_3} = 0 \) must be satisfied.

Step 2: Compute the Cross Product \(\mathbf{v_1} \times \mathbf{v_2}\)

Calculate the cross product \( \mathbf{v_1} \times \mathbf{v_2} \). Let\[ \mathbf{v_1} = (a, a, c), \quad \mathbf{v_2} = (1, 0, 1) \] The cross product is given by the determinant:\[ \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a & a & c \ 1 & 0 & 1 \end{vmatrix} \]This simplifies to:\[ \mathbf{v_1} \times \mathbf{v_2} = (a \times 1 - c \times 0) \hat{i} - (a \times 1 - c \times 1) \hat{j} + (a \times 0 - a \times 1) \hat{k} \]\[ = (a)\hat{i} - (a - c) \hat{j} - (a) \hat{k} \]\[ = (a, -(a - c), -a) \]

Step 3: Compute the Dot Product \((\mathbf{v_1} \times \mathbf{v_2}) \cdot \mathbf{v_3}\)

Now calculate the dot product between the result from Step 2 and \( \mathbf{v_3} = (c, c, b) \):\[ (\mathbf{v_1} \times \mathbf{v_2}) \cdot \mathbf{v_3} = (a, -(a - c), -a) \cdot (c, c, b) \]\[ = a \cdot c + [-(a - c)] \cdot c - a \cdot b \]\[ = ac - ac + c^2 - ab \]\[ = c^2 - ab \]

Step 4: Set the Dot Product Equal to Zero and Solve

Since the vectors are coplanar, set the equation from Step 3 equal to zero:\[ c^2 - ab = 0 \]\[ c^2 = ab \]Thus, \( c = \sqrt{ab} \), showing that \( c \) is the geometric mean of \( a \) and \( b \).

Step 5: Choose the Correct Answer

The problem gave four options for the value of \( c \). We found \( c = \sqrt{ab} \), implying that \( c \) is the geometric mean of \( a \) and \( b \). Hence, the correct choice is \( \text{(a)} \).

Key Concepts

Vector Coplanarity

Vectors are said to be coplanar if they lie in the same plane. This implies that any vector in the group can be represented as a linear combination of the others. In simpler terms, if you have three vectors, they form a plane if their spatial arrangement doesn't require a third dimension.

For vectors to be coplanar, their scalar triple product must equal zero. The scalar triple product is calculated using the formula: - \[ \mathbf{v_1} \times \mathbf{v_2} \cdot \mathbf{v_3} = 0 \]- This implies that the volume of the parallelepiped formed by these vectors is zero, meaning they are coplanar.

Understanding coplanarity is critical when analyzing vector equations, ensuring that the vectors fall within the same dimensional space without deviation.

Scalar Triple Product

The scalar triple product involves three vectors and gives a scalar result. It's used primarily to determine coplanarity of vectors. The formula for the scalar triple product is: - \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \]- It can be represented in determinant form, - \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix} \]- If this determinant is zero, the vectors are coplanar.

Hence, scalar triple product is a useful algebraic tool for checking if three vectors can lie flat on paper or a plane, needing no deviation into the third dimension.

Geometric Mean

The geometric mean is a concept borrowed from arithmetic but applied in various scientific contexts, including this vector scenario.

For two numbers, the geometric mean is the square root of their product. This is expressed as:- \[ c = \sqrt{ab} \] - Here, \( c \) is the geometric mean of \( a \) and \( b \).

Geometric mean is helpful for comparing different kinds of quantities and provides insight into proportional relationships.

In the vector context given, the condition \( c^2 = ab \) indicates that the coefficient \( c \) must indeed be the geometric mean of \( a \) and \( b \) to maintain the coplanar nature of the vectors involved.