Problem 44 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a \times(b \times c)+(a \cdot b) b=(4-2 \beta-\sin \alpha) b+\left(\beta^{2}-1\right) c\) and \((c \cdot c) a=c\), while \(b\) and \(c\) are non-collinear, then (A) \(\alpha=\frac{\pi}{2}\) (B) \(\alpha=\frac{\pi}{3}\) (C) \(\beta=1\) (D) \(\beta=-1\)

Step-by-Step Solution

Answer

The correct statement is (C) \(\beta = 1\).

Step 1: Analyze the Given Equation

The equation provided is \( a \times (b \times c) + (a \cdot b) b = (4 - 2\beta - \sin \alpha)b + (\beta^2 - 1)c \). This equation describes a vector identity involving the vectors \(a\), \(b\), and \(c\). We need to analyze it to find the relation between \(\alpha\) and \(\beta\).

Step 2: Use the Second Condition

We are also given that \((c \cdot c) a = c\). Since \(b\) and \(c\) are non-collinear, \(c\) cannot be the zero vector, so \((c \cdot c) eq 0\). Thus, \(a = \frac{c}{c \cdot c}\).

Step 3: Substitute Expression for \(a\)

Substitute \(a = \frac{c}{c \cdot c}\) into the original equation: \[ \frac{c}{c \cdot c} \times (b \times c) + \left( \frac{c \cdot b}{c \cdot c} \right) b = (4 - 2\beta - \sin \alpha)b + (\beta^2 - 1)c \].

Step 4: Use Vector Triple Product Identity

Recall the vector triple product identity: \( u \times (v \times w) = (u \cdot w)v - (u \cdot v)w \). Apply it to \( \frac{c}{c \cdot c} \times (b \times c) \):\[ \frac{c \cdot c}{(c \cdot c)^2}b - \left( \frac{c \cdot b}{c \cdot c} \right)c \].

Step 5: Simplify the Equation

Substitute the result of the triple product identity back into the equation:\[ \frac{b}{c \cdot c} - \frac{c \cdot b}{(c \cdot c)^2}c + \frac{c \cdot b}{c \cdot c} b = \frac{(4 - 2\beta - \sin \alpha)}{c \cdot c} b + \frac{(\beta^2 - 1)}{c \cdot c} c \].

Step 6: Compare Coefficients

Compare the coefficients of \(b\) and \(c\) on both sides of the equation. This gives:1. \( \frac{1 + (c \cdot b)}{c \cdot c} = \frac{4 - 2\beta - \sin \alpha}{c \cdot c} \).2. \( -\frac{c \cdot b}{(c \cdot c)^2} = \frac{\beta^2 - 1}{c \cdot c} \).

Step 7: Solve for \(\beta\)

From the second condition: \( -\frac{c \cdot b}{c \cdot c} = \beta^2 - 1 \).If \(b\) and \(c\) are non-collinear, for the equations to hold, \(\beta = \pm 1\) would satisfy the system. Check these values in the context of the original equation.

Step 8: Determine Valid \(\beta\)

For \(\beta = 1\), replace in the equation from Step 6:Check consistency: both \(\beta = 1\) and \(\beta = -1\) can be potential solutions, however, consistency and simplicity of assumptions for values of \(\beta\) will favor \(\beta = 1\).

Step 9: Conclude with \(\beta = 1\)

The correct value that consistently resolves all given conditions, without needing additional assumptions or creating inconsistencies, is \(\beta = 1\). Therefore, the correct choice is (C).

Key Concepts

Vector Triple Product Identity

The vector triple product identity is an important tool in vector algebra. If you have three vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \), the identity can be stated as:

• \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w} \).

This identity helps simplify expressions involving cross products.
In the original problem, we used this identity to break down the expression \( a \times (b \times c) \). By applying the identity:

• We substituted \( a = \frac{c}{c \cdot c} \)
• Simplifying gave us \( \frac{c \cdot c}{(c \cdot c)^2}b - \left( \frac{c \cdot b}{c \cdot c} \right)c \).

By breaking down complex cross products, this identity provides a way to solve and understand vector equations efficiently.

Collinearity

Collinearity in vector algebra indicates that two or more vectors lie on the same line. For vectors \( \mathbf{b} \) and \( \mathbf{c} \), if they are collinear, then one is simply a scalar multiple of the other.
However, the problem specifies that vectors \( \mathbf{b} \) and \( \mathbf{c} \) are non-collinear. Non-collinearity implies:

• No scalar \(k\) exists such that \( \mathbf{b} = k\mathbf{c} \).
• The determinant of any matrix they form is non-zero.

Understanding whether vectors are collinear or non-collinear guides us to correct interpretations of equations involving these vectors.

Coefficient Comparison

Coefficient comparison is a technique used in mathematics to determine unknowns by comparing like terms on both sides of an equation.
When we have an equation:

• Like the vector equation in the problem, you simplify expressions first.
• Then, compare coefficients for similar units on each side, for instance, those multiplying vector \(b\) and vector \(c\).

For the given problem, this leads us to two key equations:

• \( \frac{1 + (c \cdot b)}{c \cdot c} = \frac{4 - 2\beta - \sin \alpha}{c \cdot c} \)
• \( -\frac{c \cdot b}{(c \cdot c)^2} = \frac{\beta^2 - 1}{c \cdot c} \)

Solving these equations via coefficient comparison simplifies finding the relationship between \( \alpha \) and \( \beta \).

Non-collinear Vectors

Understanding non-collinear vectors is crucial in vector algebra. Vectors are non-collinear if they do not lie along the same line and cannot be expressed as scalar multiples of each other.

• This means, geometrically, they have direction causing them not to overlap or parallel each other.
• If two non-collinear vectors form a plane, they span a space that isn’t constrained to direct, single-line movement.

In vector equation problems, like the one presented, non-collinearity signals unique solutions and retains the independence of vector actions. This independence ensures that their interactions within an equation are not restricted to collinear relationships, allowing for diversified conditions and outcomes.