Problem 14 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a, b, c\) are non-coplanar unit vectors such that \(a \times(b \times c)=\frac{b+c}{\sqrt{2}}\), then the angle between \(a\) and \(b\) is (A) \(\frac{3 \pi}{4}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{\pi}{2}\) (D) \(\pi\)

Step-by-Step Solution

Answer

The angle between \(a\) and \(b\) is \(\frac{3\pi}{4}\).

Step 1: Using Vector Triple Product Identity

The vector triple product identity states that \( a \times (b \times c) = (a \cdot c) b - (a \cdot b) c \). Using this identity, we equate it to the given expression: \( a \times (b \times c) = \frac{b+c}{\sqrt{2}} \).

Step 2: Setting Equations

From the identity, we have: \((a \cdot c)b - (a \cdot b)c = \frac{b+c}{\sqrt{2}}\). This gives us a system of equations: \( (a \cdot c) = \frac{1}{\sqrt{2}} \) and \( (a \cdot b) = -\frac{1}{\sqrt{2}} \).

Step 3: Finding the Angle Between \(a\) and \(b\)

Given that \( a, b, c \) are unit vectors, \( a \cdot b = -\frac{1}{\sqrt{2}} \) implies \( \cos \theta = -\frac{1}{\sqrt{2}} \), where \( \theta \) is the angle between \( a \) and \( b \).

Step 4: Calculating the Angle from Cosine Value

Solving \( \cos \theta = -\frac{1}{\sqrt{2}} \) gives \( \theta = \frac{3\pi}{4} \). Thus, the angle between \( a \) and \( b \) is \( \frac{3\pi}{4} \).

Key Concepts

Unit Vectors

Unit vectors are vectors with a magnitude of 1. They are often used in physics and mathematics to denote directions in space. The notation for unit vectors is usually a lowercase letter with a caret, like \( \hat{a} \) or \( \hat{b} \).

Here are some key features of unit vectors:

• Magnitude is always 1. This simplifies calculations in many vector operations.
• Often used to represent basis directions, such as the standard unit vectors \( \hat{i}, \hat{j}, \hat{k} \) in 3D Cartesian coordinates representing the x, y, and z axes respectively.
• Any vector \( \mathbf{v} \) can be converted into a unit vector, also known as normalization, by dividing by its magnitude: \( \hat{v} = \frac{\mathbf{v}}{||\mathbf{v}||} \).

In our original problem, the vectors \( a, b, \) and \( c \) are non-coplanar unit vectors, meaning they each have a magnitude of 1 and are not all lying on the same plane, making it possible to define a unique set of directions in 3D space.

Vector Triple Product Identity

The vector triple product is an important identity in vector algebra, used to simplify expressions involving cross products. It states that for any vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), the expression \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \) can be expanded as:

\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]

This identity is often used to simplify complex vector expressions by converting cross products into dot products, primarily when solving physics problems involving torque or magnetic forces, or in mathematical proofs.

In the problem provided, this identity helps transform the complex vector equation into a system of simpler equations, allowing us to derive relationships between the dot products \( \mathbf{a} \cdot \mathbf{b} \) and \( \mathbf{a} \cdot \mathbf{c} \), necessary to solve for the angle between the vectors.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. Algebraically, the dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated as:

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

Geometrically, the dot product relates to the cosine of the angle \( \theta \) between the two vectors and is given by:

\[ \mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| \, ||\mathbf{b}|| \cos \theta \]

Where \(||\mathbf{a}||\) and \(||\mathbf{b}||\) are the magnitudes of the vectors. The result is:

• Positive if the angle is acute.
• Zero if the vectors are orthogonal (perpendicular).
• Negative if the angle is obtuse.

In the problem, finding \( \mathbf{a} \cdot \mathbf{b} = -\frac{1}{\sqrt{2}} \) gives a clue about the angle between \( a \) and \( b \). Since the dot product is negative, the angle is more than 90 degrees, specifically \( \frac{3\pi}{4} \), consistent with \( \cos \theta = -\frac{1}{\sqrt{2}} \). This shows how the dot product is instrumental in determining angles between vectors.