Problem 109 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three unit vectors such that \(\vec{a} \times(\vec{b} \times \vec{c})=\frac{\sqrt{3}}{2}(\vec{b}+\vec{c})\). if \(\vec{b}\) is not parallel to \(\vec{c}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is (A) \(\frac{5 \pi}{6}\) (B) \(\frac{3 \pi}{4}\) (C) \(\frac{\pi}{2}\) (D) \(\frac{2 \pi}{3}\)

Step-by-Step Solution

Answer

The angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{5\pi}{6} \). (Option A)

Step 1: Use the Vector Triple Product Identity

The vector triple product identity is \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \). Based on this identity, our given expression becomes \( (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c}) \).

Step 2: Compare Vectors

Split both sides to compare the coefficients of vectors \( \vec{b} \) and \( \vec{c} \). From the equation, we have: \( \vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2} \) and \( -\vec{a} \cdot \vec{b} = \frac{\sqrt{3}}{2} \). Thus, we also have \( \vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2} \).

Step 3: Use Dot Product to Find the Angle

The dot product \( \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos\theta \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). Since both are unit vectors, \( \vec{a} \cdot \vec{b} = \cos\theta = -\frac{\sqrt{3}}{2} \).

Step 4: Calculate the Angle

Using \( \cos\theta = -\frac{\sqrt{3}}{2} \), determine \( \theta \). The angle \( \theta \) whose cosine is \(-\frac{\sqrt{3}}{2}\) is \( \theta = \frac{5\pi}{6} \) or \( \theta = \frac{7\pi}{6} \), but we consider it in the range \( [0, \pi] \) which makes \( \theta = \frac{5\pi}{6} \).

Key Concepts

Vector Triple Product

The vector triple product involves three vectors where one vector is taken as a product of the other two vectors. It's written as \( \vec{a} \times (\vec{b} \times \vec{c}) \). An identity helps simplify this expression: \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \). This formula can look complex, but it essentially breaks down the interaction of these three vectors.

- The first term \((\vec{a} \cdot \vec{c}) \vec{b}\) shows how the projection of \(\vec{a}\) on \(\vec{c}\) can scale vector \(\vec{b}\).
- The second term \(-(\vec{a} \cdot \vec{b}) \vec{c}\) focuses on how the projection of \(\vec{a}\) on \(\vec{b}\) inversely scales vector \(\vec{c}\).

In our original problem, applying this identity lets us compare the coefficients of \(\vec{b}\) and \(\vec{c}\), leading to some crucial insights about the relationships between the vectors. Once you get used to the identity, it becomes an immensely powerful tool in vector calculus.

Dot Product

In vector algebra, the dot product is a way to multiply two vectors, and it results in a scalar rather than another vector. It's calculated as \( \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos\theta \), where \(\theta\) is the angle between the vectors.

Essentially, the dot product measures how much one vector extends in the direction of another. Several key aspects of the dot product include:

• Finding the angle between vectors, which was crucial in our solution.
• Determining whether vectors are orthogonal: if the dot product is zero, the vectors are perpendicular.

In this exercise, we found \( \vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2} \), leading us directly to calculate the angle \(\theta\). Thus, using the dot product not only helps in calculating angles but also provides insight into vector alignments.

Angle Between Vectors

The angle between two vectors is fundamental in understanding spatial orientation. For any two vectors \( \vec{a} \) and \( \vec{b} \), the cosine of the angle \( \theta \) between them is given by the normalized dot product: \( \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} \).

When both vectors are unit vectors, this simplifies to just \( \vec{a} \cdot \vec{b} = \cos(\theta) \). In the original problem, this understanding of the angle between vectors helped find \( \theta \) when given \( \cos\theta = -\frac{\sqrt{3}}{2} \).

To determine the actual angle, consider the properties of cosine:

• If \( \cos(\theta) \,>\, 0 \), \( \theta \) is acute.
• If \( \cos(\theta) = 0 \), the vectors are perpendicular.
• If \( \cos(\theta) \,<\, 0 \), \( \theta \) is obtuse, which was relevant in finding \(\theta = \frac{5\pi}{6}\) for our problem.

The angle tells us not only about the extent of vector separation but can also hint at directional alignment, making it invaluable in vector analysis.