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

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 \(\frac{\pi}{4}\).

Step 1: Understanding the Identity

We need to use the vector triple product identity: \( a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \). This will help us rewrite the given expression in a more manageable form.

Step 2: Applying the Identity

Plug the identity into the given equation: \((a \cdot c)b - (a \cdot b)c = \frac{b+c}{\sqrt{2}}.\)This represents the transformed expression for the vector triple product.

Step 3: Equating Components

For vectors to be equal, each corresponding component must be the same. From:\((a \cdot c)b - (a \cdot b)c = \frac{b+c}{\sqrt{2}}\), compare components:1. \((a \cdot c)b = \frac{b}{\sqrt{2}}\)2. \((a \cdot b)c = \frac{c}{\sqrt{2}}\). From this, it follows that \(a \cdot c = \frac{1}{\sqrt{2}}\) and \(a \cdot b = \frac{1}{\sqrt{2}}\).

Step 4: Calculating the Angle Between Vectors

The dot product \(a \cdot b\) for unit vectors is given by: \(a \cdot b = |a||b|\cos\theta = \cos\theta\).Substitute the value obtained earlier:\(\cos\theta = \frac{1}{\sqrt{2}}\).This implies \(\theta = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\).

Step 5: Conclusion

The calculated angle \(\frac{\pi}{4}\) corresponds to option (B) in the given choices.

Key Concepts

Vector Triple Product

In vector algebra, the vector triple product is an interesting identity that involves three vectors and two cross products. This product is represented as \(a \times (b \times c)\). It is equivalent to a linear combination of vectors \(b\) and \(c\):

• \( a \times (b \times c) = (a \cdot c) b - (a \cdot b) c \)

This formula is crucial because it allows you to simplify expressions that involve three vectors and two cross products by rewriting them in terms of dot products.
The vector triple product does not give another vector orthogonal to existing vectors, unlike a simple cross product.
It depicts how a vector perpendicular to a plane defined by two original vectors can lie in the same plane as the first vector.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It combines two vectors to yield a scalar. The dot product of two vectors \(a\) and \(b\) is represented as \(a \cdot b\).

• Mathematically, it is defined as \(a \cdot b = |a||b|\cos \theta \), where \(\theta\) is the angle between the vectors.
• The dot product is related to the projection of one vector onto another and is used to find the angle between vectors.

In our problem, evaluating the dot product helps establish equations like \((a \cdot b)c = \frac{c}{\sqrt{2}}\), which plays a crucial role in solving for vector angles.
It’s particularly useful because it simplifies to things like \(\cos \theta\) when dealing with unit vectors.

Angle Between Vectors

Determining the angle between two vectors is a common task in vector algebra. This is generally found using the dot product of the vectors, as the cosine of the angle is directly related to the dot product.

• For two vectors \(a\) and \(b\), the formula is \(\cos \theta = \frac{a \cdot b}{|a||b|} \).
• When the vectors are unit vectors, the formula simplifies to \(\cos \theta = a \cdot b\).

In practical terms, this calculation is essential for finding the orientation or relative position of two vectors in space.
In the exercise, we found \(\theta\) by calculating \(\cos \theta = \frac{1}{\sqrt{2}}\), which leads to the angle \(\theta = \frac{\pi}{4}\). This shows that the vectors are positioned at a 45-degree angle relative to each other.