Problem 24 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a, b, c\) are three non-parallel unit vectors such that \(a \times(b \times c)=\frac{1}{2} b\), then the angles which a makes with \(b\) and \(c\) are (A) \(90^{\circ}, 60^{\circ}\) (B) \(45^{\circ}, 60^{\circ}\) (C) \(30^{\circ}, 60^{\circ}\) (D) none of these

Step-by-Step Solution

Answer

The angles are \(90^{\circ}, 60^{\circ}\), which matches option (A).

Step 1: Vector Triple Product Identity

Recall the vector triple product identity: \(a \times (b \times c) = (a \cdot c)b - (a \cdot b)c\). We will use this identity to simplify the given equation \(a \times (b \times c) = \frac{1}{2} b\).

Step 2: Apply the Identity

Substitute the identity into the given expression: \((a \cdot c) b - (a \cdot b) c = \frac{1}{2} b\). This implies that the vector on the left side must only have its component along \(b\), since it should equate to \(\frac{1}{2} b\).

Step 3: Equate and Simplify

Equate the components: On comparing both sides, \((a \cdot c) b - (a \cdot b)c = \frac{1}{2}b \), we find that the term \(-(a \cdot b)c\) should vanish, which means \(a \cdot b = 0\), and so \(a\) is perpendicular to \(b\). Also, \(a \cdot c\) must equal \(\frac{1}{2}\).

Step 4: Calculate the Angles

Since \(a \cdot b = 0\), the angle between \(a\) and \(b\) is \(90^{\circ}\). From \(a \cdot c = \frac{1}{2}\), using the formula \(a \cdot c = |a||c|\cos\theta\), we get \(\cos\theta = \frac{1}{2}\). Hence, the angle between \(a\) and \(c\) is \(60^{\circ}\).

Step 5: Verify and Conclude

Since \(a\) is perpendicular to \(b\) and forms a \(60^{\circ}\) angle with \(c\), the option matching these angles is (A) \(90^{\circ}, 60^{\circ}\).

Key Concepts

Vector Cross Product

The vector cross product is a fundamental operation in vector algebra that combines two vectors to produce another vector. Its importance arises in physics and engineering, especially when dealing with rotational effects and torque. The cross product of two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), is denoted by \( \mathbf{a} \times \mathbf{b} \). This resultant vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). This perpendicular nature is often remembered through the "right-hand rule" which provides the direction of the resulting vector.
The magnitude of the cross product is given by: \[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| \, |\mathbf{b}| \, \sin(\theta) \] where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \). This means that the larger the angle up to 90 degrees, the more significant the vector result, reaching its maximum at right angles (\( 90^{\circ} \)).
As illustrated in the exercise, the vector cross product can be used in more complex identities like the vector triple product identity which simplifies vector expressions and is essential for solving many vector-related problems.

Dot Product

The dot product, also known as the scalar product, is a straightforward method to multiply two vectors, resulting in a scalar quantity. Unlike the cross product, the dot product gives us insight into the extent to which two vectors point in the same direction. Given vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is expressed as: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \, \cos(\theta) \] where \( \theta \) is the angle between the vectors.
This calculation is particularly useful when determining orthogonality. If \( \mathbf{a} \cdot \mathbf{b} = 0 \), it confirms that \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular. In the original exercise, this helped us conclude that vectors \( a \) and \( b \) form a \( 90^{\circ} \) angle with each other.
Also beneficial in many fields of study, the dot product helps simplify expressions involving the projection of one vector onto another. Therefore, mastering the dot product and its geometric interpretation provides significant advantages in understanding vector algebra.

Geometric Interpretation of Vectors

The geometric interpretation of vectors offers a visual representation of vector operations, such as the dot product and cross product, aiding in a deeper understanding of vector algebra. Vectors are visually depicted as arrows, with both magnitude (length of the arrow) and direction (orientation of the arrow).
In the realm of geometric interpretation:

• The magnitude corresponds to the "size" of the effect that vector imparts, like the speed of an object.
• The direction of a vector shows the path an object follows, crucial for determining resultant vectors in cross and dot products.

Visualizing vector products can clarify the understanding of how vectors interact. For example, using vector triple product identity as a visual tool allows students to dissect complex vector interactions into simpler component parts, enhancing comprehension.
By considering the original exercise, understanding this geometric interpretation eased the conceptual leap from equations to real-world scenarios. Therefore, practicing these visual interpretations brings abstract formulas to tangible contexts, which is invaluable in scientific and technical fields.