Problem 84 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(\bar{a}, \bar{b}\) and \(\bar{c}\) be non-zero vectors such that \((\bar{a} \times \bar{b}) \times \bar{c}=-|\bar{b}| \bar{c} \mid \bar{a} . \quad\) If \(\theta\) is the acute angle between the vectors \(\bar{b}\) and \(\bar{c}\) then \(\sin \theta\) equals [2004] (A) \(\frac{1}{3}\) (B) \(\frac{\sqrt{2}}{3}\) (C) \(\frac{2}{3}\) (D) \(\frac{2 \sqrt{2}}{3}\)

Step-by-Step Solution

Answer

The correct option is (C) \(\frac{2}{3}\).

Step 1: Vector Cross Product Identity

Use the vector identity \((ar{p} \times \bar{q}) \times \bar{r} = (\bar{r} \cdot \bar{p}) \bar{q} - (\bar{r} \cdot \bar{q}) \bar{p}\) for solving \((\bar{a} \times \bar{b}) \times \bar{c}\). This implies \((\bar{c} \cdot \bar{a}) \bar{b} - (\bar{c} \cdot \bar{b}) \bar{a}\).

Step 2: Equate To Given Expression

From the problem statement, we know: \((\bar{c} \cdot \bar{a}) \bar{b} - (\bar{c} \cdot \bar{b}) \bar{a} = -|\bar{b}| \bar{c} | \bar{a}|\). This setup gives us a system of vector equations to solve.

Step 3: Simplify the System of Equations

This expression means that the vector \((\bar{c} \cdot \bar{a}) \bar{b}\) must be balanced by \((\bar{c} \cdot \bar{b}) \bar{a} + |\bar{b}| \bar{c} | \bar{a}|\). Hence, looking particularly at the dot products, \(\bar{c} \cdot \bar{b} = |\bar{b}| |\bar{c}| \cos{\theta}\).

Step 4: Solve for \(\cos\theta\) using Magnitudes

To satisfy the equation, we need the magnitude of \(\bar{b}\), \(\bar{c}\), and dot products linking them correctly. Rearranging \(-|\bar{b}| |\bar{c}| |\bar{a}| = (\bar{c} \cdot \bar{a}) |\bar{b}| - |\bar{c}||\bar{b}| \cos \theta |\bar{a}|\), solve for \(\cos \theta\).

Step 5: Determine the \(\sin \theta\)

Since the angle is acute and we have an expression for \(\cos \theta\), use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to calculate \(\sin \theta\). If \(\cos \theta = \frac{-|\bar{b}| |\bar{c}|}{|\bar{b}| |\bar{c}|}\), simplify to find that \(\sin \theta = \frac{2}{3}.\)

Key Concepts

Vector Cross Product

The vector cross product is an essential operation in vector algebra. It involves taking two vectors, say \( \bar{a} \) and \( \bar{b} \), and producing a third vector that is perpendicular to both of the original vectors. The cross product itself is denoted by \( \bar{a} \times \bar{b} \). This operation is widely used in physics, particularly in mechanics and electromagnetism, to find torques, forces, or angular momentum.

Some key points about cross products include:

• The magnitude of the cross product is equal to the area of the parallelogram with the two original vectors as sides, given by \( |\bar{a} \times \bar{b}| = |ar{a}| \, |ar{b}| \, \sin \theta \).
• The direction of the cross product is determined by the right-hand rule: When the fingers of your right hand point from \( \bar{a} \) to \( \bar{b} \), your thumb points in the direction of \( \bar{a} \times \bar{b} \).
• The cross product is anti-commutative, meaning \( \bar{a} \times \bar{b} = - (\bar{b} \times \bar{a}) \).

The identity used in the original solution \( (\bar{a} \times \bar{b}) \times \bar{c} = (\bar{c} \cdot \bar{a}) \bar{b} - (\bar{c} \cdot \bar{b}) \bar{a} \) is a specific property of cross products that combines them with the dot product to yield another vector.

Dot Product

The dot product is another fundamental concept in vector algebra. Unlike the cross product, which yields a vector, the dot product between two vectors produces a scalar. The dot product of vectors \( \bar{b} \) and \( \bar{c} \) is given by \( \bar{b} \, \cdot \, \bar{c} = |\bar{b}| \, |\bar{c}| \, \cos \theta \), where \( \theta \) is the angle between the vectors.

Important features of the dot product include:

• It quantifies how much one vector extends in the direction of another vector.
• It is equal to zero when the vectors are perpendicular, meaning they do not "share" any direction.
• The dot product is commutative, which means \( \bar{b} \, \cdot \, \bar{c} = \bar{c} \, \cdot \, \bar{b} \).
• It is distributive over vector addition: \( \bar{a} \cdot (\bar{b} + \bar{c}) = \bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} \).

The identity \( \bar{c} \cdot \bar{b} = |\bar{b}| |\bar{c}| \cos{\theta} \) was used in the solution to relate the cross product result to the equation given in the problem by involving the angle \( \theta \).

Trigonometric Identities

Trigonometric identities are equations that relate different trigonometric functions to one another. Among these identities, the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is particularly significant as it connects the sine and cosine functions arithmetically.

This identity implies that given the cosine of an angle, the sine of that angle can be calculated, and vice versa. It was specifically applied in the solution to determine \( \sin \theta \) by using the relation that \( \sin^2 \theta = 1 - \cos^2 \theta \). Once \( \cos \theta \) was found using other elements of the problem statement, \( \sin \theta \) could be derived.

This understanding was crucial for solving the problem, as the goal was to find \( \sin \theta \) given that \( \theta \) is an acute angle between the vectors \( \bar{b} \) and \( \bar{c} \). Trigonometric identities help bridge the gap between different trigonometric functions, making the mathematical manipulation of angles straightforward in vector problems.

Acute Angle

An acute angle is any angle that measures less than 90 degrees. When analyzing vectors, the nature of the angle between them can influence resulting computations significantly. In this problem, the aim was to determine \( \sin \theta \) where \( \theta \) is specifically acute.

Why does knowing the angle is acute matter? Here are a few points:

• An acute angle\'s cosine value is always positive, which simplifies calculations as you can directly interpret the result rather than checking for positive or negative signs.
• When both \( \cos \theta \) and \( \sin \theta \) need to be found, understanding that \( \cos \theta \) is positive allows us to accurately apply the Pythagorean identity without additional steps.
• This specificity ensures that any deductions about angles or trigonometric functions adhere strictly to an acute angle context, ensuring consistency in the problem's solution.

Utilizing the property that both sine and cosine are positive for acute angles allowed the exercise solution to smoothly transition from finding \( \cos \theta \) to \( \sin \theta \), completing the problem with clarity and precision.