Problem 73 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of them are collinear and \((\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}\). If \(\theta\) is the angle between vectors \(\vec{b}\) and \(\vec{c}\), then a value of \(\sin \theta\) is : [2015] (a) \(\frac{2}{3}\) (b) \(\frac{-2 \sqrt{3}}{3}\) (c) \(\frac{2 \sqrt{2}}{3}\) (d) \(\frac{-\sqrt{2}}{3}\)

Step-by-Step Solution

Answer

The value of \(\sin \theta\) is \(\frac{2\sqrt{2}}{3}\).

Step 1: Understanding the Given Vector Triple Product

We are given \[(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}||\vec{c}| \vec{a}\]Using the vector triple product identity, \[ (\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u} \]we can apply this to yield:\[ ((\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}) = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}\]Since no two vectors are collinear, coefficients of \(\vec{a}\) and \(\vec{b}\) must independently equal zero.

Step 2: Equating Coefficient of \(\vec{a}\)

From the equation \[((\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}) = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}\]we equate coefficients to get:\[ -(\vec{b} \cdot \vec{c}) = \frac{1}{3} |\vec{b}| |\vec{c}|\]Solving for \(\vec{b} \cdot \vec{c}\), we get:\[ \vec{b} \cdot \vec{c} = -\frac{1}{3} |\vec{b}| |\vec{c}| \]

Step 3: Expressing \(\vec{b} \cdot \vec{c}\) in Terms of \(\cos \theta\)

The dot product \(\vec{b} \cdot \vec{c}\) can also be expressed as \[ |\vec{b}||\vec{c}| \cos \theta\]Using \(\vec{b} \cdot \vec{c} = -\frac{1}{3} |\vec{b}| |\vec{c}|\), we equate:\[ |\vec{b}||\vec{c}| \cos \theta = -\frac{1}{3} |\vec{b}| |\vec{c}|\]Cancel \(|\vec{b}||\vec{c}|\) to find:\[ \cos \theta = -\frac{1}{3} \]

Step 4: Using the Identity \(\sin^2 \theta + \cos^2 \theta = 1\)

We know \[ \cos \theta = -\frac{1}{3} \]Next, use the identity \[ \sin^2 \theta + \cos^2 \theta = 1 \]This becomes:\[ \sin^2 \theta + \left(-\frac{1}{3}\right)^2 = 1 \]\[ \sin^2 \theta + \frac{1}{9} = 1 \]Subtract \(\frac{1}{9}\) from both sides to obtain:\[ \sin^2 \theta = 1 - \frac{1}{9} \]\[ \sin^2 \theta = \frac{8}{9} \]

Step 5: Solving for \(\sin \theta\) and Choosing the Correct Value

Take the square root of \(\sin^2 \theta\):\[ \sin \theta = \pm \sqrt{\frac{8}{9}} \]\[ \sin \theta = \pm \frac{2\sqrt{2}}{3} \]However, since the original problem does not specify a direction or sign preference, it's generally considered positive unless context dictates otherwise. Thus, the acceptable value is \(\sin \theta = \frac{2\sqrt{2}}{3}\).

Key Concepts

cross product

The cross product of two vectors, often denoted as \(\vec{u} \times \vec{v}\), results in a vector that is perpendicular to both \(\vec{u}\) and \(\vec{v}\). This operation is defined in three-dimensional space and is not commutative, meaning \(\vec{u} \times \vec{v} eq \vec{v} \times \vec{u}\); however, it is anti-commutative, so \(\vec{u} \times \vec{v} = - (\vec{v} \times \vec{u})\). The magnitude of the cross product is given by the formula:

• \(|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}| \sin \phi\)

where \(\phi\) is the angle between \(\vec{u}\) and \(\vec{v}\). The result of this product is a vector that provides valuable information about the orientation and area of the parallelogram formed by \(\vec{u}\) and \(\vec{v}\).
In our original exercise involving the vector triple product \((\vec{a} \times \vec{b}) \times \vec{c}\), using the triple product identity simplifies the computation, helping us relate the vectors in terms of their dot products instead of their cross products.

dot product

The dot product between two vectors, written as \(\vec{u} \cdot \vec{v}\), yields a scalar quantity. This operation is commutative, i.e., \(\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}\). The dot product is calculated as:

• \(\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos \phi\)

where \(\phi\) is the angle between the two vectors. This product helps measure the extent to which two vectors point in the same direction. For perpendicular vectors, the dot product is zero since \(\cos 90^\circ = 0\).
In our exercise, knowing that \(\vec{b} \cdot \vec{c} = -\frac{1}{3} |\vec{b}||\vec{c}|\) provided insight into the angle \(\theta\) between \(\vec{b}\) and \(\vec{c}\), leading us to find \(\cos \theta = -\frac{1}{3}\).

trigonometric identities

Trigonometric identities are fundamental relationships involving the trigonometric functions, used extensively in various areas of mathematics. A key identity is:

• \(\sin^2 \theta + \cos^2 \theta = 1\)

This identity is invaluable for finding either \(\sin \theta\) or \(\cos \theta\) when one is known. In the context of our problem, knowing \(\cos \theta = -\frac{1}{3}\), we applied this identity to derive \(\sin^2 \theta = \frac{8}{9}\).
This simplification process is crucial as trigonometric identities often link angles and magnitudes in vector calculations. Once \(\sin^2 \theta\) was determined, solving for \(\sin \theta\) was straightforward by taking the square root.

angle between vectors

The angle between two vectors \(\vec{u}\) and \(\vec{v}\) provides a measure of their directional relationship. It is determined using either the dot product or the cross product. When using the dot product, the formula is:

• \(\cos \phi = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}\)

This calculation is straightforward and gives a scalar result indicating the angle. When the dot product leads to a negative cosine value, as seen in our problem with \(\cos \theta = -\frac{1}{3}\), it implies that the angle is obtuse (greater than 90 degrees but less than 180 degrees).
Understanding this angle is crucial in many physics and engineering applications where directional relationships between forces or movements must be understood.