Problem 16 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let the unit vectors \(a\) and \(b\) be perpendicular to each other and the unit vector \(c\) be inclined at an angle \(\theta\) to both \(a\) and \(b\). If \(c=x a+y b+z(a \times b)\), then (A) \(x=\cos \theta, y=\sin \theta, z=\cos 2 \theta\) (B) \(x=\sin \theta, y=\cos \theta, z=-\cos 2 \theta\) (C) \(x=y=\cos \theta, z^{2}=\cos 2 \theta\) (D) \(x=y=\cos \theta, z^{2}=-\cos 2 \theta\)

Step-by-Step Solution

Answer

The answer is (C): \(x = y = \cos \theta, z^2 = \cos 2\theta\).

Step 1: Understand the Given Vectors and Their Properties

We are given that vectors \(a\) and \(b\) are unit vectors and perpendicular to each other, meaning \(a \cdot b = 0\). The vector \(c\) is inclined at an angle \(\theta\) to both \(a\) and \(b\), meaning \(c \cdot a = \cos \theta\) and \(c \cdot b = \cos \theta\). The vector \(c\) is expressed in terms of \(a\), \(b\), and their cross product \(a \times b\).

Step 2: Express Vector c in Terms of Components

Vector \(c\) can be written as \(c = x a + y b + z (a \times b)\), where \(x, y,\) and \(z\) are coefficients to be determined. This expression implies that \(c\) lies in the plane formed by \(a\) and \(b\), plus possibly a component along the perpendicular direction \(a \times b\).

Step 3: Find Constraints on Coefficients Using Dot Product

Use the given condition \(c \cdot a = \cos \theta\) to find one constraint: \(x = c \cdot a = \cos \theta\). Similarly, use \(c \cdot b = \cos \theta\) to find another constraint: \(y = c \cdot b = \cos \theta\). Thus, \(x = y = \cos \theta\).

Step 4: Analyze Perpendicularity Conditions for z

Since \(c\) must also be a unit vector, we use the condition \(\|c\| = 1\), which leads to \(x^2 + y^2 + z^2 = 1\). Substituting \(x = y = \cos \theta\), we obtain \(2\cos^2 \theta + z^2 = 1\).

Step 5: Simplify to Solve for z

From the equation \(2\cos^2 \theta + z^2 = 1\), simplify to get \(z^2 = 1 - 2\cos^2 \theta = \cos 2\theta\). This matches condition set by option (C).

Step 6: Compare with Given Options

Given the solutions above, compare them with the options provided. We have found \(x = y = \cos \theta\) and \(z^2 = \cos 2\theta\). This matches exactly with option (C).

Key Concepts

Unit Vectors

Unit vectors are essential building blocks in vector algebra. They have a magnitude of 1 and are used to indicate direction only. Think of them like navigation arrows that show the way without concern for distance.
For example, if we have unit vectors \(a\) and \(b\), they essentially just point in particular directions. In our exercise, both \(a\) and \(b\) are unit vectors, giving us useful properties when performing calculations with them. Since unit vectors have a magnitude of 1, their length simplifies many vector operations and equations.

• Unit vectors are commonly expressed in terms of their components in coordinate systems. For instance, \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are unit vectors along the x, y, z axes, respectively.
• In our problem, knowing \(a\) and \(b\) are unit vectors helps us deduce mathematically clean results from their dot products or cross products.

This foundational concept helps because any vector can be broken down into components aligned with these unit vectors.

Perpendicular Vectors

Perpendicular vectors meet at a right angle, like the walls of a room. Mathematically, if two vectors \(a\) and \(b\) are perpendicular, their dot product is zero: \(a \cdot b = 0\). This is exactly what's happening in our original exercise.
In physics and engineering, perpendicular vectors often signify situations where forces or motions are independent of each other.

• In our case, vectors \(a\) and \(b\) being perpendicular simplifies certain calculations. By using their dot product result of zero, we see that they don't influence one another's direction in the same plane.
• Knowing vectors \(a\) and \(b\) are perpendicular guides us in understanding how \(c\) can be constructed using them.

Parts of \(c\) lying in the same plane won't be affected by the other's direction, making them easier to analyze.

Vector Cross Product

The vector cross product is a powerful tool for finding a vector that's perpendicular to two other vectors. Given vectors \(a\) and \(b\), their cross product \(a \times b\) results in a vector orthogonal to both.
This feature is immensely useful in applications where perpendicularity plays a role, like finding the normal to a surface. In our exercise, the cross product \(a \times b\) provides a component for \(c\) that's distinct from the components along \(a\) and \(b\).

• The cross product magnitude is given by \(|a \times b| = |a| |b| \sin \theta\). Since \(a\) and \(b\) are unit vectors and perpendicular, this simplifies further. The direction is set by the right-hand rule.
• In the equation \(c = x a + y b + z (a \times b)\), the \(z (a \times b)\) term captures the presence of \(c\) outside the \(a-b\) plane.

Utilizing the properties of the cross product, we enhance our understanding of how vectors interact spatially.

Dot Product Properties

The dot product of two vectors reveals geometric insights, like the angle between them. It's calculated as \(a \cdot b = |a| |b| \cos \theta\). If one vector projects onto another, this operation helps in figuring out how much it aligns with that vector.
In our exercise, the dot product is instrumental in setting the boundaries for \(x\) and \(y\) using \(c \cdot a = \cos \theta\) and \(c \cdot b = \cos \theta\). When \(a\) and \(b\) are unit vectors and \(\theta\) is the angle between them and \(c\), these dot products translate into simple trigonometric terms.

• The properties of the dot product like commutativity and distributivity facilitate breaking down complex equations.
• The normalization (where \(|c| = 1\)) provides a constraint helping solve for unknowns like \(z\).

By understanding dot product properties, we can decipher how components of \(c\) parallel to \(a\) and \(b\) are expressed as \(x = y = \cos \theta\). This makes solving geometrical problems succinct and structured.