Problem 58 - A Complete Resource Book in Mathematics for JEE Main

Question

The vectors \(a, b, c\) are of same length and taken pairwise, they form equal angles. If \(a=i+j\) and \(b=j+k\), then \(c=\) (A) \(i+k\) (B) \(j+k\) (C) \(i+k\) (D) \(-\frac{i}{3}+\frac{4}{3} j-\frac{k}{3}\)

Step-by-Step Solution

Answer

The vector \( c = -\frac{i}{3} + \frac{4}{3}j - \frac{k}{3} \), matching option D.

Step 1: Understanding the Vector Properties

The problem states that vectors \( a, b, \) and \( c \) not only are of equal length but also form the same angle pairwise. This implies that the dot products between each pair are equal provided the vectors' magnitudes are equal.

Step 2: Calculate Magnitude of Vectors

First, we calculate the magnitude of given vectors. For \( a = i + j \), the magnitude is \(|a| = \sqrt{1^2 + 1^2} = \sqrt{2}\). Similarly, for \( b = j + k \), \(|b| = \sqrt{1^2 + 1^2} = \sqrt{2}\). Thus, \(|c| = \sqrt{2}\) as well.

Step 3: Use the Dot Product Equivalence Property

Since \( a \cdot b = a \cdot c = b \cdot c\), we calculate \( a \cdot b \). The vectors \( a = i + j \) and \( b = j + k \) give the dot product \( a \cdot b = (0 \cdot 0 + 1 \cdot 1 + 0 \cdot 1) = 1 \). So, \( a \cdot c = b \cdot c = 1 \).

Step 4: Express Dot Products in Terms of \( c = xi + yj + zk \)

Express vector \( c \) as \( c = xi + yj + zk \) and equate the dot products: \( a \cdot c = (1 \cdot x + 1 \cdot y + 0 \cdot z) = x + y = 1 \). Similarly, \( b \cdot c = (0 \cdot x + 1 \cdot y + 1 \cdot z) = y + z = 1 \).

Step 5: Solve the System of Equations

We now have two equations: \( x+y=1 \) and \( y+z=1 \). From \( x+y=1 \), we can express \( x = 1 - y \). Substitute into the second equation to get \( 1-y+z=1 \), leading to \( z = y \).

Step 6: Set Magnitude Equation for \( c \)

The magnitude condition gives the equation \( x^2 + y^2 + z^2 = 2 \). Substitute \( x = 1 - y \) and \( z = y \) to get \( (1-y)^2 + y^2 + y^2 = 2 \). Simplify this to \( 1 - 2y + y^2 + 2y^2 = 2 \), further to \( 3y^2 - 2y -1 = 0 \).

Step 7: Solve the Quadratic Equation

Solve the quadratic equation \( 3y^2 - 2y - 1 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a=3, b=-2, c=-1 \), getting \( y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{6} \). Simplifying gives \( y = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \). Thus, \( y = 1 \) or \( y = -\frac{1}{3} \).

Step 8: Determine Corresponding Components of \( c \)

For \( y = 1 \), \( x = 0 \) and \( z = 1 \); giving \( c = j + k \). For \( y = -\frac{1}{3} \), \( x = 1 - \left(-\frac{1}{3}\right) = \frac{4}{3} \) and \( z = -\frac{1}{3} \); giving \( c = \frac{4}{3}j - \frac{1}{3}k - \frac{1}{3}i \).

Step 9: Choose Matching Option

Among the given options, \( c = -\frac{i}{3} + \frac{4}{3}j - \frac{k}{3} \) matches option D. Therefore, this is the correct solution.

Key Concepts

Dot Product

The dot product is a fundamental concept in vector mathematics. It's a way to multiply two vectors, resulting in a scalar value. This scalar can tell us a lot about the relationship between two vectors.
One of its key properties is that it is equal to the product of the magnitudes of the vectors and the cosine of the angle between them. Mathematically, this is expressed as:

• \( a \cdot b = |a| |b| \cos \theta \)

where \( \theta \) is the angle between the vectors.
In practical terms, if the dot product is zero, the vectors are orthogonal (they form a 90-degree angle). When the dot product is positive, the vectors point in a similar direction. If negative, they point in opposite directions. In this exercise, we established the equality of dot products for different vector pairs, which led us to solve for vector components.

Vector Magnitude

The magnitude of a vector is akin to its length or size. It is calculated using the square root of the sum of the squares of its components. For a vector \( a = xi + yj + zk \), the magnitude is given by:

• \( |a| = \sqrt{x^2 + y^2 + z^2} \)

Vector magnitude is critical in understanding the scale of the vector and comparing vectors of different sizes. For vectors that are described in terms of unit vectors, like \( i, j, \) and \( k \), calculating the magnitude involves determining the square root of the sum of the components squared.
In our exercise, knowing that the magnitudes of the vectors \( a, b, \) and \( c \) were equal allowed us to equate their expressions and proceed with solving for the variable components of \( c \). This uniformity in magnitude was crucial for solving the quadratic equation.

Quadratic Equation

Quadratic equations are polynomial equations of the second degree, typically taking the form \( ax^2 + bx + c = 0 \). They can be solved using several methods, such as factoring, completing the square, or using the quadratic formula. The quadratic formula, which is often the most straightforward method, is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The solutions to a quadratic equation are known as the roots. These roots can be real or complex numbers, and they represent the x-values where the polynomial equals zero.
In this exercise, applying the quadratic formula enabled us to find the possible values for the vector components after substituting known values into the magnitude equation of vector \( c \). Solving this quadratic equation provided us the values needed for \( y \), which further helped derive the full expression for vector \( c \).

Angle Between Vectors

The angle between vectors is a measure of their direction relative to each other. This concept is particularly significant in understanding how two vectors relate to one another geometrically and is often calculated using the dot product.
The cosine of the angle \( \theta \) between vectors \( a \) and \( b \) can be calculated using the formula:

• \( \cos \theta = \frac{a \cdot b}{|a||b|} \)

Where \( a \cdot b \) is the dot product. Knowing \( \theta \) helps in uncertain terms and projections in vector fields.
In this particular exercise, knowing that the vectors formed equal angles with each other pairwise, allowed for the simplification of calculations among the vector relationships. It helped in equating their pairwise dot products, guiding the solution toward determining equal components within the set angles.