Problem 80 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of these are collinear. If the vector \(\vec{a}+2 \vec{b}\) is collinear with \(\vec{c}\) and \(\vec{b}+3 \vec{c}\) is collinear with \(\vec{a}(\lambda\) being some non-zero scalar) then \(\vec{a}+2 \vec{b}+6 \vec{c}\) equals [2004](A) \(\lambda \vec{a}\) (B) \(\lambda \vec{b}\) (C) \(\lambda \vec{c}\) (D) 0

Step-by-Step Solution

Answer

The vector \( \vec{a}+2\vec{b}+6\vec{c} \) equals 0.

Step 1: Understanding Vector Collinearity

Two vectors are collinear if one is a scalar multiple of the other. Given that \( \vec{a} + 2\vec{b} \) is collinear with \( \vec{c} \), it means there exists a scalar \( \mu \) such that \( \vec{a} + 2\vec{b} = \mu \vec{c} \). Similarly, given that \( \vec{b} + 3\vec{c} \) is collinear with \( \vec{a} \), there exists a scalar \( u \) such that \( \vec{b} + 3\vec{c} = u \vec{a} \).

Step 2: Writing Equations for Collinearity

From the given conditions, we set up the equations: 1. \( \vec{a} + 2\vec{b} = \mu \vec{c} \) 2. \( \vec{b} + 3\vec{c} = u \vec{a} \).

Step 3: Expressing Vectors in Terms of Other Vectors

Using the first equation, express \( \vec{a} \) in terms of \( \vec{b} \) and \( \vec{c} \): \( \vec{a} = \mu \vec{c} - 2\vec{b} \). Using the second equation, express \( \vec{b} \) in terms of \( \vec{a} \) and \( \vec{c} \): \( \vec{b} = u \vec{a} - 3\vec{c} \).

Step 4: Checking for Consistency

Substitute \( \vec{a} \) as \( \mu \vec{c} - 2\vec{b} \) into \( \vec{b} = u \vec{a} - 3\vec{c} \): \( \vec{b} = u (\mu \vec{c} - 2\vec{b}) - 3\vec{c} \). Simplify to: \( \vec{b} = u \mu \vec{c} - 2u \vec{b} - 3\vec{c} \). This equation simplifies to \( \vec{b} + 2u \vec{b} = (u \mu - 3)\vec{c} \), which implies \( \vec{b} (1 + 2u) = (u \mu - 3) \vec{c} \). Since \( \vec{b} \) and \( \vec{c} \) are non-collinear, both sides must be zero: 1. \( 1 + 2u = 0 \) so \( u = -\frac{1}{2} \). 2. \( u \mu - 3 = 0 \) so \( \mu = -6 \).

Step 5: Finding the Resultant Vector

Substitute \( u = -\frac{1}{2} \) and \( \mu = -6 \) into the definition: \( \vec{a} + 2\vec{b} + 6\vec{c} = (\mu \vec{c} + 6\vec{c}) = 0 \). Hence, \( \vec{a} + 2\vec{b} + 6\vec{c} \) is equal to zero.

Key Concepts

Collinearity in Vectors

Understanding collinearity is important when working with vectors. Two vectors are collinear if they lie on the same or directly opposite lines. Essentially, this means one vector is just a scaled version of the other. For example, vector \( \vec{a} \) and vector \( \vec{b} \) are collinear if we can express \( \vec{a} = k \vec{b} \), where \( k \) is a scalar. This simplifies analysis in vector algebra, as it places vectors along the same line in space.
Given the exercise, we have \( \vec{a} + 2\vec{b} \) collinear with \( \vec{c} \). Therefore, there exists a scalar \( \mu \) so that \( \vec{a} + 2\vec{b} = \mu \vec{c} \). Likewise, if \( \vec{b} + 3\vec{c} \) is collinear with \( \vec{a} \), then \( \vec{b} + 3\vec{c} = u \vec{a} \) where \( u \) is another scalar.
These relationships allow us to explore properties of the vectors by working with scalar multiples, which is key in understanding the collinearity concept between non-parallel vectors when solving vector algebra problems.

Scalar Multiplication of Vectors

Scalar multiplication involves multiplying a vector by a scalar, changing the magnitude of the vector while keeping its direction the same. It is a fundamental operation in vector algebra. If you multiply a vector \( \vec{v} \) by a positive scalar \( c \), the vector extends in the same direction but is longer or shorter depending on \( c \).
In the exercise, the scalar multipliers \( \mu \) and \( u \) create conditions under which vectors \( \vec{a} + 2\vec{b} \) and \( \vec{b} + 3\vec{c} \) align with others forming collinear relationships. Here, \( \vec{a} \) is expressed through \( \vec{c} \) and \( \vec{b} \) in \( \vec{a} = \mu \vec{c} - 2\vec{b} \), showing how scalar multipliers control vector alignment and magnitude, simplifying their expressions.
This reveals the power of scalar multiplication in altering vector magnitudes without changing their directional properties, crucial for transforming and simplifying vector equations.

Understanding Vector Algebra

Vector algebra is the study of vectors using algebraic operations. These include addition, subtraction, and multiplication (by scalars or vectors themselves). Vectors possess both magnitude and direction, making their manipulation unique.
The exercise showcases vector algebra through combined operations and scalar multiplication. We determine the expression of vectors in relation to the collinear conditions provided. By substituting one vector expression into another, we simplify and find the parameters \( u \) and \( \mu \) which satisfy the equation. Thus, vector algebra helps navigate complex spatial relationships by transforming expressions into manageable equations.
Understanding vector algebra extends beyond basic arithmetic. It includes recognizing the geometric implications, knowing how vectors interact in space, and applying linear transformations to solve geometrical problems, all highlighted in simplifying the equations to achieve a zero result in the exercise.