Problem 24 - 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 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) 0 (b) \(\lambda \vec{b}\) (c) \(\lambda \vec{c}\) (d) \(\lambda \vec{a}\)

Step-by-Step Solution

Answer

The vector is \( \lambda\vec{c} \).

Step 1: Understanding Collinearity

The vector \( \vec{u} \) is collinear with vector \( \vec{v} \) if \( \vec{u} = k \vec{v} \) for some scalar \( k \). Given that \( \vec{a}+2\vec{b} \) is collinear with \( \vec{c} \), we can write \( \vec{a} + 2\vec{b} = k_1 \vec{c} \).

Step 2: Expressing Second Condition

For the vector \( \vec{b}+3\vec{c} \) to be collinear with \( \vec{a} \), it can be expressed as \( \vec{b} + 3\vec{c} = k_2 \vec{a} \).

Step 3: Substituting and Analyzing

Substitute expression from Step 2 into Step 1: \( \vec{a} + 2(k_2\vec{a} - 3\vec{c}) = k_1\vec{c} \). Simplify it to \( (1 + 2k_2)\vec{a} - 6\vec{c} = k_1\vec{c} \).

Step 4: Solving for Scalars

Since both sides are vectors that are equal, separate the vector components: \( (1 + 2k_2)\vec{a} = 0 \) and \( k_1\vec{c} = 6\vec{c} \). This implies \( k_1 = 6 \) and since vectors are non-zero, \( 1 + 2k_2 = 0 \).

Step 5: Calculating \( k_2 \)

Rearranging \( 1 + 2k_2 = 0 \) gives \( k_2 = -\frac{1}{2} \).

Step 6: Finding \( \vec{a} + 2 \vec{b} + 6 \vec{c} \)

Given \( \vec{a} + 2\vec{b} = 6\vec{c} \), substitute \( 6\vec{c} = \vec{a} + 2\vec{b} \) into \( \vec{a} + 2\vec{b} + 6\vec{c} \). We get \( = 6\vec{c} + 6\vec{c} = 12\vec{c} \).

Step 7: Final Conclusion

By considering the scalar multiplication, \( 12\vec{c} \) can be expressed as a multiple of \( \vec{c} \) with \( \lambda = 12 \). So, \( \vec{a} + 2 \vec{b} + 6 \vec{c} = \lambda\vec{c} \).

Key Concepts

Collinearity of Vectors

Understanding the collinearity of vectors is crucial in vector algebra. Two vectors are collinear if they lie along the same line, which means one is a scalar multiple of the other. If vector \( \vec{u} \) is collinear with vector \( \vec{v} \), this relationship can be mathematically expressed as \( \vec{u} = k \vec{v} \), where \( k \) is some scalar value. It's like saying, one vector can "stretch" to match the direction and perhaps the length of the other, by multiplying with \( k \). Collinearity becomes an essential condition when solving vector problems, as it provides us with powerful relationships that can simplify the algebra involved.

Scalar Multiplication

Scalar multiplication is a process where you multiply a vector by a scalar (a real number). This operation changes the magnitude of the vector but not its direction, unless the scalar is negative, which reverses the direction. Formally, if \( \vec{v} \) is a vector and \( k \) is a scalar, then the scalar multiple \( k \vec{v} \) is a new vector in the same or opposite direction as \( \vec{v} \), with a magnitude \( |k| \) times that of \( \vec{v} \). Here’s how scalar multiplication can simplify vector problems:

• Helps to express collinear vectors with relative ease by setting up equations based on given conditions.
• Leads to find relationships between vectors that may otherwise seem complex.
• Enables manipulation of vector components which can be useful in a variety of applications from physics to engineering.

Non-collinear Vectors

When vectors are non-collinear, they do not lie on the same line, and hence, cannot be expressed as scalar multiples of each other. In vector mathematics, non-collinear vectors provide a wealth of additional information, as they allow representations in multiple dimensions. For instance, in a 3D space, the vectors might represent three independent axes.
Key insights on non-collinear vectors:

• They ensure independence, providing more freedom and complexity to problems.
• These vectors form the basis for defining planes and volumes, expanding beyond the simplicity of line-based vector relationships.
• Understanding them is essential for higher-dimensional algebraic manipulations and geometric interpretations.

In the given problem, non-collinearity guarantees that the vectors \( \vec{a}, \vec{b}, \vec{c} \) cannot be reduced to simpler expressions, maintaining the dimensional complexity required by the conditions set within the exercise.