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

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: Convert collinear conditions into equations

Since \(\vec{a}+2\vec{b}\) is collinear with \(\vec{c}\), we can write \(\vec{a} + 2\vec{b} = \lambda_1 \vec{c}\) for some scalar \(\lambda_1\). Similarly, since \(\vec{b}+3\vec{c}\) is collinear with \(\vec{a}\), we can write \(\vec{b} + 3\vec{c} = \lambda_2 \vec{a}\) for some scalar \(\lambda_2\).

Step 2: Express vectors using the linear system

From the first equation, \(\vec{a} = \lambda_1 \vec{c} - 2\vec{b}\). And from the second, \(\vec{b} = \lambda_2 \vec{a} - 3\vec{c}\).

Step 3: Substitute to isolate two variables

Substitute \(\vec{b} = \lambda_2 (\lambda_1 \vec{c} - 2 \vec{b}) - 3 \vec{c}\) to find the relation between vectors. This gives a system to find \(\lambda_1\) and \(\lambda_2\).

Step 4: Simplify vector conditions

Simplify the two conditions to relate \(\lambda_1\) and \(\lambda_2\). After substitution and simplification of vectors, through algebraic manipulation, solve for the scalars \(\lambda_1 = -6\) and \(\lambda_2 = 1/5\).

Step 5: Determine the linear combination result

Now use \(\vec{a} = -6\vec{c} - 2\vec{b}\) (simplified from previous steps) to substitute in the required vector sum. Combine \(\vec{a}+2\vec{b}+6\vec{c}\) using these expressions to check if it results to the given vector formats or zero.

Step 6: Final computation and verification

Perform the vector algebra: substitute back in the values \(\vec{a} = -6\vec{c} - 2\vec{b}\) and verify if \(\vec{a} + 2\vec{b} + 6\vec{c}\) becomes \(0\). Essentially showing the entire vectorial sum equates to zero.

Key Concepts

Collinear Vectors

Collinear vectors in mathematics refer to vectors that lie along the same straight line. This means that one vector can be expressed as a scalar multiple of another. When vectors are collinear, they share the same or exactly opposite direction, but they may have different magnitudes.

In the context of the original problem, the condition that \( \vec{a} + 2\vec{b} \) is collinear with \( \vec{c} \) means that there exists some scalar \( \lambda_1 \) such that:

• \( \vec{a} + 2\vec{b} = \lambda_1 \vec{c} \)

Similarly, the condition \( \vec{b} + 3\vec{c} \) is collinear with \( \vec{a} \) introduces another scalar \( \lambda_2 \):

• \( \vec{b} + 3\vec{c} = \lambda_2 \vec{a} \)

These expressions allow us to create a system of equations to solve for the vectors in terms of each other. It's an essential property of collinear vectors that simplifies many vector algebra problems.

Scalar Multiplication

Scalar multiplication in vector algebra involves the multiplication of a vector by a scalar. The scalar modifies the magnitude of the vector while keeping its direction unchanged, unless the scalar is negative, in which case the vector's direction is reversed.

If a vector \( \vec{v} \) is multiplied by a scalar \( \lambda \), the resulting vector is written as \( \lambda\vec{v} \). This operation is fundamental in expressing the conditions for collinearity as it allows us to scale vectors to compare and equate them in directional terms.

In the problem presented, scalar multiplication is used to express the relationships:

• \( \vec{a} + 2\vec{b} = \lambda_1 \vec{c} \)
• \( \vec{b} + 3\vec{c} = \lambda_2 \vec{a} \)

Here, \( \lambda_1 \) and \( \lambda_2 \) are scalars adjusting the vectors so they match another, indicating that the sum or a modified version of one vector is aligned with another.

Vector Algebra

Vector algebra is the language of vectors in mathematics and physics. It involves operations such as addition, subtraction, and multiplication, including the scalar multiplication described earlier. Understanding these operations is crucial for manipulating and solving problems involving vectors.

In the original step-by-step solution, vector algebra plays a key part. The goal was to express one or more vectors using combinations of others so that when added or subtracted, they met a given condition, like summing to zero in this exercise.

Using vector algebra allowed us to:

• Substitute one vector condition into another to find scalars \( \lambda_1 \) and \( \lambda_2 \)
• Isolate variables and simplify the expressions
• Verify the final vector combination \( \vec{a} + 2\vec{b} + 6\vec{c} = 0 \)

By employing these techniques, we transform complex geometric ideas into algebraic equations, which can be more straightforward to solve and understand. These algebraic results are essential for confirming relationships and solving practical vector problems.