Let a, b and c be three non-zero vectors such that no two of these are collinear. If the vector a +

Q: Wait, why do we reject the negative root here? Or rather, why do we assume the coefficients must be zero?

That is a fantastic question! We aren't rejecting a root; we are using the fundamental property of linear independence. Because $\vec{a}$ and $\vec{c}$ are non-collinear, they cannot be expressed as multiples of each other. Therefore, the only way a linear combination like $x\vec{a} + y\vec{c} = \vec{0}$ can exist is if $x=0$ and $y=0$. It is the bedrock of vector algebra!

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm! However, this is a pure vector algebra problem. Energy conservation applies to physical systems with forces and motion. Here, we are dealing with the abstract geometry of space. Stick to the vector properties—they are the 'laws of physics' for this specific problem!

Q: Is it possible that $\vec{a} + 2\vec{b} + 6\vec{c}$ is not the zero vector?

Mathematically, no. Given the constraints that no two vectors are collinear, the system forces the target expression to collapse into the zero vector. It is a beautiful example of how constraints in a system dictate the final outcome.

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

Options:

(A) $\vec{0}$

(B) $\lambda\vec{b}$

(C) $\lambda\vec{c}$

(D) $\lambda\vec{a}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space with three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. These vectors are non-zero, and no two of them are collinear, meaning they point in unique, independent directions.

Our mission is to evaluate the vector expression:
$$\vec{V} = \vec{a} + 2\vec{b} + 6\vec{c}$$

### Decoding the Conditions

The problem provides two vital pieces of information. First, $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$. In vector language, this implies:
$$\vec{a} + 2\vec{b} = m\vec{c}$$
where $m$ is a scalar.

Second, $\vec{b} + 3\vec{c}$ is collinear with $\vec{a}$. Following the same logic, we write:
$$\vec{b} + 3\vec{c} = n\vec{a}$$
where $n$ is another scalar. These two equations are the keys to solving the system.

### The Art of Substitution

Consider our target vector $\vec{V} = \vec{a} + 2\vec{b} + 6\vec{c}$. We can substitute the first condition into this expression:
$$\vec{V} = (\vec{a} + 2\vec{b}) + 6\vec{c} = m\vec{c} + 6\vec{c} = (m + 6)\vec{c}$$

Alternatively, take the second equation $\vec{b} + 3\vec{c} = n\vec{a}$ and multiply it by $2$:
$$2\vec{b} + 6\vec{c} = 2n\vec{a}$$

Now, substitute this into the target expression $\vec{V} = \vec{a} + (2\vec{b} + 6\vec{c})$:
$$\vec{V} = \vec{a} + 2n\vec{a} = (1 + 2n)\vec{a}$$

### The Power of Linear Independence

We now have two expressions for the same vector $\vec{V}$:
$$\vec{V} = (m + 6)\vec{c} \quad 	ext{and} \quad \vec{V} = (1 + 2n)\vec{a}$$

Equating these two, we get:
$$(1 + 2n)\vec{a} = (m + 6)\vec{c} \implies (1 + 2n)\vec{a} - (m + 6)\vec{c} = \vec{0}$$

Because $\vec{a}$ and $\vec{c}$ are not collinear, they are linearly independent. A linear combination of independent vectors equals the zero vector if and only if the coefficients are zero:
$$1 + 2n = 0 \quad 	ext{and} \quad m + 6 = 0$$

### The Grand Finale

From $m + 6 = 0$, we find $m = -6$. Substituting this back into our first expression for $\vec{V}$:
$$\vec{V} = (-6 + 6)\vec{c} = 0\vec{c} = \vec{0}$$

The entire expression collapses into the **zero vector**. The final result is:
$$\vec{V} = \vec{0}$$

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be three non-zero vectors such that no two of these are collinear. If the vector $a + 2 b$ is collinear with $c$ and $b + 3 c$ is collinear with $a$ ( $λ$ being some non-zero scalar) then $a + 2 b + 6 c$ equals

Select Answer:

Visualized Solution

Visualizing the Vectors $a, b, c$

Condition 1: Collinearity with $c$

Condition 2: Collinearity with $a$

Expressing the Target Vector (Method 1)

Simplifying Method 1

Expressing the Target Vector (Method 2)

Simplifying Method 2

Equating the Two Expressions

Applying Linear Independence

Final Calculation and Result

The Sigma Insight: Addition of Vectors

Analyzing the Setup

Decoding the Conditions

The Art of Substitution

The Power of Linear Independence

The Grand Finale

Similar Questions

If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$ , then

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then:

The vector $a = α \hat{i} + 2 \hat{j} + β \hat{k}$ lies in the plane of the vectors $b = \hat{i} + \hat{j}$ and $c = \hat{j} + \hat{k}$ and bisects the angle between $b$ and $c$ . Then which one of the following gives possible values of $α$ and $β$ ?

Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

If $a$ and $b$ are unit vectors, then the greatest value of $3 ∣ a + b ∣ + ∣ a - b ∣$ is

If the position vectors of the vertices A, B and C of a $Δ A B C$ are respectively $4 \hat{i} + 7 \hat{j} + 8 \hat{k}$ , $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ , then the position vector of the point, where the bisector of $∠ A$ meets BC is :-

Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

If the vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $A B C$ , then the length of the median through $A$ is