Let a, b and c be three non-zero vectors such that b and c are non-collinear if a + 5b is collinear

Q: Wait, why do we reject the possibility that $\vec{b}$ and $\vec{c}$ could be collinear?

We don't reject it; the problem statement explicitly tells us they are non-collinear. If they were collinear, they would not form a basis for the plane, and we wouldn't be able to uniquely equate the coefficients of $\vec{b}$ and $\vec{c}$ to solve for $\lambda$ and $\mu$.

Q: Can I solve this using the cross product instead of substitution?

Absolutely! You could take the cross product of the equations with $\vec{b}$ or $\vec{c}$. However, substitution is often more direct for this type of linear system. Both methods are mathematically sound, but substitution keeps the algebra cleaner here.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear if $\vec{a} + 5\vec{b}$ is collinear with $\vec{c}$, $\vec{b} + 6\vec{c}$ is collinear with $\vec{a}$ and $\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$, then $\alpha + \beta$ is equal to

Options:

(A) 35

(B) 30

(C) -30

(D) -25

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have three non-zero vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$.

The problem provides a crucial piece of information: $\vec{b}$ and $\vec{c}$ are non-collinear. This is the anchor of our entire solution, as they define a plane and act as a coordinate system for any vector lying within that plane.

### The First Bridge: Translating Collinearity

The problem presents us with two conditions. First, $\vec{a} + 5\vec{b}$ is collinear with $\vec{c}$.

In the language of linear algebra, this translates to a scalar relationship:
$$\vec{a} + 5\vec{b} = \lambda\vec{c}$$
where $\lambda$ is some unknown scalar. This implies that the vector sum $\vec{a} + 5\vec{b}$ points in the same direction as $\vec{c}$, scaled by $\lambda$.

Similarly, the second condition states that $\vec{b} + 6\vec{c}$ is collinear with $\vec{a}$:
$$\vec{b} + 6\vec{c} = \mu\vec{a}$$
where $\mu$ is another scalar. We now have two equations that act as bridges between our three vectors.

### The Algebraic Dance: Substitution

Now, let's weave these equations together to find the relationship between $\vec{a}$, $\vec{b}$, and $\vec{c}$. From our first equation, we isolate $\vec{a}$:
$$\vec{a} = \lambda\vec{c} - 5\vec{b}$$

We take this expression for $\vec{a}$ and substitute it into our second equation:
$$\vec{b} + 6\vec{c} = \mu(\lambda\vec{c} - 5\vec{b})$$

Expanding the right side carefully, we obtain:
$$\vec{b} + 6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$$

### The Power of Linear Independence

Because $\vec{b}$ and $\vec{c}$ are non-collinear, they are linearly independent. This means that for a linear combination of $\vec{b}$ and $\vec{c}$ to equal another linear combination of the same vectors, the coefficients must match individually.

On the left side, the coefficient of $\vec{b}$ is $1$, and on the right, it is $-5\mu$. Thus:
$$1 = -5\mu \Rightarrow \mu = -\frac{1}{5}$$

Similarly, for $\vec{c}$, the coefficient on the left is $6$, and on the right, it is $\mu\lambda$. Setting these equal:
$$6 = \mu\lambda$$

Substituting our value for $\mu$, we get $6 = (-\frac{1}{5})\lambda$, which leads us to:
$$\lambda = -30$$

### The Final Revelation

We have found our scalars. Plugging $\lambda = -30$ back into our first equation:
$$\vec{a} + 5\vec{b} = -30\vec{c}$$

Rearranging this to match the target form $\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$, we get:
$$\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$$

Comparing this to the target equation, we identify $\alpha = 5$ and $\beta = 30$. The final step is to calculate the sum:
$$\alpha + \beta = 5 + 30 = \mathbf{35}$$

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be three non-zero vectors such that $b$ and $c$ are non-collinear if $a + 5 b$ is collinear with $c$ , $b + 6 c$ is collinear with $a$ and $a + α b + β c = 0$ , then $α + β$ is equal to

Select Answer:

Visualized Solution

Initial Vectors

First Collinearity Condition

Second Collinearity Condition

Substitution Strategy

Plugging into Equation 2

Expanding the Expression

Linear Independence Property

Solving for $μ$

Solving for $λ$

Final Vector Equation

Comparing with Target Form

Final Answer Calculation

The Sigma Insight: Scalar and Vector Quantities

Analyzing the Setup

The First Bridge

The Algebraic Dance

The Power of Linear Independence

The Final Revelation

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