Let a = 2i + j - k and b = i + 2j + k be two vectors. Consider a vector c = αa + βb, α, β ∈ R. If th

Q: Wait, why is the dot product of $\vec{c}$ and $\vec{a} \times \vec{b}$ zero?

That is a fantastic question! Remember that $\vec{c}$ is defined as a linear combination of $\vec{a}$ and $\vec{b}$, which means $\vec{c}$ lies entirely within the plane spanned by $\vec{a}$ and $\vec{b}$. By definition, the cross product $\vec{a} \times \vec{b}$ produces a vector that is perpendicular to that plane. Since $\vec{c}$ is in the plane and $\vec{a} \times \vec{b}$ is perpendicular to it, they are orthogonal. The dot product of any two orthogonal vectors is always zero.

Q: Do I really need to calculate the cross product $\vec{a} \times \vec{b}$?

Not at all! In fact, calculating it would be a waste of precious time in an exam. As we discussed, the geometric property of the scalar triple product allows us to see that the term $(\vec{a} \times \vec{b}) \cdot \vec{c}$ vanishes entirely. Trust the geometry; it is often faster than the algebra.

Q: How do I know when to substitute $\beta = 2 - \alpha$?

You should look for this substitution whenever you have a constraint equation involving your variables. We derived $\alpha + \beta = 2$ from the projection condition. Once you have that, you have a system with one degree of freedom. Reducing the expression to a single variable (like $\alpha$) is the standard way to minimize a quadratic function. It turns a multivariable problem into a simple single-variable calculus or algebraic problem.

Question

Let $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ be two vectors. Consider a vector $\vec{c} = \alpha \vec{a} + \beta \vec{b}, \alpha, \beta \in \mathbb{R}$. If the projection of $\vec{c}$ on the vector $(\vec{a} + \vec{b})$ is $3\sqrt{2}$, then the minimum value of $(\vec{c} - (\vec{a} 	imes \vec{b})) \cdot \vec{c}$ equals

Sigma Percentile · Accepted Answer

### The Geometry of the Plane: Understanding Linear Combinations

Imagine you are standing in a three-dimensional space, looking at two vectors, $\vec{a}$ and $\vec{b}$. When we define $\vec{c} = \alpha \vec{a} + \beta \vec{b}$, we are not just writing an equation; we are defining a world.

This vector $\vec{c}$ is a linear combination, which means it is trapped, in the most beautiful way, within the plane formed by $\vec{a}$ and $\vec{b}$. This is the fundamental geometric reality of our problem.

Before we dive into the algebra, take a moment to visualize this blue plane. Every possible value of $\alpha$ and $\beta$ just moves $\vec{c}$ around within this flat surface.

### The Projection: Finding the Constraint

The problem gives us a specific condition: the projection of $\vec{c}$ on the vector $(\vec{a} + \vec{b})$ is $3\sqrt{2}$. Let us break this down.

First, we calculate the sum:
$$\vec{a} + \vec{b} = (2+1)\hat{i} + (1+2)\hat{j} + (-1+1)\hat{k} = 3\hat{i} + 3\hat{j}$$

The magnitude of this vector is:
$$|\vec{a} + \vec{b}| = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$$

The projection formula is given by:
$$\frac{\vec{c} \cdot (\vec{a} + \vec{b})}{|\vec{a} + \vec{b}|}$$

Setting this equal to $3\sqrt{2}$, we find that $\vec{c} \cdot (\vec{a} + \vec{b}) = 18$. By substituting $\vec{c} = \alpha \vec{a} + \beta \vec{b}$ and using the dot products $|\vec{a}|^2 = 6$, $|\vec{b}|^2 = 6$, and $\vec{a} \cdot \vec{b} = 3$, we arrive at the elegant relation:
$$9\alpha + 9\beta = 18 \Rightarrow \alpha + \beta = 2$$

This is our key to the kingdom.

### The Scalar Triple Product: The Great Simplification

Now, we face the target expression: $(\vec{c} - (\vec{a} 	imes \vec{b})) \cdot \vec{c}$. It looks intimidating, but let us expand it:
$$\vec{c} \cdot \vec{c} - (\vec{a} 	imes \vec{b}) \cdot \vec{c}$$

Here is the moment of truth. Because $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, and $\vec{a} 	imes \vec{b}$ is the normal to that plane, their dot product is zero.

The expression collapses into just $|\vec{c}|^2$. We have successfully stripped away the complexity, leaving us with a simple task: minimize the magnitude squared of $\vec{c}$.

### The Final Minimization: Completing the Square

We expand $|\vec{c}|^2 = |\alpha \vec{a} + \beta \vec{b}|^2$ to get:
$$|\vec{c}|^2 = 6\alpha^2 + 6\beta^2 + 6\alpha\beta$$

Substituting $\beta = 2 - \alpha$, we get:
$$6(\alpha^2 + (2 - \alpha)^2 + \alpha(2 - \alpha))$$

Simplifying this quadratic, we find:
$$6(\alpha^2 - 2\alpha + 4)$$

To find the minimum, we complete the square:
$$6((\alpha - 1)^2 + 3)$$

The minimum value is clearly $6 	imes 3 = 18$. You have navigated the geometry, simplified the expression, and solved the quadratic. The final answer is **18**.

Enter Numerical Value:

Visualized Solution

Given Vectors

Calculate $a + b$

Projection Formula

Simplify Projection

Basic Dot Products

Substitute $c$

Expand and Simplify

The Target Expression

Scalar Triple Product

Simplified Target

Substitute Known Values

Eliminate $β$

Simplify the Quadratic

Complete the Square

The Sigma Insight: Scalar (Dot) Product

The Geometry of the Plane

The Projection

The Scalar Triple Product

The Final Minimization

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