For three unit vectors a, b, c satisfying |a - b|² + |b - c|² + |c - a|² = 9 and |2a + kb + kc| = 3,

Q: Wait, why does $|\vec{a} - \vec{b}|^2$ expand to $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$?

This is a fundamental property of the dot product! Remember that the square of a vector's magnitude is the dot product of the vector with itself: $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. So, $|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$. By distributing the dot product, you get $\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$. Since the dot product is commutative ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$), this simplifies perfectly to $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$. It is a powerful tool you will use constantly in JEE vector problems!

Q: How do we know $\vec{a} + \vec{b} + \vec{c} = \vec{0}$? Is that always true for unit vectors?

No, it is not always true! It is only true because of the specific condition given in this problem. When we calculated the sum of the dot products to be $-\frac{3}{2}$ and plugged it into the identity $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$, we found the result was $1 + 1 + 1 + 2(-\frac{3}{2}) = 3 - 3 = 0$. Since the squared magnitude of a vector is zero, the vector itself must be the zero vector. It is a beautiful consequence of the constraints provided!

Q: Why do we get two values for $k$ and why do we reject the negative one?

When you solve the absolute value equation $|2 - k| = 3$, you are looking for all numbers $k$ whose distance from 2 on the number line is 3. This leads to two possibilities: $2 - k = 3$ (which gives $k = -1$) and $2 - k = -3$ (which gives $k = 5$). The problem explicitly asks for the 'positive value of $k$', so we must discard $k = -1$. Always read the question carefully to see if there are constraints on your final answer!

Question

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9$ and $|2\vec{a} + k\vec{b} + k\vec{c}| = 3$, the positive value of k is

Options:

(A) 4

(B) 2

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the center of a unit sphere in three-dimensional space. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, all originating from where you stand and reaching out to touch the surface of this sphere.

Because they are unit vectors, their magnitudes are all exactly one:
$$|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$$

The problem presents us with a fascinating condition:
$$|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9$$

At first glance, this looks like a daunting algebraic mess. However, complexity is often just a mask for underlying elegance. Let us peel back that mask.

### The Power of Expansion

To decode this, we reach for our most reliable tool: the vector expansion formula. We know that for any two vectors, the square of their difference is $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$.

When we apply this to each of the three terms in our condition, we expand the expression as follows:
$$(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a}) = 9$$

Notice how each squared magnitude appears twice. We can group them:
$$2(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$$

Since we know $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$, the first part of this expression is simply $2(1 + 1 + 1) = 6$. Our equation simplifies to:
$$6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$$

With a quick rearrangement, we find the sum of the pairwise dot products:
$$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$$

### The Hidden Zero

This dot product sum is the missing piece of a much larger puzzle. Consider the identity for the square of the sum of three vectors:
$$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$$

Substituting our known values into this identity, we get:
$$|\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 2\left(-\frac{3}{2}\right) = 3 - 3 = 0$$

This is the "Aha!" moment. If the magnitude of a vector is zero, the vector itself must be the null vector:
$$\vec{a} + \vec{b} + \vec{c} = \vec{0}$$

Geometrically, this means that if you place these three vectors head-to-tail, they form a closed triangle. They perfectly balance each other out.

### The Final Transformation

Now, we turn our attention to the second condition: $|2\vec{a} + k\vec{b} + k\vec{c}| = 3$. We can factor $k$ out of the terms involving $b$ and $c$:
$$|2\vec{a} + k(\vec{b} + \vec{c})| = 3$$

From our previous deduction, we know that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, which implies $\vec{b} + \vec{c} = -\vec{a}$. Substituting this into our equation:
$$|2\vec{a} + k(-\vec{a})| = 3 \Rightarrow |(2 - k)\vec{a}| = 3$$

Using the property that the magnitude of a scalar times a vector is the absolute value of the scalar times the magnitude of the vector, and knowing $|\vec{a}| = 1$, we get:
$$|2 - k| \cdot |\vec{a}| = 3 \Rightarrow |2 - k| = 3$$

Solving this absolute value equation gives us two possibilities: $2 - k = 3$ (yielding $k = -1$) or $2 - k = -3$ (yielding $k = 5$). Since the problem asks for the positive value of $k$, we arrive at our final answer:

**$k = 5$**

Animated Solution for Mathematics - Vector Algebra: For three unit vectors $a, b, c$ satisfying $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 9$ and $∣2 a + k b + k c ∣ = 3$ , the positive value of k is

Select Answer:

Visualized Solution

Understanding Unit Vectors

Expanding the First Condition

Applying the Expansion

Substituting Unit Magnitudes

Finding the Sum of Dot Products

The "Aha!" Moment: Sum of Vectors

Substituting Known Values

Deducing the Zero Vector

Using the Second Condition

Substituting the Vector Sum

Simplifying the Equation

Applying Unit Magnitude Again

Solving for k

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Power of Expansion

The Hidden Zero

The Final Transformation

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