Let three vectors a, b and c be such that c is coplanar with a and b, a . c = 7 and b is perpendicul

Q: Wait, why do we use the vector triple product? Can't we just say $\vec{c} = x\vec{a} + y\vec{b}$ and solve for $x$ and $y$?

That is a brilliant alternative approach! You absolutely can set $\vec{c} = x\vec{a} + y\vec{b}$. Since $\vec{b} \cdot \vec{c} = 0$, you would get $x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b}) = 0$. This gives you a relationship between $x$ and $y$. Then, using $\vec{a} \cdot \vec{c} = 7$, you get another equation. Both methods are perfectly valid. The Vector Triple Product is often preferred in JEE because it is a direct, one-step construction that avoids solving a system of linear equations.

Q: Why is $\vec{b} \times (\vec{a} \times \vec{b})$ guaranteed to be in the plane of $\vec{a}$ and $\vec{b}$?

Think about the cross product $\vec{a} \times \vec{b}$. By definition, this vector is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. Now, when you take the cross product of $\vec{b}$ with this new vector, you are creating a vector that is perpendicular to both $\vec{b}$ and the vector $(\vec{a} \times \vec{b})$. Since $(\vec{a} \times \vec{b})$ is the normal to the plane, any vector perpendicular to it must lie within the plane itself. It is a beautiful geometric consequence!

Q: Why do we need the scalar $\lambda$ in our expression for $\vec{c}$?

The Vector Triple Product $\vec{b} \times (\vec{a} \times \vec{b})$ only gives us the correct *direction* for $\vec{c}$. However, a vector is defined by both direction and magnitude. The problem gives us the condition $\vec{a} \cdot \vec{c} = 7$, which acts as a constraint on the magnitude. The scalar $\lambda$ is our degree of freedom that allows us to scale the vector to satisfy this specific condition.

Question

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}, \vec{a} \cdot \vec{c} = 7$ and $\vec{b}$ is perpendicular to $\vec{c}$, where $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{k}$, then the value of $2|\vec{a} + \vec{b} + \vec{c}|^2$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the heart of vector algebra. Today, we are not just solving a problem; we are navigating the architecture of three-dimensional space.

We are given two vectors, $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{k}$. We are tasked with finding a third vector, $\vec{c}$, that satisfies two specific constraints: it must be coplanar with $\vec{a}$ and $\vec{b}$, and it must be perpendicular to $\vec{b}$.

Imagine these vectors as physical rods in space. The plane formed by $\vec{a}$ and $\vec{b}$ is our stage. Vector $\vec{c}$ must lie flat on this stage, yet it must stand at a perfect ninety-degree angle to $\vec{b}$.

### The Magic of the Vector Triple Product

How do we construct such a vector? We invoke the Vector Triple Product:
$$\vec{c} = \lambda(\vec{b} 	imes (\vec{a} 	imes \vec{b}))$$

Why this specific combination? Because $\vec{a} 	imes \vec{b}$ creates a vector perpendicular to the plane of $\vec{a}$ and $\vec{b}$. When we cross this result with $\vec{b}$ again, we are essentially rotating back into the plane.

To make this calculation manageable, we use the expansion formula, often remembered as the 'BAC minus CAB' rule:
$$\vec{A} 	imes (\vec{B} 	imes \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$$

Applying this to our expression, we get:
$$\vec{c} = \lambda((\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b})$$

### The Algebraic Unfolding

Now, let us perform the arithmetic with precision. First, we calculate the dot products:
$$\vec{b} \cdot \vec{b} = (2)^2 + (0)^2 + (1)^2 = 5$$
$$\vec{b} \cdot \vec{a} = (2)(-1) + (0)(1) + (1)(1) = -1$$

Substituting these back, our expression for $\vec{c}$ becomes:
$$\vec{c} = \lambda(5\vec{a} - (-1)\vec{b}) = \lambda(5\vec{a} + \vec{b})$$

Substituting the components of $\vec{a}$ and $\vec{b}$:
$$\vec{c} = \lambda(5(-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k})) = \lambda(-3\hat{i} + 5\hat{j} + 6\hat{k})$$

We are given the constraint $\vec{a} \cdot \vec{c} = 7$. Substituting our expression for $\vec{c}$:
$$(-\hat{i} + \hat{j} + \hat{k}) \cdot \lambda(-3\hat{i} + 5\hat{j} + 6\hat{k}) = 7$$

This simplifies to $\lambda(3 + 5 + 6) = 7$, which means $14\lambda = 7$, or $\lambda = \frac{1}{2}$.

### The Final Synthesis

With $\lambda = \frac{1}{2}$, our vector $\vec{c}$ is fully revealed:
$$\vec{c} = -\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$$

Now, we calculate the sum $\vec{a} + \vec{b} + \vec{c}$. Adding the components:
$$\vec{a} + \vec{b} = \hat{i} + \hat{j} + 2\hat{k}$$
$$\vec{a} + \vec{b} + \vec{c} = (1 - \frac{3}{2})\hat{i} + (1 + \frac{5}{2})\hat{j} + (2 + 3)\hat{k} = -\frac{1}{2}\hat{i} + \frac{7}{2}\hat{j} + 5\hat{k}$$

The final step is to find $2|\vec{a} + \vec{b} + \vec{c}|^2$. The magnitude squared is:
$$|\vec{a} + \vec{b} + \vec{c}|^2 = (-\frac{1}{2})^2 + (\frac{7}{2})^2 + 5^2 = \frac{1}{4} + \frac{49}{4} + 25 = \frac{50}{4} + 25 = 12.5 + 25 = 37.5$$

Multiplying by 2, we arrive at our destination:
$$2 	imes 37.5 = \mathbf{75}$$

Animated Solution for Mathematics - Vector Algebra: Let three vectors $a, b$ and $c$ be such that $c$ is coplanar with $a$ and $b, a \cdot c = 7$ and $b$ is perpendicular to $c$ , where $a = - \hat{i} + \hat{j} + \hat{k}$ and $b = 2 \hat{i} + \hat{k}$ , then the value of $2∣ a + b + c ∣^{2}$ is

Enter Numerical Value:

Visualized Solution

Identify Given Vectors

Coplanarity and Perpendicularity

Vector Triple Product Tool

Expanding the VTP

Calculate Dot Products

Substitute into Vector Expression

Simplify Vector $c$

Find $λ$ using Dot Product

Solve for $λ$

Final Vector $c$

Calculate Vector Sum

Magnitude Squared

Final Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Magic of the Vector Triple Product

The Algebraic Unfolding

The Final Synthesis

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