If a = 2i + j + 3k, b = 3i + 3j + k and c = c1i + c2j + c3k are coplanar vectors and a · c = 5, b ⊥

Q: Why is the scalar triple product zero for coplanar vectors?

The scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$ represents the signed volume of the parallelepiped formed by the three vectors. If the vectors are coplanar, they lie on a flat surface, meaning they cannot 'heighten' into the third dimension. Thus, the volume is zero.

Q: Can I use the cross product instead of the scalar triple product?

Yes, you could find the normal vector to the plane using $\vec{n} = \vec{a} \times \vec{b}$ and then state that $\vec{c}$ must be perpendicular to $\vec{n}$ (i.e., $\vec{c} \cdot \vec{n} = 0$). This is mathematically equivalent to the scalar triple product condition, but the determinant method is often faster and less prone to calculation errors.

Q: Why does the question ask for $122(c_1 + c_2 + c_3)$?

In JEE Advanced problems, you will often encounter coefficients that seem arbitrary, like 122. This is a deliberate design choice by the examiners. It ensures that even if your individual components $c_1, c_2, c_3$ are fractions, their sum will likely have a denominator that cancels out with the multiplier, leading to a clean, integer answer. It is a hint that your algebraic path is correct.

Question

If $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}, \vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ are coplanar vectors and $\vec{a} \cdot \vec{c} = 5, \vec{b} \perp \vec{c}$, then $122(c_1 + c_2 + c_3)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system. You have two fixed vectors, $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}$.

Now, introduce a third, mysterious vector $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. The problem states that these three vectors are coplanar.

This is a geometric constraint that defines the soul of the problem. If three vectors are coplanar, they lie on the same flat surface and cannot form a 3D volume.

Mathematically, this means the volume of the parallelepiped formed by them is zero. We express this using the scalar triple product: $[\vec{a}, \vec{b}, \vec{c}] = 0$.

### The Determinant: Our Algebraic Key

To translate this geometric reality into algebra, we use the determinant. We place the components of $\vec{a}$ in the first row, $\vec{b}$ in the second, and $\vec{c}$ in the third:

$$
\begin{vmatrix} 2 & 1 & 3 \ 3 & 3 & 1 \ c_1 & c_2 & c_3 \end{vmatrix} = 0
$$

Expanding this determinant along the third row is the most efficient path. For $c_1$, we calculate the minor $(1 - 9) = -8$.

For $c_2$, we apply the checkerboard sign rule, giving $-(2 - 9) = 7$. For $c_3$, we get $(6 - 3) = 3$.

This yields our first linear equation: $-8c_1 + 7c_2 + 3c_3 = 0$.

### The Power of Dot Products

We have three unknowns, so we need three equations. The problem provides two more constraints.

First, $\vec{a} \cdot \vec{c} = 5$. The dot product is the sum of the products of corresponding components: $2c_1 + c_2 + 3c_3 = 5$.

Second, $\vec{b} \perp \vec{c}$. When two vectors are perpendicular, their dot product is zero: $\vec{b} \cdot \vec{c} = 0$. This gives us $3c_1 + 3c_2 + c_3 = 0$.

Now, we have a system of three linear equations:
1) $-8c_1 + 7c_2 + 3c_3 = 0$
2) $2c_1 + c_2 + 3c_3 = 5$
3) $3c_1 + 3c_2 + c_3 = 0$

### The Final Resolution

Solving this system requires patience. From the third equation, we can express $c_3$ as $c_3 = -3c_1 - 3c_2$.

Substituting this into the first two equations allows us to eliminate $c_3$ and solve for $c_1$ and $c_2$. After careful algebraic manipulation, we find:

$$
c_1 = \frac{10}{122}, \quad c_2 = -\frac{85}{122}, \quad c_3 = \frac{225}{122}
$$

The sum of these components is:

$$
c_1 + c_2 + c_3 = \frac{10 - 85 + 225}{122} = \frac{150}{122}
$$

The question asks for $122(c_1 + c_2 + c_3)$. Substituting our sum, we get:

$$
122 	imes \frac{150}{122} = 150
$$

The beauty of this problem lies in how the complex geometry and the seemingly messy fractions resolve into a clean, elegant integer. The final answer is **150**.

Animated Solution for Mathematics - Vector Algebra: If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant

Using the Dot Product Condition

Forming the Second Equation

The Orthogonality Condition

Forming the Third Equation

The System of Equations

Solving the System

Calculating the Sum

Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Determinant

The Power of Dot Products

The Final Resolution

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