If the vectors a = λi + μj + 4k, b = -2i + 4j - 2k and c = 2i + 3j + k are coplanar and the projecti

Q: Why do we use the absolute value in the projection formula?

The projection length is a physical distance, which must be non-negative. The dot product $\vec{a} \cdot \vec{b}$ can be negative if the angle between the vectors is obtuse. The absolute value ensures we are talking about the magnitude of the shadow, not its direction.

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

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ geometrically represents the volume of the parallelepiped formed by the three vectors. If the vectors are coplanar, they lie on a flat plane, meaning the 'height' of this parallelepiped is zero. Thus, the volume must be zero.

Q: Is it possible for the projection to be negative?

No, the projection length is a scalar quantity representing distance. If the problem asks for the 'projection of $\vec{a}$ on $\vec{b}$' as a value, it implies the magnitude of the vector projection. Always treat it as a positive length.

Question

If the vectors $\vec{a} = \lambda \hat{i} + \mu \hat{j} + 4\hat{k}, \vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k}$ and $\vec{c} = 2\hat{i} + 3\hat{j} + \hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is $\sqrt{54}$ units, then the sum of all possible values of $\lambda + \mu$ is equal to

Options:

(A) 0

(B) 6

(C) 24

(D) 18

Sigma Percentile · Accepted Answer

### The Geometry of Flatness: A Journey into Coplanarity

Imagine you are standing in a 3D coordinate system with three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$. The problem states that these vectors are coplanar, meaning they lie perfectly flat on a single plane with no volume between them.

In the context of JEE Advanced, this geometric "flatness" is translated into the **scalar triple product**. Since the volume of the parallelepiped formed by these vectors is zero, we set the determinant of their components to zero:

$$
\begin{vmatrix}
\lambda & \mu & 4 \
-2 & 4 & -2 \
2 & 3 & 1
\end{vmatrix} = 0
$$

Expanding this determinant along the first row, we calculate:

$$
\lambda(4 - (-6)) - \mu(-2 - (-4)) + 4(-6 - 8) = 0
$$

This simplifies to $10\lambda - 2\mu - 56 = 0$. Dividing by two, we arrive at our first elegant constraint:

$$
5\lambda - \mu = 28
$$

### The Shadow on the Wall: Understanding Projections

Next, we consider the condition that the projection of $\vec{a}$ on $\vec{b}$ is $\sqrt{54}$. The formula for the projection of $\vec{a}$ onto $\vec{b}$ is given by:

$$
\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \sqrt{54}
$$

First, we find the magnitude of $\vec{b}$:

$$
|\vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}
$$

The dot product $\vec{a} \cdot \vec{b}$ is calculated as $-2\lambda + 4\mu - 8$. Substituting these into the projection formula, we get:

$$
\frac{|-2\lambda + 4\mu - 8|}{\sqrt{24}} = \sqrt{54}
$$

Cross-multiplying yields $|-2\lambda + 4\mu - 8| = \sqrt{54 	imes 24} = \sqrt{1296} = 36$.

### The Fork in the Road: Solving the System

The absolute value equation $|-2\lambda + 4\mu - 8| = 36$ splits our path into two distinct possibilities:

1. $-2\lambda + 4\mu - 8 = 36 \Rightarrow \lambda - 2\mu = -22$
2. $-2\lambda + 4\mu - 8 = -36 \Rightarrow \lambda - 2\mu = 14$

We now solve these alongside our first constraint, $5\lambda - \mu = 28$.

**Case 1:** Solving the system $\lambda - 2\mu = -22$ and $5\lambda - \mu = 28$:
Multiplying the second equation by 2 gives $10\lambda - 2\mu = 56$. Subtracting the first from this, we get $9\lambda = 78$, so $\lambda = \frac{26}{3}$. Substituting back, we find $\mu = \frac{46}{3}$, yielding $\lambda + \mu = 24$.

**Case 2:** Solving the system $\lambda - 2\mu = 14$ and $5\lambda - \mu = 28$:
Multiplying the second equation by 2 gives $10\lambda - 2\mu = 56$. Subtracting the first from this, we get $9\lambda = 42$, so $\lambda = \frac{14}{3}$. Substituting back, we find $\mu = -\frac{14}{3}$, yielding $\lambda + \mu = 0$.

### Final Calculation

The problem asks for the sum of all possible values of $\lambda + \mu$. Adding our two results, $24 + 0$, we arrive at the final answer:

**24**

Select Answer:

Visualized Solution

Define the Given Vectors $a, b, c$

Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant

Projection of $a$ on $b$

Calculating Magnitude and Dot Product

Setting up the Projection Equation

Handling the Absolute Value

Solving Case 1

Solving Case 2

Final Sum of Values

The Sigma Insight: Scalar Triple Product

The Geometry of Flatness

The Shadow on the Wall

The Fork in the Road

Final Calculation

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