If the vectors p = (a+1)i + aj + ak, q = ai + (a+1)j + ak and r = ai + aj + (a+1)k (a in R) are copl

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

The scalar triple product $[\vec{p} \vec{q} \vec{r}]$ represents the volume of the parallelepiped formed by the three vectors. If they are coplanar, they lie on the same flat surface, meaning the 'height' of this parallelepiped is zero. Thus, the volume is zero.

Q: Why can't I just calculate the cross product $\vec{p} \times \vec{q}$ directly?

You absolutely can, but it is a massive time-sink. In JEE Advanced, time is your most precious resource. Lagrange's Identity is a powerful tool that connects the dot and cross products, allowing you to bypass the tedious calculation of the cross product components.

Q: How did you know to add all rows in the determinant?

Look at the structure of the matrix. Each row has the same elements, just shifted. Adding all rows creates a row of identical terms ($3a+1$), which can then be factored out. This is a standard pattern in determinant problems—always look for symmetry!

Question

If the vectors $\vec{p} = (a+1)\hat{i} + a\hat{j} + a\hat{k}$, $\vec{q} = a\hat{i} + (a+1)\hat{j} + a\hat{k}$ and $\vec{r} = a\hat{i} + a\hat{j} + (a+1)\hat{k}$ ($a \in \mathbb{R}$) are coplanar and $3(\vec{p} \cdot \vec{q})^2 - \lambda |\vec{p} 	imes \vec{q}|^2 = 0$, then value of $\lambda$ is:

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D space, holding three vectors: $\vec{p}$, $\vec{q}$, and $\vec{r}$. In the general case, these three vectors would reach out into the void, spanning a volume.

But here, we are told they are coplanar. They are trapped, forced to lie on a single, flat 2D surface.

This geometric constraint is not just a description; it is a powerful mathematical key. When vectors are coplanar, the parallelepiped they form has no height, and therefore, its volume is zero.

This is the physical soul of the scalar triple product, which we express as $[\vec{p} \vec{q} \vec{r}] = 0$. This is our starting point, the first step in our journey to find the value of $a$.

### The Determinant Dance

To translate this geometric reality into algebra, we construct a determinant using the components of our vectors. We place the components of $\vec{p}$, $\vec{q}$, and $\vec{r}$ into the rows of a $3 	imes 3$ matrix:

$$
\begin{vmatrix} 
a+1 & a & a \ 
a & a+1 & a \ 
a & a & a+1 
\end{vmatrix} = 0
$$

Now, a novice might rush to expand this directly, but that is a trap. Look at the symmetry!

If we perform the row operation $R_1 	o R_1 + R_2 + R_3$, the first row becomes $(3a+1, 3a+1, 3a+1)$. We can then factor out $(3a+1)$, leaving us with a much simpler determinant.

This is the elegance of linear algebra—finding the path of least resistance. Solving $(3a+1) = 0$ gives us $a = -\frac{1}{3}$.

### The Lagrange Masterstroke

With $a = -\frac{1}{3}$ in hand, we find the components of our vectors. Substituting this value, we get $\vec{p} = \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{1}{3}\hat{k}$ and $\vec{q} = -\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}$.

Our goal is to solve the equation $3(\vec{p} \cdot \vec{q})^2 - \lambda |\vec{p} 	imes \vec{q}|^2 = 0$. We could calculate the cross product $\vec{p} 	imes \vec{q}$ directly, but that is tedious.

Instead, we use Lagrange's Identity:

$$
|\vec{p} 	imes \vec{q}|^2 = |\vec{p}|^2 |\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2
$$

This identity is a master stroke, connecting the dot product and the cross product magnitude. We calculate the dot product $\vec{p} \cdot \vec{q} = -\frac{1}{3}$ and the squared magnitudes $|\vec{p}|^2 = |\vec{q}|^2 = \frac{2}{3}$.

### The Final Calculation

Now, we substitute these values back into our equation. The squared magnitude of the cross product becomes:

$$
|\vec{p} 	imes \vec{q}|^2 = \left( \frac{2}{3} ight) \left( \frac{2}{3} ight) - \left( -\frac{1}{3} ight)^2 = \frac{4}{9} - \frac{1}{9} = \frac{1}{3}
$$

Finally, we plug everything into the given equation: $3(-\frac{1}{3})^2 - \lambda(\frac{1}{3}) = 0$. This simplifies to:

$$
3 \left( \frac{1}{9} ight) - \frac{\lambda}{3} = 0 \Rightarrow \frac{1}{3} - \frac{\lambda}{3} = 0
$$

The result is immediate and satisfying: **$\lambda = 1$**. We have navigated the geometry, simplified the algebra, and used a powerful identity to reach the truth.

Enter Numerical Value:

Visualized Solution

Understanding Coplanarity

Condition for Coplanarity

Setting up the Determinant

Simplifying the Determinant

Factoring out $(3 a + 1)$

Solving for $a$

Finding Vector Components

Calculating $p \cdot q$

Using Lagrange's Identity

Calculating Magnitudes

Finding $∣ p \times q ∣^{2}$

Final Equation for $λ$

Solving for $λ$

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Determinant Dance

The Lagrange Masterstroke

The Final Calculation

Similar Questions

If $P = (a + 1) \hat{i} + a \hat{j} + a \hat{k}$ , $Q = a \hat{i} + (a + 1) \hat{j} + a \hat{k}$ , $R = a \hat{i} + a \hat{j} + (a + 1) \hat{k}$ and $P, Q, R$ are coplanar vectors and $3 (P \cdot Q)^{2} - λ ∣ R \times Q ∣^{2} = 0$ then value of $λ$ is

Let $a, b$ and $c$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $a - b + c, λ a - 3 b + 4 c, - a + 2 b - 3 c$ and $2 a - 4 b + 6 c$ respectively. If AB, AC and AD are coplanar, then $λ$ is:

Let $a, b$ and $c$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $a - b + c$ , $λ a - 3 b + 4 c$ , $- a + 2 b - 3 c$ and $2 a - 4 b + 6 c$ respectively. If AB, AC and AD are coplanar, then $λ$ is :

The number of distinct real values of $λ$ , for which the vectors $- λ^{2} \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} - λ^{2} \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - λ^{2} \hat{k}$ are coplanar, is

If the vectors $a \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ ( $a \neq = b \neq = c \neq = 1$ ) are coplanar, then the value of $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$

If $a, b, c$ are non coplanar vectors and $λ$ is a real number then $[λ (a + b) λ^{2} b λ c] = [a b + c b]$ for

If $[a \times b b \times c c \times a] = λ [a b c]^{2}$ then $λ$ is equal to

If $a, b, c$ are non-coplanar vectors and $λ$ is a real number, then the vectors $a + 2 b + 3 c$ , $λ b + 4 c$ and $(2 λ - 1) c$ are non coplanar for

Let a, b and c be distinct positive numbers. If the vectors $a \hat{i} + a \hat{j} + c \hat{k}, \hat{i} + \hat{k}$ and $c \hat{i} + c \hat{j} + b \hat{k}$ are co-planar, then c is equal to:

Enter Numerical Value:

Visualized Solution

Understanding Coplanarity

Condition for Coplanarity

Setting up the Determinant

Simplifying the Determinant

Factoring out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (3a+1)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding Vector Components

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​⋅q​

Using Lagrange's Identity

Calculating Magnitudes

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣p​×q​∣2

Final Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ