If (1, 5, 35), (7, 5, 5), (1, λ, 7) and (2λ, 1, 2) are coplanar, then the sum of all possible values

Q: Wait, why do we use the scalar triple product? Can't we just find the equation of the plane?

You absolutely could find the equation of the plane, but that is a much longer path! The scalar triple product is a powerful shortcut because it directly relates the geometric condition of coplanarity to the algebraic condition of a zero determinant. It is the most efficient tool in your arsenal for this type of problem.

Q: Does it matter which point I choose as the origin for the vectors?

Great question! No, it does not matter at all. Whether you choose $P, Q, R,$ or $S$ as your starting point, the volume of the parallelepiped formed by the vectors will always be zero if the points are coplanar. Choosing $P$ was just a convenient choice to simplify the arithmetic.

Q: Why not solve for $\lambda$ using the quadratic formula? Is the sum of roots formula really necessary?

You could solve for $\lambda$ using the quadratic formula, but it is prone to calculation errors, especially with square roots. Using the sum of roots formula, $-\frac{b}{a}$, is not just faster; it is more elegant and less likely to lead to a silly mistake. In JEE Advanced, efficiency and accuracy are your best friends.

Question

If $(1, 5, 35), (7, 5, 5), (1, \lambda, 7)$ and $(2\lambda, 1, 2)$ are coplanar, then the sum of all possible values of $\lambda$ is:

Options:

(A) $-\frac{44}{5}$

(B) $\frac{39}{5}$

(C) $-\frac{39}{5}$

(D) $\frac{44}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have four points floating in this space: $P(1, 5, 35)$, $Q(7, 5, 5)$, $R(1, \lambda, 7)$, and $S(2\lambda, 1, 2)$.

We are told they are coplanar, meaning they all rest on the same flat sheet of paper. This is a rigid geometric constraint, and our goal is to find the sum of all possible values of $\lambda$ that satisfy this condition.

### The Vectorial Bridge

To solve this, we use the scalar triple product as a bridge between geometry and algebra. If we anchor ourselves at point $P$ and draw vectors to $Q, R,$ and $S$, we obtain the vectors $\vec{PQ}$, $\vec{PR}$, and $\vec{PS}$.

If these points are coplanar, the volume of the parallelepiped they form must be zero. Mathematically, this is expressed as the determinant of these three vectors being zero.

We construct the vectors as follows:

$$\vec{PQ} = (7-1, 5-5, 5-35) = (6, 0, -30)$$

$$\vec{PR} = (1-1, \lambda-5, 7-35) = (0, \lambda-5, -28)$$

$$\vec{PS} = (2\lambda-1, 1-5, 2-35) = (2\lambda-1, -4, -33)$$

### The Determinant Dance

Now, we set up our master equation by calculating the determinant:

$$\begin{vmatrix} 6 & 0 & -30 \ 0 & \lambda - 5 & -28 \ 2\lambda - 1 & -4 & -33 \end{vmatrix} = 0$$

Expanding this along the first row, we obtain:

$$6[(\lambda-5)(-33) - (-4)(-28)] - 0 + (-30)[0 - (\lambda-5)(2\lambda-1)] = 0$$

The first part simplifies to $6[-33\lambda + 165 - 112]$, which is $6[-33\lambda + 53]$. The second part simplifies to $30[2\lambda^2 - 11\lambda + 5]$.

### The Final Algebraic Flourish

Combining these expressions, we have:

$$6[-33\lambda + 53] + 30[2\lambda^2 - 11\lambda + 5] = 0$$

Dividing the entire equation by $6$ gives:

$$(-33\lambda + 53) + 5(2\lambda^2 - 11\lambda + 5) = 0$$

Expanding this results in $-33\lambda + 53 + 10\lambda^2 - 55\lambda + 25 = 0$, which simplifies to the quadratic equation:

$$10\lambda^2 - 88\lambda + 78 = 0$$

Dividing by $2$ yields the simplified quadratic:

$$5\lambda^2 - 44\lambda + 39 = 0$$

Instead of solving for $\lambda$ directly, we use Vieta's formulas. The sum of the roots is given by $-\frac{b}{a}$:

$$	ext{Sum} = -\frac{-44}{5} = \frac{44}{5}$$

The sum of all possible values of $\lambda$ is **$\frac{44}{5}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If $(1, 5, 35), (7, 5, 5), (1, λ, 7)$ and $(2 λ, 1, 2)$ are coplanar, then the sum of all possible values of $λ$ is:

Select Answer:

Visualized Solution

Visualizing Points $P, Q, R, S$

Condition for Coplanarity

Calculating Vectors $P Q, P R, P S$

The Determinant Equation

Expanding the Determinant

Simplifying the Expression

Forming the Quadratic Equation

Sum of the Roots Formula

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Vectorial Bridge

The Determinant Dance

The Final Algebraic Flourish

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (1,5,35),(7,5,5),(1,λ,7) and (2λ,1,2) are coplanar, then the sum of all possible values of λ is: