Let the lines (x-1)/λ = (y-2)/1 = (z-3)/2 and (x+26)/-2 = (y+18)/3 = (z+28)/λ be coplanar and P be t

Q: Why do we use the Scalar Triple Product here? Can't we just find the intersection point?

Finding the intersection is a valid path, but it is often algebraically messier. The Scalar Triple Product is the elegant, JEE-Advanced way to enforce coplanarity without needing to solve for the intersection point first. It directly utilizes the geometric property that the three vectors must be linearly dependent, which is a much more powerful tool in your arsenal.

Q: How do I know which point to use for the plane equation?

Great question! Since both lines lie on the plane, any point on either line is a valid point on the plane. You can pick point $A(1, 2, 3)$ from the first line or point $B(-26, -18, -28)$ from the second. The resulting plane equation will be identical because both points satisfy the same linear constraint.

Q: What if the cross product of the direction vectors is zero?

That would imply the lines are parallel. If they are parallel, the cross product of direction vectors is indeed zero, and you would need a different approach—specifically, using the vector connecting the two points and one of the direction vectors to find the normal. But here, the lines are skew-intersecting, so the cross product is non-zero.

Question

Let the lines $\frac{x-1}{\lambda} = \frac{y-2}{1} = \frac{z-3}{2}$ and $\frac{x+26}{-2} = \frac{y+18}{3} = \frac{z+28}{\lambda}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on $P$?

Options:

(A) $(0, -2, -2)$

(B) $(-5, 0, -1)$

(C) $(3, -1, 0)$

(D) $(0, 4, 5)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

From the given symmetric equations:
$$\frac{x-1}{\lambda} = \frac{y-2}{1} = \frac{z-3}{2}$$
$$\frac{x+26}{-2} = \frac{y+18}{3} = \frac{z+28}{\lambda}$$

We identify the points and direction vectors for the two lines:
Line $L_1$ passes through $A(1, 2, 3)$ with direction $\vec{d_1} = \langle \lambda, 1, 2 angle$.
Line $L_2$ passes through $B(-26, -18, -28)$ with direction $\vec{d_2} = \langle -2, 3, \lambda angle$.

### The Condition for Coplanarity

For the lines to be coplanar, the vector $\vec{AB}$ connecting the two lines must lie in the same plane as the direction vectors $\vec{d_1}$ and $\vec{d_2}$. This implies that the scalar triple product of these three vectors must be zero:
$$[\vec{AB} \quad \vec{d_1} \quad \vec{d_2}] = 0$$

First, we calculate the vector $\vec{AB}$:
$$\vec{AB} = B - A = \langle -26-1, -18-2, -28-3 angle = \langle -27, -20, -31 angle$$

### Solving for the Parameter $\lambda$

We set up the determinant equation:
$$\begin{vmatrix} -27 & -20 & -31 \ \lambda & 1 & 2 \ -2 & 3 & \lambda \end{vmatrix} = 0$$

Expanding along the first row:
$$-27(\lambda - 6) + 20(\lambda^2 + 4) - 31(3\lambda + 2) = 0$$
$$-27\lambda + 162 + 20\lambda^2 + 80 - 93\lambda - 62 = 0$$
$$20\lambda^2 - 120\lambda + 180 = 0$$

Dividing by 20, we obtain:
$$\lambda^2 - 6\lambda + 9 = 0 \Rightarrow (\lambda - 3)^2 = 0$$
Thus, we find **$\lambda = 3$**.

### Determining the Plane Equation

With $\lambda = 3$, the direction vectors are $\vec{d_1} = \langle 3, 1, 2 angle$ and $\vec{d_2} = \langle -2, 3, 3 angle$. The normal vector $\vec{n}$ to the plane is the cross product of these directions:
$$\vec{n} = \vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ -2 & 3 & 3 \end{vmatrix} = -3\hat{i} - 13\hat{j} + 11\hat{k}$$

Using the point-normal form with point $A(1, 2, 3)$:
$$-3(x - 1) - 13(y - 2) + 11(z - 3) = 0$$
$$-3x + 3 - 13y + 26 + 11z - 33 = 0$$

The final equation of the plane $P$ is:
**$3x + 13y - 11z + 4 = 0$**

### Identifying the Outsider

To identify the point that does not lie on the plane, we substitute the coordinates of the given point $(0, 4, 5)$ into the plane equation:
$$3(0) + 13(4) - 11(5) + 4 = 0 + 52 - 55 + 4 = 1$$

Since $1 
eq 0$, the point $(0, 4, 5)$ does not satisfy the equation of the plane. Therefore, **$(0, 4, 5)$ is the outsider.**

Select Answer:

Visualized Solution

Visualizing the Coplanar Lines

Extracting Line Parameters

The Condition for Coplanarity

Setting up the Determinant

Solving for $λ$

Finding the Normal Vector $n$

Calculating the Normal Vector

Constructing the Plane Equation

Verifying the Points

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Condition for Coplanarity

Solving for the Parameter $λ$

Determining the Plane Equation

Identifying the Outsider

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Select Answer:

Visualized Solution

Visualizing the Coplanar Lines

Extracting Line Parameters

The Condition for Coplanarity

Setting up the Determinant

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

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Calculating the Normal Vector

Constructing the Plane Equation

Verifying the Points

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Condition for Coplanarity

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

Determining the Plane Equation

Identifying the Outsider

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