Consider the planes 3x - 6y - 2z = 15 and 2x + y - 2z = 5.
STATEMENT-1 : The parametric equations of

Q: Wait, why do we reject the negative root here? Or rather, why is Statement-1 false if the direction vector matches?

Great question! A line is defined by two things: its direction and a point it passes through. Statement-2 correctly identifies the direction vector. However, Statement-1 provides a specific line. Even if the direction is correct, the line must actually lie on the planes. By testing a point from the parametric equation, we proved it doesn't satisfy the plane equations, meaning it's a parallel line, not the intersection line.

Q: Why can't we just use energy conservation or some other shortcut?

In 3D geometry, there are no 'energy' shortcuts, but there are geometric ones! The cross product is the most powerful tool here because it naturally finds the vector perpendicular to two surfaces. Always rely on the fundamental definitions of normal vectors and cross products for plane intersection problems.

Q: Does the order of the cross product matter? Like, $\vec{n_1} \times \vec{n_2}$ versus $\vec{n_2} \times \vec{n_1}$?

Excellent observation. The order changes the sign of the vector (e.g., $\vec{v}$ vs $-\vec{v}$), but for a line, both vectors are parallel. Since the direction is the same, the line remains the same. So, the order doesn't affect the final truth value of the statements.

Question

Consider the planes $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$.
STATEMENT-1 : The parametric equations of the line of intersection of the given planes are $x = 3 + 14t, y = 1 + 2t, z = 15t$. because
STATEMENT-2 : The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of given planes.

Options:

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of the intersection of two planes is defined by the line where they meet. We are given two planes:
$P_1: 3x - 6y - 2z = 15$
$P_2: 2x + y - 2z = 5$

Our objective is to verify the validity of the provided statements regarding the line of intersection.

### The DNA of a Plane

Every plane has a unique fingerprint known as its normal vector, which acts as a compass pointing directly away from its surface. For $P_1$, the coefficients of $x, y,$ and $z$ provide the normal vector:
$$\vec{n_1} = 3\hat{i} - 6\hat{j} - 2\hat{k}$$

Similarly, for $P_2$, we identify the normal vector:
$$\vec{n_2} = 2\hat{i} + \hat{j} - 2\hat{k}$$

These vectors are the fundamental keys to unlocking the geometry of the intersection.

### The Bridge: The Cross Product

The line of intersection must lie on both planes simultaneously. Consequently, it must be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. To find a vector $\vec{v}$ that is perpendicular to both, we calculate the cross product:
$$\vec{v} = \vec{n_1} 	imes \vec{n_2}$$

We set up the determinant as follows:
$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -6 & -2 \ 2 & 1 & -2 \end{vmatrix}$$

Expanding this determinant, we obtain:
$$\vec{v} = \hat{i}((-6)(-2) - (-2)(1)) - \hat{j}((3)(-2) - (-2)(2)) + \hat{k}((3)(1) - (-6)(2))$$

Simplifying the components, we arrive at:
$$\vec{v} = 14\hat{i} + 2\hat{j} + 15\hat{k}$$

This vector represents the direction of the line of intersection. Since Statement-2 claims this specific vector is parallel to the line, we conclude that **Statement-2 is True**.

### The Trap: Direction vs. Position

We now evaluate Statement-1, which provides the parametric equations:
$x = 3 + 14t$
$y = 1 + 2t$
$z = 15t$

The direction vector $(14, 2, 15)$ matches our calculated vector. However, a line is defined by both its direction and a specific point through which it passes.

If we set $t=0$, the line passes through the point $P(3, 1, 0)$. For this line to be the intersection, point $P$ must satisfy the equations of both planes.

Testing $P$ in $P_1$:
$$3(3) - 6(1) - 2(0) = 9 - 6 = 3$$

The equation for $P_1$ requires the result to be $15$. Since $3 
eq 15$, the point $P$ does not lie on the plane.

The line described in Statement-1 is parallel to the intersection, but it is not the intersection itself. Therefore, **Statement-1 is False**.

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

Normal Vectors of the Planes

Direction of the Line of Intersection

Setting up the Cross Product

Expanding the Determinant

Calculating the Direction Vector

Verifying Statement-2

Testing Statement-1

Extracting a Point

Checking Point P on Plane 1

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The DNA of a Plane

The Bridge

The Trap

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