The plane containing the line (x-3)/2 = (y+2)/-1 = (z-1)/3 and also containing its projection on the

Q: Wait, why is the normal of $P_1$ parallel to $P_2$? I thought normals were always perpendicular to their own planes.

That is a fantastic observation! You are absolutely right that a normal is perpendicular to its own plane. However, think about the geometry: if $P_2$ is the plane containing the projection of $L$ onto $P_1$, then $P_2$ must be perpendicular to $P_1$. If two planes are perpendicular, the normal vector of one plane must lie within the other plane (or be parallel to it). Imagine a wall ($P_2$) standing on a floor ($P_1$). The floor's normal points straight up, which is parallel to the wall's surface. That is the key!

Q: Can't we just find the projection line $L'$ first and then find the plane containing $L$ and $L'$? It seems more direct.

You certainly could, and mathematically, it would lead you to the same answer! However, finding the projection line $L'$ requires finding the intersection of $P_1$ and the plane containing $L$ that is perpendicular to $P_1$. That is a multi-step process involving finding the intersection of two planes. Using the cross product property is the 'JEE shortcut'—it saves you time and reduces the chance of calculation errors. In the exam, efficiency is just as important as accuracy.

Q: Why do we use the cross product here? Is there any other way to find the normal vector?

The cross product is the most powerful tool in your 3D geometry toolkit for this exact scenario. We need a vector that is perpendicular to two known vectors: the direction of the line $\vec{d}$ and the normal of the plane $\vec{n_1}$. The cross product $\vec{d} \times \vec{n_1}$ is defined specifically to produce a vector perpendicular to both. While you could technically use the dot product and solve a system of linear equations to find the components of the normal vector, the cross product is elegant, fast, and designed for this purpose.

Question

The plane containing the line $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z-1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$, contains which one of the following points ?

Options:

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

(B) $(-2, 2, 2)$

(C) $(0, -2, 2)$

(D) $(2, 2, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room. The floor is the plane $P_1$, defined by the equation $2x + 3y - z = 5$. Now, imagine a line $L$ floating in the air above this floor, defined by the symmetric form:

$$\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z-1}{3}$$

The problem asks us to find the equation of a plane $P_2$ that contains both the original line $L$ and its shadow $L'$ on the floor. Think of this plane $P_2$ as a vertical wall erected on the floor $P_1$, passing exactly through the line $L$ and its shadow.

Because this wall is built by dropping perpendiculars from $L$ to $P_1$, the wall itself must be perpendicular to the floor. This is the **'Aha!' moment**: if $P_2$ is perpendicular to $P_1$, then the normal vector of $P_1$, denoted as $\vec{n_1}$, must be parallel to the plane $P_2$.

### The Toolkit of Vectors

To define a plane, we need a point on the plane and a normal vector. Since our plane $P_2$ contains the line $L$, any point on $L$ is also on $P_2$. From the equation of the line, we extract the point $A(3, -2, 1)$ and the direction vector $\vec{d} = (2, -1, 3)$.

Next, we determine the normal vector of $P_2$, which we call $\vec{n_2}$. We know that $P_2$ contains the direction vector $\vec{d}$ of the line $L$ and is parallel to the normal vector of the floor, $\vec{n_1} = (2, 3, -1)$.

Therefore, $\vec{n_2}$ must be perpendicular to both $\vec{d}$ and $\vec{n_1}$. We find this using the cross product: $\vec{n_2} = \vec{d} 	imes \vec{n_1}$.

### The Calculation

We set up the determinant to compute the cross product:

$$\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 2 & 3 & -1 \end{vmatrix}$$

Expanding this determinant, we obtain:

$$\vec{n_2} = \hat{i}((-1)(-1) - (3)(3)) - \hat{j}((2)(-1) - (3)(2)) + \hat{k}((2)(3) - (-1)(2))$$

$$\vec{n_2} = \hat{i}(1 - 9) - \hat{j}(-2 - 6) + \hat{k}(6 + 2)$$

$$\vec{n_2} = -8\hat{i} + 8\hat{j} + 8\hat{k}$$

We can simplify this vector by dividing by $-8$, yielding a cleaner normal vector: $\vec{n_2} = (1, -1, -1)$.

### Final Equation

Using the point-normal form of a plane equation, $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$, we substitute our point $A(3, -2, 1)$ and our normal $\vec{n_2}(1, -1, -1)$:

$$1(x-3) - 1(y+2) - 1(z-1) = 0$$

Expanding and simplifying the expression:

$$x - 3 - y - 2 - z + 1 = 0$$

$$x - y - z - 4 = 0 \implies \mathbf{x - y - z = 4}$$

The final equation of the plane is **$x - y - z = 4$**.

Animated Solution for Mathematics - Three Dimensional Geometry: The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points ?

Select Answer:

Visualized Solution

Visualizing the Setup

The Projection of Line $L$

Defining the Target Plane $P_{2}$

Extracting Point and Direction from $L$

The Geometric Insight

Finding the Normal to $P_{2}$

Setting up the Cross Product

Computing the Cross Product

Simplifying the Normal Vector

Applying the Point-Normal Form

Expanding the Equation

Final Equation of Plane $P_{2}$

Testing the Options

Verifying the Correct Point

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Toolkit of Vectors

The Calculation

Final Equation

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