Let the plane ax + by + cz = d pass through (2, 3, -5) and is perpendicular to the planes 2x + y - 5

Q: Wait, why do we use the cross product here?

Great question! The cross product is the perfect tool because it produces a vector that is orthogonal (perpendicular) to both input vectors. Since our required plane is perpendicular to two given planes, its normal vector must be perpendicular to the normal vectors of those two planes. The cross product gives us exactly that direction.

Q: What if the cross product results in a vector with a common factor?

The problem specifies that $\gcd(|a|, |b|, |c|, d) = 1$. If your cross product calculation results in a vector like $(36, -2, 14)$, you should divide the entire equation by the greatest common divisor (in this case, 2) to simplify it to $(18, -1, 7)$ before proceeding. This ensures you meet the GCD condition.

Q: Why did we multiply by -1 at the end?

The problem explicitly requires the equation to be in the form $ax + by + cz = d$ with $d > 0$. If your initial calculation leads to $d = -2$, the condition is not met. Multiplying the entire equation by $-1$ changes the sign of $d$ to $+2$ while keeping the plane itself unchanged. It is a simple but crucial step to satisfy the problem's constraints.

Question

Let the plane $ax + by + cz = d$ pass through $(2, 3, -5)$ and is perpendicular to the planes $2x + y - 5z = 10$ and $3x + 5y - 7z = 12$. If $a, b, c, d$ are integers $d > 0$ and $\gcd(|a|, |b|, |c|, d) = 1$, then the value of $a + 7b + c + 20d$ is equal to

Options:

(A) 18

(B) 20

(C) 24

(D) 22

Sigma Percentile · Accepted Answer

### The Geometry of Planes: A 3D Adventure

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving a problem; we are constructing a plane in 3D space.

Imagine you are standing in a room, and you are given two walls (planes) that intersect. You need to find a third wall that is perpendicular to both of them. This is the essence of our challenge.

### Phase 1: The Normal Vector - Our Compass

Every plane has a 'normal vector'—a vector that points straight out from its surface, defining its orientation.

For the planes $P_1: 2x + y - 5z = 10$ and $P_2: 3x + 5y - 7z = 12$, the coefficients of $x, y,$ and $z$ are our keys. They give us the normal vectors directly: $\vec{n_1} = (2, 1, -5)$ and $\vec{n_2} = (3, 5, -7)$.

These vectors are the 'DNA' of our planes.

### Phase 2: The Magic of the Cross Product

We need a new plane that is perpendicular to both $P_1$ and $P_2$. Geometrically, this means the normal vector of our new plane, $\vec{n}$, must be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$.

How do we find a vector perpendicular to two others? The cross product is our magic wand! We calculate $\vec{n} = \vec{n_1} 	imes \vec{n_2}$ using a determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -5 \ 3 & 5 & -7 \end{vmatrix}$$

Expanding this, we get $\vec{n} = \hat{i}(1(-7) - (-5)(5)) - \hat{j}(2(-7) - (-5)(3)) + \hat{k}(2(5) - 1(3))$.

Simplifying this, we find $\vec{n} = 18\hat{i} - 1\hat{j} + 7\hat{k}$. This vector $(18, -1, 7)$ is the normal to our required plane.

### Phase 3: Anchoring the Plane

A normal vector tells us the orientation, but not the position. To fix the plane in space, we need a point. We are given the point $(2, 3, -5)$.

Using the point-normal form, $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$, we substitute our values:

$$18(x - 2) - 1(y - 3) + 7(z - (-5)) = 0$$

Expanding this, we get $18x - 36 - y + 3 + 7z + 35 = 0$, which simplifies to $18x - y + 7z + 2 = 0$, or $18x - y + 7z = -2$.

### Phase 4: The Final Polish

The problem demands $d > 0$. Currently, $d = -2$.

We simply multiply the entire equation by $-1$ to get $-18x + y - 7z = 2$. Now, $a = -18, b = 1, c = -7,$ and $d = 2$.

The GCD condition $\gcd(18, 1, 7, 2) = 1$ is satisfied. Finally, we calculate $a + 7b + c + 20d$:

$$a + 7b + c + 20d = (-18) + 7(1) + (-7) + 20(2)$$

$$= -18 + 7 - 7 + 40 = 22$$

We have arrived at our destination! The elegance of this result, **22**, is a testament to the beauty of 3D geometry.

Select Answer:

Visualized Solution

Identify Plane Normals

Condition for Perpendicularity

The Cross Product Setup

Expanding the Determinant

Calculating the Normal Vector

The Point on the Plane

The Point-Normal Form

Substituting Values

Expanding the Equation

Standard Form $a x + b y + cz = d$

Applying the Constraint $d > 0$

Identifying Parameters

Setting up the Final Expression

Calculating the Final Answer

The Sigma Insight: Equation of a Plane

The Geometry of Planes

Phase 1

Phase 2

Phase 3

Phase 4

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