A plane E is perpendicular to the two planes 2x - 2y + z = 0 and x - y + 2z = 4, and passes through

Q: Wait, why do we use the cross product here? Can't we just use the dot product?

Great question! The dot product tells us if two vectors are perpendicular (if the result is zero), but the cross product is the tool that generates a new vector perpendicular to two existing ones. Since our plane $E$ is perpendicular to both given planes, its normal must be perpendicular to both of their normals. The cross product is the perfect geometric tool to find that common perpendicular direction.

Q: Why did we divide the normal vector by $-3$? Is that allowed?

Absolutely! A plane is defined by its normal direction, not its magnitude. Any vector parallel to the normal works just as well. By dividing $(-3, -3, 0)$ by $-3$, we get $(1, 1, 0)$, which is parallel to the original normal but much cleaner to work with. It simplifies the algebra significantly without changing the plane itself.

Q: Why does the $z$ component disappear in the plane equation?

It happens because the $z$ component of our normal vector $\vec{n}$ turned out to be $0$. When we plug this into the plane equation $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$, the term $C(z - z_1)$ becomes $0(z - 1)$, which is just $0$. This tells us that the plane is parallel to the $z$-axis, meaning its orientation doesn't change as you move along the $z$-direction.

Question

A plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$, and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$, then $(PQ)^2$ is equal to

Options:

(A) 9

(B) 12

(C) 21

(D) 33

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every plane is defined by its normal vector—the arrow that points straight out of its surface. For the first plane, $2x - 2y + z = 0$, the normal vector is $\vec{n}_1 = (2, -2, 1)$.

For the second plane, $x - y + 2z = 4$, the normal is $\vec{n}_2 = (1, -1, 2)$.

Because our target plane $E$ is perpendicular to both of these, its normal vector $\vec{n}$ must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$. This is the definition of the cross product.

### Finding the Orientation

We calculate $\vec{n} = \vec{n}_1 	imes \vec{n}_2$ using the determinant:

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

Expanding this, we get $\vec{n} = \hat{i}(-4 - (-1)) - \hat{j}(4 - 1) + \hat{k}(-2 - (-2))$, which simplifies to $\vec{n} = -3\hat{i} - 3\hat{j} + 0\hat{k}$.

To make our calculations easier, we can scale this vector by dividing by $-3$, giving us the simplified direction ratios $(1, 1, 0)$.

### Constructing the Plane

Now that we have the normal vector $(1, 1, 0)$ and we know the plane passes through the point $P(1, -1, 1)$, we use the point-normal form: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.

Substituting our values, we get $1(x - 1) + 1(y - (-1)) + 0(z - 1) = 0$.

Expanding this, we find $x - 1 + y + 1 = 0$, which simplifies beautifully to **$x + y = 0$**. This is the equation of our plane $E$.

### The Distance Constraint

We are told that the distance from point $Q(a, a, 2)$ to this plane is $3\sqrt{2}$. The perpendicular distance formula is:

$$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Plugging in our values, we get:

$$3\sqrt{2} = \frac{|a + a + 0(2)|}{\sqrt{1^2 + 1^2 + 0^2}}$$

This simplifies to $3\sqrt{2} = \frac{|2a|}{\sqrt{2}}$. Multiplying both sides by $\sqrt{2}$, we get $|2a| = 3 	imes 2 = 6$, so $|a| = 3$. Squaring this, we find **$a^2 = 9$**.

### Final Calculation

We need to find $(PQ)^2$. Using the distance formula between $P(1, -1, 1)$ and $Q(a, a, 2)$, we have:

$$(PQ)^2 = (a - 1)^2 + (a - (-1))^2 + (2 - 1)^2$$

Expanding this, we get $(PQ)^2 = (a^2 - 2a + 1) + (a^2 + 2a + 1) + 1$.

The $-2a$ and $+2a$ terms cancel out, leaving us with $(PQ)^2 = 2a^2 + 3$. Substituting $a^2 = 9$, we get:

$$(PQ)^2 = 2(9) + 3 = 18 + 3 = \mathbf{21}$$

Select Answer:

Visualized Solution

Identify Plane Normals

Logic for Normal of Plane $E$

Setup Cross Product

Calculate Cross Product

Simplify Direction Ratios

Equation of Plane Formula

Substitute Point and Normal

Simplify Plane Equation

Distance from Point $Q$ to Plane $E$

Setup Distance Equation

Solve for $a^{2}$

Distance Formula for $(P Q)^{2}$

Substitute Coordinates of $P$ and $Q$

Expand and Simplify

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Finding the Orientation

Constructing the Plane

The Distance Constraint

Final Calculation

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Select Answer:

Visualized Solution

Identify Plane Normals

Logic for Normal of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Setup Cross Product

Calculate Cross Product

Simplify Direction Ratios

Equation of Plane Formula

Substitute Point and Normal

Simplify Plane Equation

Distance from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Setup Distance Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (PQ)2

Substitute Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Expand and Simplify

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Finding the Orientation

Constructing the Plane

The Distance Constraint

Final Calculation

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Logic for Normal of Plane $E$

Distance from Point $Q$ to Plane $E$

Solve for $a^{2}$

Distance Formula for $(P Q)^{2}$

Substitute Coordinates of $P$ and $Q$

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