Let the points on the plane P be equidistant from the points (-4, 2, 1) and (2, -2, 3). Then the acu

Q: Why is the normal vector of the plane the same as the vector $\vec{AB}$?

Think about the definition of a perpendicular bisector. The plane is perpendicular to the segment $AB$. Therefore, the direction of the segment $AB$ is the direction of the normal to the plane.

Q: Why do we take the absolute value in the dot product formula?

The question asks for the *acute* angle. The dot product can be negative, which would give an obtuse angle. The absolute value ensures we get the acute one.

Q: Can I simplify the normal vector $\vec{n}_1 = (6, -4, 2)$ to $(3, -2, 1)$?

Absolutely! Direction ratios are just ratios. Scaling them down makes your arithmetic cleaner and reduces the chance of calculation errors.

Question

Let the points on the plane $P$ be equidistant from the points $(-4, 2, 1)$ and $(2, -2, 3)$. Then the acute angle between the plane $P$ and the plane $2x + y + 3z = 1$ is

Options:

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{4}$

(C) $\frac{\pi}{3}$

(D) $\frac{5\pi}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty 3D coordinate system. You have two fixed points, $A(-4, 2, 1)$ and $B(2, -2, 3)$.

You are tasked with finding the locus of all points that are equidistant from these two markers. In 3D space, the set of all points equidistant from two fixed points forms a plane, specifically the **perpendicular bisector plane**.

This plane acts as a mirror, perfectly dividing the space between $A$ and $B$.

### Finding the DNA of the Plane

To define any plane in 3D space, we need two things: a point on the plane and a normal vector perpendicular to it.

First, let's find the point. Since our plane is the perpendicular bisector, it must pass through the midpoint $M$ of the segment $AB$.

Using the midpoint formula, we calculate:

$$M = \left( \frac{-4+2}{2}, \frac{2-2}{2}, \frac{1+3}{2} ight) = (-1, 0, 2)$$

Now, for the normal vector. Because the plane is perpendicular to the segment $AB$, the vector $\vec{AB}$ itself serves as the normal vector $\vec{n}_1$.

We calculate:

$$\vec{AB} = (2 - (-4))\hat{i} + (-2 - 2)\hat{j} + (3 - 1)\hat{k} = 6\hat{i} - 4\hat{j} + 2\hat{k}$$

To make our calculations easier, we can scale this vector down by dividing by $2$, giving us a simplified normal vector:

$$\vec{n}_1 = 3\hat{i} - 2\hat{j} + \hat{k}$$

### The Angle of Intersection

We are given a second plane, $P'$, defined by $2x + y + 3z = 1$. From this equation, we can immediately extract its normal vector:

$$\vec{n}_2 = 2\hat{i} + 1\hat{j} + 3\hat{k}$$

The angle $	heta$ between two planes is identical to the angle between their normal vectors. We use the elegant dot product formula:

$$\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1|| |\vec{n}_2||}$$

Let's compute the dot product:

$$\vec{n}_1 \cdot \vec{n}_2 = (3)(2) + (-2)(1) + (1)(3) = 6 - 2 + 3 = 7$$

Next, we find the magnitudes:

$$||\vec{n}_1|| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14}$$

$$||\vec{n}_2|| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}$$

Substituting these into our formula, we get:

$$\cos 	heta = \frac{7}{\sqrt{14} \cdot \sqrt{14}} = \frac{7}{14} = \frac{1}{2}$$

Since $\cos 	heta = \frac{1}{2}$, the acute angle $	heta$ is **$\frac{\pi}{3}$**. You have successfully navigated the 3D landscape to find the intersection of these two planes.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the points on the plane $P$ be equidistant from the points $(- 4, 2, 1)$ and $(2, - 2, 3)$ . Then the acute angle between the plane $P$ and the plane $2 x + y + 3 z = 1$ is

Select Answer:

Visualized Solution

Locus of Equidistant Points

The Perpendicular Bisector Plane

Coordinates of Midpoint $M$

Normal Vector of Plane $P$

Simplifying the Normal Vector

Introducing the Second Plane

Angle Between Two Planes

Calculating the Dot Product

Evaluating the Dot Product

Calculating Magnitudes

Substituting into Cosine Formula

Final Angle Calculation

The Sigma Insight: Angle Between Two Planes

Analyzing the Setup

Finding the DNA of the Plane

The Angle of Intersection

Similar Questions

The acute angle between the planes $P_{1}$ and $P_{2}$ , when $P_{1}$ and $P_{2}$ are the planes passing through the intersection of the planes $5 x + 8 y + 13 z - 29 = 0$ and $8 x - 7 y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$ , respectively, is

Let $P_{1} : 2 x + y - z = 3$ and $P_{2} : x + 2 y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

Let the acute angle bisector of the two planes $x - 2 y - 2 z + 1 = 0$ and $2 x - 3 y - 6 z + 1 = 0$ be the plane $P$ . Then which of the following points lies on $P$ ?

A plane which bisects the angle between the two given planes $2 x - y + 2 z - 4 = 0$ and $x + 2 y + 2 z - 2 = 0$ , passes through the point:

The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $π /4$ with plane $x + y = 3$ are

A tetrahedron has vertices $P (1, 2, 1)$ , $Q (2, 1, 3)$ , $R (- 1, 1, 2)$ and $O (0, 0, 0)$ . The angle between the faces $O P Q$ and $P QR$ is :

A tetrahedron has vertices at $O (0, 0, 0), A (1, 2, 1), B (2, 1, 3)$ and $C (- 1, 1, 2)$ . Then the angle between the faces OAB and ABC will be