Problem 96 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A tetrahedron has vertices at \(\mathrm{O}(0,0,0), \mathrm{A}(1,2,1) \mathrm{B}(2,1,3)\) and \(\mathrm{C}(-1,1,2)\). Then the angle between the faces \(\mathrm{OAB}\) and ABC will be (a) \(90^{\circ}\) (b) \(\cos ^{-1}\left(\frac{19}{35}\right)\) (c) \(\cos ^{-1}\left(\frac{17}{31}\right)\) (d) \(30^{\circ}\)

Step-by-Step Solution

Answer

The angle is \( \cos^{-1}\left(\frac{19}{35}\right) \).

Step 1: Finding Normal Vectors

First, we need to find the normal vectors to the planes OAB and ABC. The normal vector can be found using the cross product of any two non-parallel vectors in the plane.

Step 2: Vectors in Plane OAB

In the plane OAB, the vectors OA and OB are given by: \( \vec{OA} = \langle 1-0, 2-0, 1-0 \rangle = \langle 1, 2, 1 \rangle \) and \( \vec{OB} = \langle 2-0, 1-0, 3-0 \rangle = \langle 2, 1, 3 \rangle \).

Step 3: Cross Product for Plane OAB

Find the normal vector of plane OAB by taking the cross product of \( \vec{OA} \) and \( \vec{OB} \):\[ \vec{n_1} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} \] which simplifies to \( \langle 5, -1, -3 \rangle \).

Step 4: Vectors in Plane ABC

In the plane ABC, the vectors AB and AC are given by: \( \vec{AB} = \langle 2-1, 1-2, 3-1 \rangle = \langle 1, -1, 2 \rangle \) and \( \vec{AC} = \langle -1-1, 1-2, 2-1 \rangle = \langle -2, -1, 1 \rangle \).

Step 5: Cross Product for Plane ABC

Find the normal vector of plane ABC by taking the cross product of \( \vec{AB} \) and \( \vec{AC} \):\[ \vec{n_2} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} \] which simplifies to \( \langle 1, -5, -3 \rangle \).

Step 6: Angle Between Two Planes

The angle between the planes is the angle between their normal vectors \( \vec{n_1} \) and \( \vec{n_2} \). This can be calculated using the dot product formula:\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{\|\vec{n_1}\| \|\vec{n_2}\|} \]

Step 7: Dot Product of Normal Vectors

Compute the dot product: \( \vec{n_1} \cdot \vec{n_2} = 5 \times 1 + (-1) \times (-5) + (-3) \times (-3) = 5 + 5 + 9 = 19 \).

Step 8: Magnitude of Normal Vectors

Compute the magnitudes: \( \|\vec{n_1}\| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{35} \) and \( \|\vec{n_2}\| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{35} \).

Step 9: Calculate the Cosine of the Angle

Substitute the values into the formula: \[ \cos \theta = \frac{19}{35} \].

Step 10: Determine the Angle

The angle between the faces is \( \cos^{-1}\left(\frac{19}{35}\right) \).

Key Concepts

Vectors and Geometry

Understanding vectors and their role in geometry is essential to solve problems related to 3D shapes, like tetrahedrons. A vector in geometry is a quantity having both magnitude and direction, often represented as an arrow. In three-dimensional space, vectors can be used to represent points, lines, and planes by describing the direction and the position of these geometric figures.

In the problem at hand, a tetrahedron is a 3-dimensional shape with triangular faces, and finding angles between these faces is a common challenge in geometry. Each face of the tetrahedron can be represented by vectors that form a plane. For example, using the vertices given, we can express vectors like \( \vec{OA} \) and \( \vec{OB} \) based on their coordinates. These vectors help us find the orientation and boundary of the plane or face they belong to.

The calculations in our problem require moving from vertices to planes via vectors, using mathematical tools like cross products and dot products to determine relationships such as the angle between these planes.

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors in 3D space. This perpendicular vector is also known as the normal vector of a plane formed by the given vectors.

In the exercise, to find the normal vectors of the faces OAB and ABC of the tetrahedron, the cross product is used with pairs of vectors from each plane. For instance, the cross product of \( \vec{OA} = \langle 1, 2, 1 \rangle \) and \( \vec{OB} = \langle 2, 1, 3 \rangle \) gives the normal to plane OAB as \( \langle 5, -1, -3 \rangle \). This vector is perpendicular to both OA and OB and thus serves as the normal vector of plane OAB.

Performing a cross product involves solving a determinant, and it results in a vector that provides crucial information about the orientation of a plane in three-dimensional space.

Dot Product

The dot product is another operation involving two vectors, but unlike the cross product, it results in a scalar value rather than a vector. This scalar value is indicative of the angle between the vectors. If the dot product is zero, the vectors are perpendicular.

In the given problem, calculating the dot product between the normal vectors of planes OAB and ABC helps find the angle between them. Given the normal vectors \( \vec{n_1} = \langle 5, -1, -3 \rangle \) and \( \vec{n_2} = \langle 1, -5, -3 \rangle \), the dot product is calculated as \( 5 \times 1 + (-1) \times (-5) + (-3) \times (-3) = 19 \).

The result, when paired with the magnitudes of the vectors, allows us to determine the cosine of the angle between the planes using the formula \( \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{\|\vec{n_1}\| \|\vec{n_2}\|} \). The dot product thus serves as a critical step in understanding geometric relationships.

Angle Between Planes

To determine the angle between two planes, we use the normal vectors of these planes. These normal vectors are perpendicular to each plane and the angle between the normals is the same as the angle between the planes themselves.

Using the dot product of the normal vectors between the planes OAB and ABC, we calculate the cosine of the angle \( \theta \) between them. The formula \( \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{\|\vec{n_1}\| \|\vec{n_2}\|} \) is used, where \( \vec{n_1} \) and \( \vec{n_2} \) are the normal vectors for planes OAB and ABC respectively, and \( \|\vec{n_1}\| \) and \( \|\vec{n_2}\| \) are their magnitudes.

This calculation leads to \( \cos \theta = \frac{19}{35} \), giving us the precise measure of the angle between the two planes as \( \cos^{-1}\left(\frac{19}{35}\right) \). Understanding the geometric and algebraic methods to find angles in three-dimensional space is fundamental in vector geometry.