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Chapter 11: Problem 18

a. Find the area of the triangle determined by the points \(P, Q\) and \(R\) b. Find a unit vector perpendicular to plane \(P Q R\) $$P(-2,2,0), \quad Q(0,1,-1), \quad R(-1,2,-2)$$

Short Answer

Expert verified
a. The area of the triangle is \(\frac{\sqrt{14}}{2}\). b. The unit vector perpendicular to the plane is \(\left( \frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right)\).

Step by step solution

01

Determine Vectors PQ and PR

Calculate the vectors \( \vec{PQ} \) and \( \vec{PR} \) using the given points.\[ \vec{PQ} = Q - P = (0 - (-2), 1 - 2, -1 - 0) = (2, -1, -1) \]\[ \vec{PR} = R - P = (-1 - (-2), 2 - 2, -2 - 0) = (1, 0, -2) \]
02

Compute the Cross Product

Find the cross product \( \vec{PQ} \times \vec{PR} \), which gives a vector perpendicular to the plane determined by \(P, Q, R\).Using the formula for the cross product:\[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & -1 \ 1 & 0 & -2 \end{vmatrix} = (-2)\hat{i} - (-3)\hat{j} + 1\hat{k} = (-2, -3, 1) \]
03

Calculate Magnitude of Cross Product

Find the magnitude of the cross product to determine the area of the triangle.\[ |\vec{PQ} \times \vec{PR}| = \sqrt{(-2)^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]
04

Find Area of the Triangle

The area of the triangle is half the magnitude of the cross product.\[ \text{Area} = \frac{1}{2} \times \sqrt{14} \]
05

Compute a Unit Vector

The unit vector perpendicular to the plane is the cross product vector divided by its magnitude.The magnitude of vector \((-2, -3, 1)\) is \(\sqrt{14}\), so the unit vector is:\[ \vec{u} = \left( \frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Cross Product
The cross product is a fundamental operation in vector calculus used to find a vector that is perpendicular to two given vectors. For two vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), the cross product \( \vec{a} \times \vec{b} \) is calculated using the determinant of a matrix:
  • The first row consists of the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) corresponding to the x, y, and z axes respectively.
  • The second row is the components of \( \vec{a} \).
  • The third row is the components of \( \vec{b} \).
Using these rows, the cross product is:\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]After evaluating the determinant, this operation results in a new vector that's perpendicular to both \( \vec{a} \) and \( \vec{b} \). This property makes it particularly useful in 3D geometry and physics.
Magnitude Calculation
Calculating the magnitude of a vector is an essential skill in 3D geometry, as it gives the length or size of a vector. For a vector \( \vec{v} = (v_1, v_2, v_3) \), the magnitude is determined as follows:
  • Square each of the vector's components: \( v_1^2, v_2^2, v_3^2 \).
  • Add those squares together: \( v_1^2 + v_2^2 + v_3^2 \).
  • Take the square root of the sum: \( \sqrt{v_1^2 + v_2^2 + v_3^2} \).
This calculation is vital when you need to normalize vectors or when working with physical quantities like force or velocity, where knowing the size is important.
Unit Vector
A unit vector is a vector with a magnitude of 1, pointing in a specific direction. It is obtained by dividing the vector \( \vec{v} \) by its magnitude \( |\vec{v}| \). The formula for finding a unit vector \( \vec{u} \) in the direction of vector \( \vec{v} = (v_1, v_2, v_3) \) is:\[\vec{u} = \left( \frac{v_1}{|\vec{v}|}, \frac{v_2}{|\vec{v}|}, \frac{v_3}{|\vec{v}|} \right)\]
  • This process is known as normalization.
  • It is frequently used in graphics and physics to represent direction without considering length.
A unit vector is particularly significant when you need to determine a direction based solely on orientation, such as when defining coordinate axes in space.
Triangle Area in 3D
The area of a triangle in 3D space can be computed using the cross product of two of its defining vectors. For a triangle defined by points \( P, Q, R \), the vectors \( \vec{PQ} \) and \( \vec{PR} \) are first determined. Then the cross product \( \vec{PQ} \times \vec{PR} \) is calculated to find a vector perpendicular to the plane of the triangle, but its magnitude also relates to the area.
  • The area of the triangle is given by half the magnitude of this cross product: \( \frac{1}{2} |\vec{PQ} \times \vec{PR}| \).
  • This method is effective because it accounts for the orientation and placement of the triangle in 3D space.
This approach allows for precise calculations irrespective of the triangle's rotation and tilt in 3D space.
3D Geometry
3D geometry involves understanding shapes and figures in a three-dimensional space, comprising x, y, and z coordinates. Unlike 2D geometry, 3D involves an additional dimension that adds depth.
  • Points in 3D space are specified by three coordinates \( (x, y, z) \).
  • Understanding vectors and their operations, such as the cross product, is crucial in 3D geometry.
  • Using vectors allows for the calculation of distances, angles, and areas more systematically.
3D geometry is fundamental in numerous applications, from computer graphics to physical simulations, and is characterized by the additional complexity and richness that three dimensions bring.

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