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Chapter 11: Problem 27

In each part, use a scalar triple product to determine whether the vectors lie in the same plane. $$ \begin{array}{l}{\text { (a) } \mathbf{u}=\langle 1,-2,1\rangle, \mathbf{v}=\langle 3,0,-2\rangle, \mathbf{w}=\langle 5,-4,0\rangle} \\ {\text { (b) } \mathbf{u}= 5 \mathbf{i}-2 \mathbf{j}+\mathbf{k}, \mathbf{v}=4 \mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{w}=\mathbf{i}-\mathbf{j}} \\ {\text { (c) } \mathbf{u}=\langle 4,-8,1\rangle, \mathbf{v}=\langle 2,1,-2\rangle, \mathbf{w}=\langle 3,-4,12\rangle}\end{array} $$

Short Answer

Expert verified
(a) Not coplanar; (b) Coplanar; (c) Not coplanar.

Step by step solution

01

Initial Setup

To determine if the vectors lie in the same plane, calculate the scalar triple product: \( ext{STP}( extbf{u}, extbf{v}, extbf{w}) = extbf{u} \cdot ( extbf{v} \times extbf{w}) \). If the result is zero, the vectors are coplanar.
02

Part (a): Compute Cross Product

Compute the cross product \( \textbf{v} \times \textbf{w} \) for \( \textbf{v} = \langle 3,0,-2 \rangle \) and \( \textbf{w} = \langle 5,-4,0 \rangle \). Use the determinant formula:\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \3 & 0 & -2 \5 & -4 & 0\end{vmatrix} = \mathbf{i}(0 \cdot 0 - (-4) \cdot (-2)) - \mathbf{j}(3 \cdot 0 - 5 \cdot (-2)) + \mathbf{k}(3 \cdot (-4) - 5 \cdot 0)\)Giving \( \langle 8, 10, -12 \rangle \).
03

Part (a): Compute Scalar Triple Product

Calculate \( \text{STP} = \textbf{u} \cdot (\textbf{v} \times \textbf{w}) \) using \( \textbf{u} = \langle 1, -2, 1 \rangle \) and \( \langle 8, 10, -12 \rangle \).\( 1\cdot8 - 2\cdot10 + 1\cdot(-12) = 8 - 20 - 12 = -24 \).Since \( -24 eq 0 \), vectors \( \textbf{u}, \textbf{v}, \textbf{w} \) are not coplanar.
04

Part (b): Compute Cross Product

Compute \( \textbf{v} \times \textbf{w} \) for \( \textbf{v} = \langle 4,-1,1 \rangle \) and \( \textbf{w} = \langle 1,-1,0 \rangle \).\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \4 & -1 & 1 \1 & -1 & 0\end{vmatrix} = \mathbf{i}(0 - (-1)) - \mathbf{j}(4 \cdot 0 - 1 \cdot 1) + \mathbf{k}(4 \cdot (-1) - 1 \cdot (-1))\)Giving \( \langle 1, 1, -3 \rangle \).
05

Part (b): Compute Scalar Triple Product

Compute \( \text{STP} = \textbf{u} \cdot (\textbf{v} \times \textbf{w}) \) with \( \textbf{u} = \langle 5, -2, 1 \rangle \) and \( \langle 1, 1, -3 \rangle \).\( 5\cdot1 + (-2)\cdot1 + 1\cdot(-3) = 5 - 2 - 3 = 0 \).Since \( 0 \), vectors \( \textbf{u}, \textbf{v}, \textbf{w} \) are coplanar.
06

Part (c): Compute Cross Product

Compute \( \textbf{v} \times \textbf{w} \) for \( \textbf{v} = \langle 2,1,-2 \rangle \) and \( \textbf{w} = \langle 3,-4,12 \rangle \).\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \2 & 1 & -2 \3 & -4 & 12\end{vmatrix} = \mathbf{i}(1 \cdot 12 - (-4) \cdot (-2)) - \mathbf{j}(2 \cdot 12 - 3 \cdot (-2)) + \mathbf{k}(2 \cdot (-4) - 3 \cdot 1)\)Giving \( \langle -8, 30, -11 \rangle \).
07

Part (c): Compute Scalar Triple Product

Calculate \( \text{STP} = \textbf{u} \cdot (\textbf{v} \times \textbf{w}) \) with \( \textbf{u} = \langle 4, -8, 1 \rangle \) and \( \langle -8, 30, -11 \rangle \).\( 4\cdot(-8) + (-8)\cdot30 + 1\cdot(-11) = -32 - 240 - 11 = -283 \).Since \( -283 eq 0 \), vectors \( \textbf{u}, \textbf{v}, \textbf{w} \) are not coplanar.

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Key Concepts

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

Coplanar Vectors
Vectors that lie on the same plane are known as coplanar vectors. To determine if three vectors are coplanar, we use a special formula called the scalar triple product. The scalar triple product involves taking the dot product of one vector with the cross product of the other two vectors. If the result is zero, then the vectors are coplanar. In simpler terms, if three vectors all lie flat on the same imaginary sheet of paper, they are coplanar. The calculation shows if they can align in such a manner.
To clarify:
  • If the scalar triple product is zero, the vectors are coplanar.
  • If the scalar triple product is not zero, the vectors are not coplanar and form a volume.
This method is useful for problems in physics and engineering where the idea of parallel planes and forces comes into play.
Cross Product
The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. It results in another vector that is perpendicular to both of the original vectors. The cross product is particularly useful in physics and mathematics for finding a vector orthogonal to a plane. The result of a cross product can be used in calculating the scalar triple product.
The cross product is calculated using a determinant-based method, and the result is a new vector showing the perpendicular direction.
  • For vectors \( \textbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \textbf{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \textbf{a} \times \textbf{b} \) is given by:
  • \[ \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = \mathbf{i}(a_2b_3 - a_3b_2) - \mathbf{j}(a_1b_3 - a_3b_1) + \mathbf{k}(a_1b_2 - a_2b_1) \]
Understanding this helps identify if vectors share the same plane or space.
Determinant Method
Determinants can be used to solve various problems in linear algebra. One such application is in the computation of the cross product via determinants, which is part of finding whether vectors are coplanar through the scalar triple product.
The determinant is a special number that can be calculated from a square matrix. For a 3x3 matrix, the determinant helps find the orthogonal vector needed for the cross product.
  • To calculate a determinant for a 3x3 matrix, consider substituting the rows and columns with their respective vectors.
  • The matrix looks like this: \[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ v_1 & v_2 & v_3 \ w_1 & w_2 & w_3 \end{vmatrix} \]
  • The components involve multiplying across the diagonals and subtracting accordingly to find the cross product result.
Understanding the determinant method is crucial for performing operations like the scalar triple product effortlessly.

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