Find x so that the dot product of the vectors u and v below is 23.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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To solve the problem of finding \( x \) so that the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \) is 23, we are given the following vectors:

\[
\mathbf{u} = \begin{bmatrix} 3 \\ 2 \\ x \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 5 \\ -2 \\ 2 \end{bmatrix}
\]

The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated as:

\[
\mathbf{u} \cdot \mathbf{v} = (3)(5) + (2)(-2) + (x)(2)
\]

Simplifying the expression, we get:

\[
15 - 4 + 2x = 23
\]

Further simplification:

\[
11 + 2x = 23
\]

Subtracting 11 from both sides:

\[
2x = 12
\]

Dividing by 2:

\[
x = 6
\] 

However, it seems the solution mentioned in the image states \( x = 0 \), which is incorrect based on the dot product being equal to 23. By recalculating correctly, \( x = 6 \).
Transcribed Image Text:To solve the problem of finding \( x \) so that the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \) is 23, we are given the following vectors: \[ \mathbf{u} = \begin{bmatrix} 3 \\ 2 \\ x \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 5 \\ -2 \\ 2 \end{bmatrix} \] The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated as: \[ \mathbf{u} \cdot \mathbf{v} = (3)(5) + (2)(-2) + (x)(2) \] Simplifying the expression, we get: \[ 15 - 4 + 2x = 23 \] Further simplification: \[ 11 + 2x = 23 \] Subtracting 11 from both sides: \[ 2x = 12 \] Dividing by 2: \[ x = 6 \] However, it seems the solution mentioned in the image states \( x = 0 \), which is incorrect based on the dot product being equal to 23. By recalculating correctly, \( x = 6 \).
Expert Solution
Step 1

dot product will be 3*5+2*(-2)+2x

so this must be  23

 

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