Notes

Questions & Answers

\(\textbf{1)}\) Let \(\vec{u}=2\vec{i}-3\vec{j}\) and \(\vec{v}=4\vec{i} +2\vec{j}\).
Find the dot product \( \vec{u} \cdot \vec{v} \)

Show Answer

The answer is \( \vec{u} \cdot \vec{v}=(2)(4)+(-3)(2)=2 \)

 

\(\textbf{2)}\) Let \(\vec{u}=2\vec{i}-3\vec{j}\) and \(\vec{v}=4\vec{i} +2\vec{j}\).
Find the angle between \( \vec{u} \) and \( \vec{v} \)

Show Answer

The answer is \( cos^{-1} (\displaystyle\frac{2}{\sqrt{13}\sqrt{20}})\approx82.87^{\circ} \)

 

\(\textbf{3)}\) Let \(\vec{r}=2\vec{i}+5\vec{j}-1\vec{k}\) and \(\vec{s}=3\vec{i}-4\vec{j}+6\vec{k}\).
Find the dot product \( \vec{r} \cdot \vec{s} \)

Show Answer

The answer is \( \vec{r} \cdot \vec{s} =(2)(3)+(5)(-4)+(-1)(6)=-20 \)

 

\(\textbf{4)}\) Let \(\vec{r}=2\vec{i}+5\vec{j}-1\vec{k}\) and \(\vec{s}=3\vec{i}-4\vec{j}+6\vec{k}\).
Find the angle between \( \vec{r} \) and \( \vec{s} \)

Show Answer

The answer is \( \cos^{-1} \left(\displaystyle\frac{-20}{\sqrt{30}\sqrt{61}}\right)\approx117.87^{\circ} \)

 

\(\textbf{5)}\) \( |\vec{n}|=4, |\vec{c}|=8, \) and the angle between the two vectors when placed tail to tail is \( 52^{\circ}. \) Find the Dot Product \( \vec{n} \cdot \vec{c} \)

Show Answer

The answer is \( \vec{n} \cdot \vec{c}=(4)(8)\cos(52)\approx19.7 \)

 

\(\textbf{6)}\) Find the dot product of the two vectors in the picture below.

Show Answer

The answer is \( (8)(10)\cos(45)=40\sqrt{2}\approx56.6 \)

 

\(\textbf{7)}\) Find k so that \( \vec{u}=(2,3,4) \) and \( \vec{v}=(-5,k,1) \) are orthogonal.

Show Hint

Orthogonal means the vectors meet at \(90^{\circ}\), which means the dot product \(=0\)

Show Work

\(\vec{u} \cdot \vec{v}=(2)(-5)+(3)(k)+(4)(1)=-10+3k+4=-6+3k\)
\(-6+3k=0\)
\(3k=6\)
\(k=2\)

Show Answer

The answer is \(k=2\)

 

 

See Related Pages\(\)

\(\bullet\text{ Displacement Vectors}\)
\(\,\,\,\,\,\,\,\,(x_2-x_1)\vec{i}+(y_2-y_1)\vec{j}…\)

\(\bullet\text{ Magnitude, Direction, and Unit Vectors}\)
\(\,\,\,\,\,\,\,\,|\vec{u}|=\sqrt{a^2+b^2}…\)

\(\bullet\text{ Dot Product}\)
\(\,\,\,\,\,\,\,\,a \cdot b=x_1 x_2+ y_1 y_2…\)

\(\bullet\text{ Parallel and Perpendicular Vectors}\)
\(\,\,\,\,\,\,\,\,⟨8,2⟩ \text{ and } ⟨−4,−1⟩…\)

\(\bullet\text{ Scalar and Vector Projections}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{a \cdot b}{|b|^2} \, \vec{b}…\)

\(\bullet\text{ Cross Product}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Equation of a Plane}\)
\(\,\,\,\,\,\,\,\,Ax+By+Cz=D…\)

\(\bullet\text{ Andymath Homepage}\)

 

In Summary

The dot product, aka the scalar product, aka the inner product, is a mathematical operation that takes two vectors and returns a scalar value. It is a fundamental concept in linear algebra and is often used in physics and engineering to calculate the projection of one vector onto another. The dot product can be calculated two different ways.

Algebraic Formula:

If we have two vectors in three-dimensional space: \(A = (a_1, a_2, a_3) and B = (b_1, b_2, b_3)\). The dot product of A and B is calculated using the following algebraic formula:
\(A \cdot B = (a_1)(b_1) + (a_2)(b_2) + (a_3)(b_3)\)

Geometric Formula:

Another way to calculate the dot product is based on the magnitudes of the vectors and the angle \(\theta\) between them. If \(|A|\) represents the magnitude (length) of vector \(A\), and \(|B|\) represents the magnitude of vector \(B\), and \(\theta\) is the angle between the two vectors, then the dot product can be calculated using the following formula:
\(A \cdot B = |A| |B| \cos(\theta)\)

Both formulas yield the same result for the dot product of two vectors. Depending on the context and available information, you can choose the most suitable formula for your calculations. Or use both equations to find a missing piece of information in one of the formulas.