Normal Vectors

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AP Calculus BC › Normal Vectors

Questions 1 - 10

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 9\-5 \4 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\ 2\ -2\end{bmatrix}$ , are orthogonal or not.

The two vectors are orthogonal.

The two vectors are not orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\overrightarrow{a}=\begin{bmatrix} 9\-5 \4 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\ 2\ -2\end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=18-10-8=0$

The two vectors are orthogonal.

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 5\ 3\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -6\-11 \end{bmatrix}$ , are orthogonal or not.

The two vectors are not orthogonal.

The two vectors are orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2$

For our vectors, $\overrightarrow{a}=\begin{bmatrix} 5\ 3\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -6\-11 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=-30-33=-63$

The two vectors are not orthogonal.

Find the Unit Normal Vector to the given plane.

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

$V_n=(\frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}} )$

$V_n=(\frac{3}{{34}}, \frac{3}{{34}}, \frac{4}{{34}} )$

$V_n=(\frac{1}{{34}}, \frac{1}{{34}}, \frac{1}{{34}} )$

Explanation

Recall the definition of the Unit Normal Vector.

$V_n=\frac{V}{||V||}$

Let

$||V||=\sqrt{3^2+3^2+4^2}=\sqrt{34}$

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

Find the Unit Normal Vector to the given plane.

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

$V_n=(\frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}}, \frac{1}{\sqrt{34}} )$

$V_n=(\frac{3}{{34}}, \frac{3}{{34}}, \frac{4}{{34}} )$

$V_n=(\frac{1}{{34}}, \frac{1}{{34}}, \frac{1}{{34}} )$

Explanation

Recall the definition of the Unit Normal Vector.

$V_n=\frac{V}{||V||}$

Let

$||V||=\sqrt{3^2+3^2+4^2}=\sqrt{34}$

$V_n=(\frac{3}{\sqrt{34}}, \frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}} )$

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 3\7 \ 5\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\3 \-6 \end{bmatrix}$ , are orthogonal or not.

The two vectors are orthogonal.

The two vectors are not orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\overrightarrow{a}=\begin{bmatrix} 3\7 \ 5\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\3 \-6 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=9+21-30=0$

The two vectors are orthogonal.

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix}2 \0 \5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}10 \7 \-2 \end{bmatrix}$ , are orthogonal or not.

The two vectors are not orthogonal.

The two vectors are orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\overrightarrow{a}=\begin{bmatrix}2 \0 \5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}10 \7 \-2 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=20+0-10=10$

The two vectors are not orthogonal.

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 0\3 \-5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\15 \9 \end{bmatrix}$ , are orthogonal or not.

The two vectors are orthogonal.

The two vectors are not orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\overrightarrow{a}=\begin{bmatrix} 0\3 \-5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\15 \9 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=0+45-45=0$

The two vectors are orthogonal.

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 4\ 1\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -3\12 \end{bmatrix}$ , are orthogonal or not.

The two vectors are orthogonal.

The two vectors are not orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2$

For our vectors, $\overrightarrow{a}=\begin{bmatrix} 4\ 1\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -3\12 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=-12+12=0$

The two vectors are orthogonal.

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 8\ -7\ 3\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\1 \2 \end{bmatrix}$ , are orthogonal or not.

The two vectors are not orthogonal.

The two vectors are orthogonal.

Explanation

Vectors can be said to be orthogonal, that is to say perpendicular or normal, if their dot product amounts to zero:

$\overrightarrow{a}\cdot\overrightarrow{b}=0\rightarrow \overrightarrow{a}\perp\overrightarrow{b}$

To find the dot product of two vectors given the notation

$\overrightarrow{a}\begin{bmatrix} a_1\a_2 \ a_3 \end{bmatrix};\overrightarrow{b}\begin{bmatrix} b_1\b_2 \ b_3 \end{bmatrix}$

Simply multiply terms across rows:

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3$

For our vectors, $\overrightarrow{a}=\begin{bmatrix} 8\ -7\ 3\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\1 \2 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=32-7+6=31$

The two vectors are not orthogonal.

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 7\11 \13 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 3\4 \ -5\end{bmatrix}$ , are orthogonal or not.