Normal Vectors - AP Calculus BC

Card 0 of 485

Question

Find the Unit Normal Vector to the given plane.

.

Answer

Recall the definition of the Unit Normal Vector.

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

Let

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

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

Compare your answer with the correct one above

Question

Find the unit normal vector of $v(t)=t^{2}\hat{i}+3t^{2}\hat{j}$ .

Answer

To find the unit normal vector, you must first find the unit tangent vector. The equation for the unit tangent vector, , is

$T(t)=\frac{r(t)}{\left | r(t)\right |}$

where is the vector and $\left | r(t)\right |$ is the magnitude of the vector.

The equation for the unit normal vector,, is

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}$

where is the derivative of the unit tangent vector and $\left | T'(t)\right |$ is the magnitude of the derivative of the unit vector.

For this problem

$v(t)=t^{2}\hat{i}+3t^{2}\hat{j}$

$\left | v(t)\right |=\sqrt{\left (t^{2} \right )^{2}+\left (3t^{2} \right )^{2}}=\sqrt{t^{4}+9t^{4}}=\sqrt{10t^{4}}=t^{2}\sqrt{10}$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{t^{2}\hat{i}+3t^{2}\hat{j}}{t^{2}\sqrt{10}}=\frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j}$

$T'(t)=\frac{d}{dt}\left ( \frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j} \right )=0$

$\left | T'(t)\right |=\sqrt{\left (0\right )^{2}}=0$

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}=\frac{0}{0}=DNE$

There is no unit normal vector of $v(t)=t^{2}\hat{i}+3t^{2}\hat{j}$ .

Compare your answer with the correct one above

Question

Find the unit normal vector of $v(t)=sin(t)\hat{i}+cos(t)\hat{j}$ .

Answer

To find the unit normal vector, you must first find the unit tangent vector. The equation for the unit tangent vector, , is

$T(t)=\frac{r(t)}{\left | r(t)\right |}$

where is the vector and $\left | r(t)\right |$ is the magnitude of the vector.

The equation for the unit normal vector,, is

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}$

where is the derivative of the unit tangent vector and $\left | T'(t)\right |$ is the magnitude of the derivative of the unit vector.

For this problem

$v(t)=sin(t)\hat{i}+cos(t)\hat{j}$

$\left | v(t)\right |=\sqrt{\left (sin(t) \right )^{2}+\left (cos(t) \right )^{2}}=\sqrt{1}=1$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{sin(t)\hat{i}+cos(t)\hat{j}}{1}=sin(t)\hat{i}+cos(t)\hat{j}$

$T'(t)=\frac{d}{dt}\left (sin(t)\hat{i}+cos(t)\hat{j} \right )=cos(t)\hat{i}-sin(t)\hat{j}$

$\left | T'(t)\right |=\sqrt{\left (cos(t) \right )^{2}+\left (sin(t) \right )^{2}}=\sqrt{1}=1$

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}=\frac{cos(t)\hat{i}-sin(t)\hat{j}}{1}=cos(t)\hat{i}-sin(t)\hat{j}$

Compare your answer with the correct one above

Question

Find the unit normal vector of $v(t)=e^{t}\hat{i}+2e^{t}\hat{j}$ .

Answer

To find the unit normal vector, you must first find the unit tangent vector. The equation for the unit tangent vector, , is

$T(t)=\frac{r(t)}{\left | r(t)\right |}$

where is the vector and $\left | r(t)\right |$ is the magnitude of the vector.

The equation for the unit normal vector,, is

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}$

where is the derivative of the unit tangent vector and $\left | T'(t)\right |$ is the magnitude of the derivative of the unit vector.

For this problem

$v(t)=e^{t}\hat{i}+2e^{t}\hat{j}$

$\left | v(t)\right |=\sqrt{\left (e^{t} \right )^{2}+\left (2e^{t} \right )^{2}}=\sqrt{e^{2t}+4e^{2t}}=\sqrt{5e^{2t}}=e^{t}\sqrt{5}$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{e^{t}\hat{i}+2e^{t}\hat{j}}{e^{t}\sqrt{5}}=\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}$

$T'(t)=\frac{d}{dt}\left (\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j} \right )=0$

$\left | T'(t)\right |=0$

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}=\frac{0}{0}=DNE$

The normal vector of $v(t)=e^{t}\hat{i}+2e^{t}\hat{j}$ does not exist.

Compare your answer with the correct one above

Question

Find the unit normal vector of $v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$ .

Answer

To find the unit normal vector, you must first find the unit tangent vector. The equation for the unit tangent vector, , is

$T(t)=\frac{r(t)}{\left | r(t)\right |}$

where is the vector and $\left | r(t)\right |$ is the magnitude of the vector.

The equation for the unit normal vector,, is

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}$

where is the derivative of the unit tangent vector and $\left | T'(t)\right |$ is the magnitude of the derivative of the unit vector.

For this problem

$v(t)=5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}$

$\left | v(t)\right |=\sqrt{\left (5cos(t) \right )^{2}+\left (5sin(t) \right )^{2}+(2)^{2}}=\sqrt{25+4}=\sqrt{29}$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{5cos(t)\hat{i}+5sin(t)\hat{j}+2\hat{k}}{\sqrt{29}}$

$T(t)=\frac{v(t)}{\left | v(t)\right |}=\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k}$

$T'(t)=\frac{d}{dt}\left (\frac{5}{\sqrt{29}}cos(t)\hat{i}+\frac{5}{\sqrt{29}}sin(t)\hat{j}+\frac{2}{\sqrt{29}}\hat{k} \right )$

$T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}+0\hat{k}$

$T'(t)=\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}$

$\left | T'(t)\right |=\sqrt{\left (\frac{-5}{\sqrt{29}}sin(t) \right )^{2}+\left (\frac{5}{\sqrt{29}}cos(t) \right )^{2}}$

$\left | T'(t)\right |=\sqrt{\frac{25}{29}sin^2(t)+\frac{25}{29}cos^2(t)}=\sqrt{\frac{25}{29}}=\frac{5}{\sqrt{29}}$

$N(t)=\frac{T'(t)}{\left | T'(t)\right |}=\frac{\frac{-5}{\sqrt{29}}sin(t)\hat{i}+\frac{5}{\sqrt{29}}cos(t)\hat{j}}{\frac{5}{\sqrt{29}}}=-sin(t)\hat{i}+cos(t)\hat{j}$

Compare your answer with the correct one above

Question

Find a normal vector $\vec{n}$ that is perpendicular to the plane given below.

Answer

Derived from properties of plane equations, one can simply pick off the coefficients of the cartesian coordinate variable to give a normal vector $\vec{n}$ that is perpendicular to that plane. For a given plane, we can write

$ax+by+cz=C, \vec{n}= a \hat{x}+b\hat{y}+c\hat{z}$ .

From this result, we find that for our case,

$\vec{n}= -3 \hat{x}+5\hat{y}-\hat{z}$ .

Compare your answer with the correct one above

Question

Which of the following is FALSE concerning a vector normal to a plane (in -dimensional space)?

Answer

These are all true facts about normal vectors to a plane. (If the surface is not a plane, then a few of these no longer hold.)

Compare your answer with the correct one above

Question

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.

Answer

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.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} -3\4 \6 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}2 \3 \-1 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} -3\4 \6 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}2 \3 \-1 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=-6+12-6=0$

The two vectors are orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix}7 \1 \3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}0 \-5 \ 2\end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix}7 \1 \3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}0 \-5 \ 2\end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=0-5+6=1$

The two vectors are not orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 5\0 \7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -14\11 \10 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 5\0 \7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -14\11 \10 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=-70+0+70=0$

The two vectors are orthogonal.

Compare your answer with the correct one above

Question

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

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 7\7 \3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}1 \-1 \2 \end{bmatrix}$

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

The two vectors are not orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 11\8 \-10 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\3 \4 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 11\8 \-10 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\3 \4 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=22+24-40=6$

The two vectors are not orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 13\5 \12 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\2 \-3 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 13\5 \12 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\2 \-3 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=26+10-36=0$

The two vectors are orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 11\-8 \2 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\ 1\ -7\end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 11\-8 \2 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\ 1\ -7\end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=22-8-14=0$

The two vectors are orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 3\0 \3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\0 \1 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 3\0 \3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\0 \1 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=0+0+3=3$

The two vectors are not orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 4\ 9\ 0\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\ 0\-6 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 4\ 9\ 0\end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\ 0\-6 \end{bmatrix}$

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

The two vectors are orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 14\1 \15 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}1 \1 \-1 \end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 14\1 \15 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}1 \1 \-1 \end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=14+1-15=0$

The two vectors are orthogonal.

Compare your answer with the correct one above

Question

Determine whether the two vectors, $\overrightarrow{a}=\begin{bmatrix} 6\ 2\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 1\ -4\ 1\end{bmatrix}$ , are orthogonal or not.

Answer

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, $\overrightarrow{a}=\begin{bmatrix} 6\ 2\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 1\ -4\ 1\end{bmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b}=6-8+3=1$

The two vectors are not orthogonal.

Compare your answer with the correct one above

Question

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.

Answer

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.

Compare your answer with the correct one above

Tap the card to reveal the answer