Wize University Linear Algebra Textbook > Products of Vectors

Projection (Proj) and Perpendicular (Perp)

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Projection and Perpendicular
The projection of u⃗\vec{u}u onto v⃗\vec{v}v is denoted by projv⃗u⃗\text{proj}_{\vec{v}}\vec{u}projv​u and is obtained by calculating:
projv⃗u⃗=(u⃗⋅v⃗v⃗⋅v⃗)v⃗\boxed{\quad
\text{proj}_{\colorTwo{\vec v} } \colorOne{\vec u}
=
\left(
\dfrac{
\colorOne{\vec u} \cdot \colorTwo{\vec v}}
{\colorTwo{\vec v} \cdot \colorTwo{\vec v}}
\right)
\colorTwo{\vec v}
\quad}projv​u=(v⋅vu⋅v​)v​
Notes
The projection of u⃗\vec{u}u onto v⃗\vec{v}v is the "shadow" that u⃗\colorOne{\vec u}u casts onto v⃗\colorTwo{\vec v}v
projv⃗u⃗\text{proj}_{\vec v}\vec uprojv​u is a scalar multiple of v⃗\colorTwo{\vec v}v
projv⃗u⃗\text{proj}_{\vec v}\vec uprojv​u is also called the vector component of u⃗\colorFour{\vec{u}}u parallel to v⃗\colorFour{\vec{v}}v
The perpendicular of u⃗\vec{u}u onto v⃗\vec{v}v is denoted by perpv⃗u⃗\text{perp}_{\vec{v}}\vec{u}perpv​u and is obtained by calculating:
perpv⃗u⃗ = u⃗−projv⃗u⃗\boxed{\quad
\text{perp}_{\vec v}\vec u
\ =\ 
\vec u - 
\text{proj}_{\vec v}\vec u
\quad}perpv​u = u−projv​u​
The perpendicular is the vector component of u⃗\colorFour{\vec{u}}u orthogonal to v⃗\colorFour{\vec{v}}v

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Example: Projection
Let u⃗=⟨3,−1,4⟩\vec{u}=\lang 3,-1,4 \rangu=⟨3,−1,4⟩ and v⃗=⟨9,0,2⟩\vec{v}=\lang 9,0,2 \rangv=⟨9,0,2⟩.

Part A)
Find projv⃗ u⃗\text{proj}_{\vec{v}}\ \vec{u}projv​ u.

projv⃗u⃗=u⃗⋅v⃗v⃗⋅v⃗v⃗=⟨3,−1,4⟩⋅⟨9,0,2⟩⟨9,0,2⟩⋅⟨9,0,2⟩⟨9,0,2⟩=(3)(9)+(−1)(0)+(4)(2)92+02+22⟨9,0,2⟩=3585⟨9,0,2⟩=717⟨9,0,2⟩=⟨6317, 0, 1417⟩\begin{aligned}
\text{proj}_{\vec{v}}\vec{u} &=
\dfrac{\vec{u}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\vec{v}\\[1.5em]

&=
\dfrac{\lang 3,-1,4\rang\cdot\lang 9,0,2\rang}{\lang 9,0,2\rang \cdot\lang 9,0,2\rang}\lang 9,0,2\rang \\[1.5em]

&=
\dfrac{(3)(9)+(-1)(0)+(4)(2)}{9^2+0^2+2^2}\lang 9,0,2 \rang\\[1.5em]

&=
\dfrac{35}{85}\lang 9,0,2\rang \\[1.5em]

&=
\dfrac{7}{17}\lang 9,0,2 \rang\\[1.5em]

&=
\left\lang \dfrac{63}{17}, \ 0,\ \dfrac{14}{17}\right\rang \\[1.5em]
\end{aligned}

projv​u​=v⋅vu⋅v​v=⟨9,0,2⟩⋅⟨9,0,2⟩⟨3,−1,4⟩⋅⟨9,0,2⟩​⟨9,0,2⟩=92+02+22(3)(9)+(−1)(0)+(4)(2)​⟨9,0,2⟩=8535​⟨9,0,2⟩=177​⟨9,0,2⟩=⟨1763​, 0, 1714​⟩​

Part B)
Find proju⃗ v⃗\text{proj}_{\vec{u}}\ \vec{v}proju​ v.

proju⃗v⃗=v⃗⋅u⃗u⃗⋅u⃗u⃗=⟨9,0,2⟩⋅⟨3,−1,4⟩⟨3,−1,4⟩⋅⟨3,−1,4⟩⟨3,−1,4⟩=(9)(3)+(0)(−1)+(2)(4)32+(−1)2+42⟨3,−1,4⟩=3526⟨3,−1,4⟩=3526⟨3,−1,4⟩=⟨10526,−3526,7013⟩\begin{aligned}
\text{proj}_{\vec{u}}\vec{v} &= 
\dfrac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\vec{u}\\[1.5em]

&= 
\dfrac{\lang 9,0,2\rang \cdot \lang 3,-1,4\rang }{\lang 3,-1,4\rang \cdot \lang 3,-1,4\rang }\lang 3,-1,4\rang \\[1.5em]

&= 
\dfrac{(9)(3)+(0)(-1)+(2)(4)}{3^2+(-1)^2+4^2}\lang 3,-1,4\rang\\[1.5em]

&= 
\dfrac{35}{26}\lang 3,-1,4\rang\\[1.5em]

&= 
\dfrac{35}{26}\lang 3,-1,4\rang\\[1.5em]

&= 
\left\lang \dfrac{105}{26},-\dfrac{35}{26},\dfrac{70}{13} \right\rang\\[1.5em]
\end{aligned}proju​v​=u⋅uv⋅u​u=⟨3,−1,4⟩⋅⟨3,−1,4⟩⟨9,0,2⟩⋅⟨3,−1,4⟩​⟨3,−1,4⟩=32+(−1)2+42(9)(3)+(0)(−1)+(2)(4)​⟨3,−1,4⟩=2635​⟨3,−1,4⟩=2635​⟨3,−1,4⟩=⟨26105​,−2635​,1370​⟩​

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Example: Projection and Perpendicular
Let u⃗=[42], v⃗=[−31]∈R2\vec u 
=
\begin{bmatrix}
4\\2
\end{bmatrix},\ 

\vec v 
=
\begin{bmatrix}
-3\\1
\end{bmatrix} \in \reals^2u=[42​], v=[−31​]∈R2.
Part A)
Sketch vectors  u⃗\vec uu  and  v⃗\vec vv and guess whether proju⃗ v⃗\text{proj}_{\vec u}\ \vec vproju​ v is in the same direction or the opposite direction of u⃗\vec uu.
Calculate proju⃗ v⃗\text{proj}_{\vec u}\ \vec vproju​ v to check your guess.

We can see that projecting v⃗\vec vv onto u⃗\vec uu will require us to extend u⃗\vec uu in the opposite direction.

proju⃗v⃗=v⃗⋅u⃗u⃗⋅u⃗ u⃗=[−31]⋅[42][42]⋅[42][42]=(−3)(4)+(1)(2)(4)2+22[42]=−1020[42]=−12[42]=[−2−1]\begin{aligned}
\text{proj}_{\vec u} \vec v 
&=
\dfrac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\ \vec{u}\\[1em]
&= 
\frac{
\begin{bmatrix}
-3\\1
\end{bmatrix}
\cdot
\begin{bmatrix}
4\\2
\end{bmatrix}
}{
\begin{bmatrix}
4\\2
\end{bmatrix}
\cdot
\begin{bmatrix}
4\\2
\end{bmatrix}
}
\begin{bmatrix}
4\\2
\end{bmatrix}\\[3em]
&=
\dfrac{
(-3)(4) + (1)(2)}
{(4)^2 + 2^2}
\begin{bmatrix}
4\\2
\end{bmatrix}\\[1.5em]
&=
\dfrac{-10}
{20}
\begin{bmatrix}
4\\2
\end{bmatrix}\\[1.5em]
&=
-\dfrac{1}
{2}
\begin{bmatrix}
4\\2
\end{bmatrix}\\[1.5em]
&=
\boxed{
\begin{bmatrix}
-2\\
-1
\end{bmatrix}
}
\end{aligned}proju​v​=u⋅uv⋅u​ u=[42​]⋅[42​][−31​]⋅[42​]​[42​]=(4)2+22(−3)(4)+(1)(2)​[42​]=20−10​[42​]=−21​[42​]=[−2−1​]​​
Since the scalar multiple of u⃗\vec uu is negative (−12)(-\frac{1}{2})(−21​), we have confirmed that  proju⃗v⃗\text{proj}_{\vec u} \vec vproju​v is in the opposite direction of u⃗\vec uu.

PAGE BREAK

Part B)
Calculate perpu⃗v⃗\text{perp}_{\vec u} \vec vperpu​v.

We can use our answer from Part A):
proju⃗v⃗=[−2−1]\text{proj}_{\vec u} \vec v  = \begin{bmatrix}
-2\\
-1
\end{bmatrix}proju​v=[−2−1​]
So calculating perpu⃗v⃗\text{perp}_{\vec u} \vec vperpu​v is as simple as:
perpu⃗v⃗=v⃗−proju⃗v⃗=[−31]−[−2−1]=[−12]\begin{aligned}
\text{perp}_{\vec{u}}\vec{v}
&=\vec v - \text{proj}_{\vec{u}}\vec{v}\\
&=
\begin{bmatrix}
  -3\\ 1
\end{bmatrix}
-
\begin{bmatrix}
-2\\
-1
\end{bmatrix}\\
&=
\boxed{
\begin{bmatrix}
  -1\\
  2
\end{bmatrix}
}
\end{aligned}perpu​v​=v−proju​v=[−31​]−[−2−1​]=[−12​]​​
Notice that, as usual, this result is a position vector: if the vector were translated to start at the origin, it would go left 1 and up 2.

Practice: Projection
Find the projection of a⃗=⟨−2,3,6⟩\vec a=\lang -2,3,6 \ranga=⟨−2,3,6⟩ onto  b⃗=⟨−5,−1,4⟩\vec b=\lang -5,-1,4 \rangb=⟨−5,−1,4⟩.

⟨−6249,9349,18649⟩\left\lang-\dfrac{62}{49},\dfrac{93}{49},\dfrac{186}{49}\right\rang⟨−4962​,4993​,49186​⟩

⟨−15549,−3149,12449⟩\left\lang -\dfrac{155}{49},-\dfrac{31}{49},\dfrac{124}{49}\right\rang⟨−49155​,−4931​,49124​⟩

⟨−356,−76,143⟩\left\lang -\dfrac{35}{6},-\dfrac{7}{6},\dfrac{14}{3}\right\rang⟨−635​,−67​,314​⟩

⟨−15542,−3142,6221⟩\left\lang-\dfrac{155}{42},-\dfrac{31}{42},\dfrac{62}{21}\right\rang⟨−42155​,−4231​,2162​⟩

1742⟨−5,−1,4⟩\dfrac{1}{7\sqrt{42}}\lang-5,-1,4\rang742​1​⟨−5,−1,4⟩

I don't know

Practice: Projection and Perpendicular
Let u⃗=[−2−140]\vec{u}=
\begin{bmatrix}
  -2\\ -1\\ 4\\ 0
\end{bmatrix}u=​−2−140​​, v⃗=[−3014]\vec v=
\begin{bmatrix}
  -3\\ 0\\ 1\\ 4
\end{bmatrix}v=​−3014​​, and w⃗=[0011]\vec{w}=
\begin{bmatrix}
  0\\ 0\\ 1\\ 1
\end{bmatrix}w=​0011​​

Find perpv⃗ − w⃗ (u⃗+v⃗)\text{perp}_{\vec v \ -\  \vec w} \ (\vec u + \vec v)perpv − w​ (u+v).
(Enter your answer as a column vector)

I don't know

Practice: Projection
True or False
There exist vectors  u⃗,v⃗∈Rn\vec u, \vec v \in \reals^nu,v∈Rn  such that  projv⃗ u⃗=proju⃗ v⃗\text{proj}_{\vec{v}}\ \vec{u}
=
\text{proj}_{\vec{u}}\ \vec{v}projv​ u=proju​ v.

True

False

I don't know

Extra Practice

Example: Dot Product (2)

Find the scalar and vector projection of b⃗=⟨2,1,3⟩\vec{b}=\langle 2,1,3\rangleb=⟨2,1,3⟩onto a⃗=⟨−1,4,2⟩\vec{a}=\langle -1,4,2\ranglea=⟨−1,4,2⟩.

Find the vector projection of  [3,2][3,2][3,2] onto the vector [3,6]\left[3,6\right][3,6]

133 - FML 3 - 18.1W - e.g. 9

Find the vector-component ı^∣∣d⃗ \ihat_{||_{\vd}} \,^∣∣d​​ of ı^\bcb{\boldsymbol{ \ihat}}^ in the direction of the line given by y=mx\bcb{\boldsymbol{ y=mx}}y=mx. Repeat for ȷ^∣∣d⃗ \jhat_{||{\vd}} \, ^​∣∣d​, the vector-component of ȷ^\bcb{\boldsymbol{ \jhat }}^​ in the direction of the line.

133 - FML 3 - 18.1W e.g. 58.3

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0)
Find the point on the line L\bcb{L}L closest to the point P2\bcb{P_2}P2​.

133 - FML 3 - 18.1W e.g. 58.2

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0)
Find the minimum distance of P2\bcb{P_2}P2​ to the line L\bcb{L}L found above. 

133 - FML 3 - 18.1W e.g. 48

 Find a unit vector in the direction of the projection of u⃗=<102,112,−56>\bcb{\vec{u} = \left< 102, 112, -56 \right>}u=⟨102,112,−56⟩ onto v⃗=<0,3,4>\bcb{\vec{v} = \left< 0, 3, 4 \right>}v=⟨0,3,4⟩.

133 - FML 3 - 18.1W e.g. 43

 Given u⃗=<8,2,2>\bcb{\vec{u} = \left< 8, 2, 2 \right>}u=⟨8,2,2⟩ and v⃗=<2,−1,−1>\bcb{\vec{v} = \left< 2, -1, -1 \right>}v=⟨2,−1,−1⟩, find the vector projection of u⃗\bcb{\vec{u}}u onto v⃗\bcb{\vec{v}}v. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41

Let P=(−3,−3,−4)\bcb{P = (-3, -3, -4)}P=(−3,−3,−4), Q=(−5,−6,−3)\bcb{Q = (-5,-6,-3)}Q=(−5,−6,−3), R=(−6,−4,−6)\bcb{R = (-6,-4,-6)}R=(−6,−4,−6) and S=(−2,−3,−4)\bcb{S = (-2, -3, -4)}S=(−2,−3,−4). Let l1\bcb{l_1}l1​ be the line passing through P\bcb{P}P and Q\bcb{Q}Q, and let l2\bcb{l_2}l2​ be the line passing through R\bcb{R}R and S\bcb{S}S. Find the distance between R\bcb{R}R and l1\bcb{l_1}l1​. 

133 - FML 3 - 18.1W e.g. 36

For two lines having directional vectors v1\mathbf{v}_1v1​ and v2\mathbf{v}_2v2​, respectively, the shortest distance between them can be found by creating a vector that connects the two lines (by subtracting a point on one line from a point on the other), i.e. (r2−r1)\left(\mathbf{r}_2 - \mathbf{r}_1\right)(r2​−r1​), and projecting it onto the unit normal vector connecting the two lines. The unit normal is easily found by crossing their respective directionals and dividing by the magnitude of the resulting vector, as illustrated in Fig. 5. i.e.
s=∣projn^(r2−r1)∣=∣(r2−r1)⋅v1×v2∣v1×v2∣∣
 	s = \left| \text{proj}_{\mathbf{\hat{n}}} \left(\mathbf{r}_2 - \mathbf{r}_1\right)\right| = \left| \left(\mathbf{r}_2 - \mathbf{r}_1\right) \cdot \frac{\mathbf{v}_1 \times \mathbf{v}_2}{\left|\mathbf{v}_1 \times \mathbf{v}_2\right|}\right|
 s=∣projn^​(r2​−r1​)∣=​(r2​−r1​)⋅∣v1​×v2​∣v1​×v2​​​
Find the equations of the line passing through the point (2,3,−4)\bcb{(2, 3, -4)}(2,3,−4) parallel to the vector i−4k\bcb{\mathbf{i} - 4\mathbf{k}}i−4k ; express it  in vector-parametric, scalar-parametric and standard form. 
 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3

 Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)}P1​=(0,0,1) and  P2=(1,−2,3)\bcb{P_2 = (1,-2,3)}P2​=(1,−2,3) and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}v=⟨2,−3,1⟩, find the point Po\bcb{P_o}Po​ on the plane Π: 2x−3y+z=1\bcb{\Pi: \ 2x - 3y + z = 1}Π: 2x−3y+z=1 that is closest to the point P2\bcb{P_2}P2​.

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 48

 Find a unit vector in the direction of the projection of u⃗=<102,112,−56>\bcb{\vec{u} = \left< 102, 112, -56 \right>}u=⟨102,112,−56⟩ onto v⃗=<0,3,4>\bcb{\vec{v} = \left< 0, 3, 4 \right>}v=⟨0,3,4⟩.

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 43

 Given u⃗=<8,2,2>\bcb{\vec{u} = \left< 8, 2, 2 \right>}u=⟨8,2,2⟩ and v⃗=<2,−1,−1>\bcb{\vec{v} = \left< 2, -1, -1 \right>}v=⟨2,−1,−1⟩, find the vector projection of u⃗\bcb{\vec{u}}u onto v⃗\bcb{\vec{v}}v. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0), find the minimum distance of P2\bcb{P_2}P2​ to the line L\bcb{L}L that is parallel to v⃗\bcb{\vec{v}}v that passes through P1\bcb{P_1}P1​. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.2

Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)}P1​=(0,0,1) and  P2=(1,−2,3)\bcb{P_2 = (1,-2,3)}P2​=(1,−2,3) and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}v=⟨2,−3,1⟩
Find the minimum distance between the point P2\bcb{P_2}P2​ and the plane Π: 2x−3y+z=1\bcb{\Pi:\ 2x-3y + z = 1}Π: 2x−3y+z=1. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.3

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0), find the point on 
the line L\bcb{L}L (which is parallel to v⃗\bcb{\vec{v}}v that passes through P1\bcb{P_1}P1​) that is closest to the point P2\bcb{P_2}P2​.

Wize University Linear Algebra Textbook > Products of Vectors

Projection (Proj) and Perpendicular (Perp)

Popular Courses

Projection and Perpendicular

Example: Projection

Part A)

Part B)

Example: Projection and Perpendicular

Part A)

Part B)

Practice: Projection

Practice: Projection and Perpendicular

Practice: Projection

True or False

Extra Practice

Example: Dot Product (2)

133 - FML 3 - 18.1W - e.g. 9

133 - FML 3 - 18.1W e.g. 58.3

133 - FML 3 - 18.1W e.g. 58.2

133 - FML 3 - 18.1W e.g. 48

133 - FML 3 - 18.1W e.g. 43

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41

133 - FML 3 - 18.1W e.g. 36

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 48

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 43

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.2

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.3