Orthogonal projection of a vector onto another vector, the result is a vector. Meanwhile, the length of an orthogonal vector projection of a vector onto another vector always has a positive real number/scalar value.

In this article we will discuss vector projections, especially orthogonal (perpendicular) vector projections. Orthogonal projection of a vector onto another vector, the result is a vector. Meanwhile, the length of the projection of an orthogonal vector on another vector is always a positive real number/scalar. vector-projection The orthogonal projection of vector u⃗\vec{u} on vector v⃗\vec{v} can be denoted by u⃗v⃗{\vec{u}}_{\vec{v}} or p⃗\vec{p} and is defined by the following argument.

Proposition:

  1. Projeksi scalar orthogonal u⃗\vec{u} falls on v⃗\vec{v} adalah ∣∣p⃗∣∣=u⃗⋅v⃗∣v⃗∣\left| \left| \vec{p}\right| \right|=\frac{\vec{u}\cdot\vec{v}}{\left| \vec{v}\right|}
  2. The projection of the vector u⃗\vec{u} onto the vector v⃗\vec{v} is the vector u⃗v⃗=(u⃗⋅v⃗∣.v⃗∣2)v⃗ atau p⃗=u⃗∣v⃗∣∣∣p⃗∣∣{\vec{u}}_{\vec{v}}=\left( \frac{\vec{u } \cdot \vec{v}}{\left|. \vec{v} \right|^2} \right)\vec{v}\text{ atau } \vec{p}=\frac{\vec{u }}{\left| \vec{v} \right|} \left| \left| \vec{p} \right| \right|
  3. Panjang project vector u⃗\vec{u} to vector v⃗\vec{v} adalah ∣u⃗v⃗∣=∣u⃗⋅ev⃗∣ \lvert \vec{u}_{\vec{v}}\rvert=\left| \vec{u}\cdot e_{\vec{v}} \right| dengan ev⃗e_{\vec{v}} adalah vector satuan ke arah v⃗\vec{v} atau ∣u⃗v⃗∣=∣u⃗.v⃗∣v⃗∣∣\left| \vec{u}_{\vec{v}} \right|=\left| \frac{\vec{u}.\vec{v}}{\left| \vec{v} \right|} \right| .

Example 1:

Given a⃗=2i^−6j^−3k^\vec{a}=2\widehat{i}-6\widehat{j}-3\widehat{k} and b⃗=4i^+2j^−4k^\vec{b}=4\widehat{i}+2\widehat{j} -4\widehat{k}. Determine:

  1. The length of the vector projection a⃗\vec{a} on b⃗\vec{b}
  2. Orthogonal projection of vector a⃗\vec{a} on b⃗\vec{b}
  3. Orthogonal projection of vector b⃗\vec{b} on a⃗\vec{a}

Alternative Solutions:

  1. The length of the vector projection a⃗\vec{a} on b⃗\vec{b} ∣a⃗b⃗∣=∣a⃗.b⃗∣b⃗∣∣=∣(2i^−6j^−3k^)⋅(4i^+2j^−4k^)42+22+(−4)2∣=∣(2)(4)+(−6)(2)+(−3)(−4)16+4+16∣=∣8−12+1236∣=∣86∣\begin{align*} \left| \vec{a}_{\vec{b}} \right|=\left| \frac{\vec{a}.\vec{b}}{\left| \vec{b} \right|} \right|&=\left| \frac{(2\widehat{i}-6\widehat{j}-3\widehat{k})\cdot (4\widehat{i}+2\widehat{j}-4\widehat{k})}{\sqrt{{{4}^{2}}+{{2}^{2}}+{{(-4)}^{2}}}} \right| \\ &=\left| \frac{(2)(4)+(-6)(2)+(-3)(-4)}{\sqrt{16+4+16}} \right| \\ &=\left| \frac{8-12+12}{\sqrt{36}} \right|=\left| \frac{8}{6} \right| \end{align*} ∴∣a⃗b⃗∣=43 \therefore \left| \vec{a}_{\vec{b}} \right|=\frac{4}{3}

  2. Orthogonal projection of vector a⃗\vec{a} on b⃗\vec{b}
    a⃗b⃗=∣∣a⃗b⃗∣∣b⃗∣b⃗∣\vec{a}_{\vec{b}}=\left| \left| \vec{a}_{\vec{b}} \right| \right|\frac{\vec{b}}{\left| \vec{b} \right|}, karena ∣b⃗∣=6\left| \vec{b} \right|=6 dan ∣∣a⃗b⃗∣∣=43\left| \left| {{\vec{a}}_{\vec{b}}} \right| \right|=\frac{4}{3}
    a⃗b⃗=43.4i^+2j^−4k^6=89i^+49j^−89k^\begin{align*} \vec{a}_{\vec{b}}&=\frac{4}{3}.\frac{4\widehat{i}+2\widehat{j}-4\widehat{k}}{6} \\ & =\frac{8}{9}\widehat{i}+\frac{4}{9}\widehat{j}-\frac{8}{9}\widehat{k} \end{align*}

  3. Orthogonal projection of vector b⃗\vec{b} on a⃗\vec{a}
    b⃗a⃗=(b⃗⋅a⃗∣a⃗∣2)a⃗=(4i^+2j^−4k^)⋅(2i^−6j^−3k^)(22+(−6)2+(−3)2)2(2i^−6j^−3k^)=(4)(2)+(2)(−6)+(−4)(−3)22+(−6)2+(−3)2(2i^−6j^−3k^)=849(2i^−6j^−3k^)b⃗a⃗=1649i^−4849j^−2449k^)\begin{align*}\vec{b}_{\vec{a}}&=\left( \frac{\vec{b}\cdot \vec{a}}{{{\left| \vec{a} \right|}^{2}}} \right)\vec{a} \\ &=\frac{(4\widehat{i}+2\widehat{j}-4\widehat{k})\cdot (2\widehat{i}-6\widehat{j}-3\widehat{k})}{{{\left( \sqrt{{{2}^{2}}+{{(-6)}^{2}}+{{(-3)}^{2}}} \right)}^{2}}}(2\widehat{i}-6\widehat{j}-3\widehat{k}) \\ &=\frac{(4)(2)+(2)(-6)+(-4)(-3)}{{{2}^{2}}+{{(-6)}^{2}}+{{(-3)}^{2}}}(2\widehat{i}-6\widehat{j}-3\widehat{k}) \\ &=\frac{8}{49}(2\widehat{i}-6\widehat{j}-3\widehat{k}) \\ \vec{b}_{\vec{a}}&=\frac{16}{49}\widehat{i}-\frac{48}{49}\widehat{j}-\frac{24}{49}\widehat{k})\end{align*}

example 2

Given the vectors u⃗=(−1,1,−4) \vec{u} = (-1,1,-4) and v⃗=(2,−1,3) \vec{v} = ( 2, -1,3) . Determine the scalar projection and vector projection (2u⃗+3v⃗) (2\vec{u} + 3\vec{v}) falls −2v⃗ -2\vec{v} !
Solution:
for example:
a⃗=(2u⃗+3v⃗)=(−2,2,−8)+(6,−3,9)=(4,−1,1) \vec{a} = (2\vec{u} + 3\vec{v}) = (-2,2,-8) + ( 6, -3,9) = (4, -1 , 1)
b⃗=−2v⃗=(−4,2,−6) \vec{b} = -2 \vec{v} = (-4, 2,-6)

  • Determine the scalar projection a⃗ \vec{a} on b⃗ \vec{b}
    Projective scalar =a⃗.b⃗∣b⃗∣=4.(−4)+(−1).2+1.(−6)(−4)2+22+(−6)2=−16−2−6 sqrt16+4+36=−2456=−245656 &=−3756\begin{align*}\text{Projective scalar } &= \frac{\vec{a}.\vec{b}}{|\vec{b}|} \\&= \frac{4 .(-4) + (-1). 2 + 1. (-6)}{\sqrt{(-4)^2 + 2^2 + (-6)^2} } \\&= \frac{-16 - 2 - 6}{\ sqrt{16 + 4 + 36 } } \\&= \frac{-24}{\sqrt{56} } \\&= \frac{-24}{56} \sqrt{56} \ \&= -\frac{3}{7} \sqrt{56}\end{align*}
    ∴\therefore so the scalar projection is −3756 -\frac{3}{7} \sqrt{56} .
  • Determine the projection of the vector a⃗ \vec{a} on b⃗ \vec{b}
    Proyeksi vector =(a⃗.b⃗∣b⃗∣2)b⃗=(−24(56)2)(−4,2,−6)=(−2456)(−4,2,−6)=(−37)(−4,2,−6)=(127,−67,187)\begin{align*}\text{Proyeksi vector } &= \left( \frac{\vec{a}.\vec{b}}{|\vec{b}|^2} \right) \vec{b}\\&= \left( \frac{-24}{(\sqrt{56})^2} \right) ( -4, 2,-6)\\&= \left( \frac{-24}{56} \right) ( -4, 2,-6)\\ &= \left( -\frac{3}{7} \right) ( -4, 2,-6)\\&= \left( \frac{12}{7}, -\frac{6}{7}, \frac{18}{7} \right)\end{align*}
    ∴\therefore so, the vector projection is (127,−67,187) \left( \frac{12}{7}, -\frac{6}{7}, \frac{18}{7} \right) .

Example 3

Diketahui vector p⃗=2i⃗+j⃗+2k⃗ \vec{p} = 2\vec{i}+\vec{j} +2\vec{k} dan q⃗=3i⃗+bj⃗+k⃗ \vec{q} = 3\vec{i} + b\vec{j} + \vec{k} . If ∣r⃗∣ |\vec{r}| adalah panjang projeksi vector q⃗ \vec{q} on p⃗ \vec{p} and ∣r⃗∣=4 |\vec{r}| = 4 , then determine the value of b b !
Solution:
Given the vectors p⃗=(2,1,2) \vec{p} = (2, 1, 2) and q⃗=(3,b,1) \vec{q} = (3, b, 1) .

  • Determining the value of b b by orthogonal projection q⃗ \vec{q} on p⃗ \vec{p} :
    Panjang proyeksi =∣q⃗.p⃗∣p⃗∣∣∣r⃗∣=∣q⃗.p⃗∣p⃗∣∣4=∣2.3+1.b+2.122+12+22∣4=∣b+89∣4=∣b+83∣∣b+8∣=12b=4∨b=−20\begin{align*}\text{Panjang proyeksi } &= \left| \frac{\vec{q}.\vec{p}}{|\vec{p}|} \right|\\|\vec{r}| &= \left| \frac{\vec{q}.\vec{p}}{|\vec{p}|} \right|\\4 &= \left| \frac{ 2.3 + 1.b + 2.1 }{ \sqrt{2^2 + 1^2 + 2^2 } } \right|\\4 &= \left| \frac{ b + 8 }{ \sqrt{9} } \right|\\4 &= \left| \frac{ b + 8 }{ 3 } \right| \\| b + 8 | &= 12 \\ b &= 4 \vee b = -20 \end{align*}
    So, the possible values ​​of b b are b=−20 b = -20 or b=4 b = 4 .

Example 4

Determine the projection of the vector a⃗=(2,0,1) \vec{a} = (2,0,1) on the vector b⃗ \vec{b} which is parallel and equal in length but in the opposite direction to vector c⃗=(0,2,−2) \vec{c} = (0, 2, -2 ) !
Solution:
Diketahui vector b⃗=−c⃗=−(0,2,−2)=(0,−2,2) \vec{b} = - \vec{c} = -(0, 2, -2) = (0, -2, 2) .

  • Determine the projection of the vector a⃗ \vec{a} on b⃗ \vec{b} :
    Proyeksi vector =(a⃗.b⃗∣b⃗∣2)b⃗=(2.0+0.(−2)+1.2(02+(−2)2+22)2)(0,−2,2)=(2(8)2)(0,−2,2)=(28)(0,−2,2)=(14)(0,−2,2)=(0,−12,12)\begin{align*}\text{Proyeksi vector } &= \left( \frac{\vec{a}.\vec{b}}{|\vec{b}|^2} \right) \vec{b} \\&= \left( \frac{2.0 + 0. (-2) + 1.2}{(\sqrt{0^2 + (-2)^2 + 2^2 })^2 } \right) (0, -2, 2)\\&= \left( \frac{2}{(\sqrt{8 })^2 } \right) (0, -2, 2)\\ &= \left( \frac{2}{8 } \right) (0, -2, 2) \\&= \left( \frac{1}{4 } \right) (0, -2, 2)\\&= \left( 0, -\frac{1}{2} , \frac{1}{2} \right) \end{align*}
    So, the resulting vector projection is (0,−12,12) \left( 0, -\frac{1}{2} , \frac{1}{2} \right) .