A Projection is like a shadow casted on a vector. The definition of scalar and vector projections are shown below.
of \(\vec{a}\text{ on }\vec{b}=\vert\text{proj}\left(\vec{a}\text{ on }\vec{b}\right)\vert= \dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert} \)
of \(\vec{b}\text{ on }\vec{a}=\vert\text{proj}\left(\vec{b}\text{ on }\vec{a}\right)\vert=\dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{a}\vert} \)
Vector projection on b will result in a vector in the direction of \(\vec{b} \).
proj\(\left(\vec{a}\text{ on }\vec{b}\right)=\dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert} \hat{b} \)
\(=\left(\dfrac{\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}}\right)\vec{b}=\dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert^2}\vec{b} \)
Enter in two vectors below. Then, Click the + button to toggle between addition and subtraction.
First, we can determine the scalar projection of \(\vec{u}\text{ onto }\vec{v}\):
\(\vert\text{proj}\left(\vec{u}\text{ on }\vec{v}\right)\vert=\dfrac{\vec{u}\cdot\vec{v}}{\vert\vec{v}\vert}\)
\(=\dfrac{5(1)+6(4)-3(5)}{\sqrt{1^2+4^2+5^2}}\)
\(=\dfrac{14}{\sqrt{42}}\)
\(=\dfrac{14\sqrt{42}}{42}=\dfrac{\sqrt{42}}{3}\)
Therefore, we can determine the scalar projection of \(\vec{u}\text{ onto }\vec{v}\) is \(\boldsymbol{\dfrac{\sqrt{42}}{3}}\).
Next, we can determine the vector projection of \(\vec{u}\) onto \(\vec{v}\):
\(\text{proj}\)\(\left(\vec{u}\text{ on }\vec{v}\right)=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\)
\(=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\)
\(=\dfrac{\sqrt{42}}{3}(1, 4, 5)\)
\(=\left(\dfrac{\sqrt{42}}{3}, \dfrac{4\sqrt{42}}{3}, \dfrac{5\sqrt{42}}{3}\right)\)
Therefore, we can determine the vector projection of \(\vec{u}\text{ onto }\vec{v}\) is \(\boldsymbol{\left(\dfrac{\sqrt{42}}{3}, \dfrac{4\sqrt{42}}{3}, \dfrac{5\sqrt{42}}{3}\right)} \).
First, we can determine the scalar projection of \(\vec{v}\text{ onto }\vec{u}\):
\(\vert\text{proj}\left(\vec{v}\text{ on }\vec{u}\right)\vert=\dfrac{\vec{v}\cdot\vec{u}}{\vert\vec{u}\vert} \)
\(=\dfrac{1(5)+4(6)+5(-3)}{\sqrt{5^2+6^2+3^2}}\)
\(=\dfrac{14}{\sqrt{70}}\)
Therefore, we can determine the scalar projection of \(\vec{v}\text{ onto }\vec{u}\) is \(\boldsymbol{\dfrac{14}{\sqrt{70}}}\).
Next, we can determine the vector projection of \(\vec{v}\text{ onto }\vec{u}\):
proj \(\left(\vec{v}\text{ on }\vec{u}\right)=\dfrac{\vec{v}\cdot\vec{u}}{|\vec{u}|^2}\vec{u} \)
\(=\dfrac{14}{70}(5, 6, -3)\)
\(=\left(1, \dfrac{6}{5}, \dfrac{-3}{5}\right)\)
If \(\text{proj} \left(\vec{u}\text{ on }\vec{v}\right)=\vec{0}\)
\(\left(\vert\vec{u}\vert\cos{θ}\right)\hat{v}=\vec{0} \)
\(\vert\vec{u}\vert{≠0}\text{ }\vert\hat{v}\vert{≠0 }\text{ }∴\cos{θ}=0 \)
\(θ = 90°\)
Yes. since if \(\vec{u}\perp\vec{v}\text{ then }\vec{v}\perp\vec{u} \)
First, we can determine \(\vec{PQ}\):
\(\vec{PQ}=\left(-1-2, 2-3, 5-5\right)\)
\(\vec{PQ}=(-3, -1, 0)\)
\(Proj \left(\vec{PQ} \text{ on } \hat{i}\right)=\dfrac{\vec{PQ}\cdot\hat{i}}{\hat{i}\cdot\hat{i}}\hat{i}=\dfrac{(-3)(1)+0+0 }{1}(1, 0, 0)=(3, 0, 0)\)
Proj\(\left(\vec{PQ} \text{ on } \hat{j}\right)=(0, -1, 0)\)
Proj\(\left(\vec{PQ} \text{on} \hat{k}\right)=(0, 0, 0)\)