A Projection is like a shadow casted on a vector. The definition of scalar and vector projections are shown below.
Scalar Projection
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
Vector projection on b will result in a vector in the direction of \(\vec{b} \).
proj\(\left(\vec{a}\text{ on }\vec{b}\right)=\dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert}
\hat{b}
\)
\(=\left(\dfrac{\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}}\right)\vec{b}=\dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert^2}\vec{b}
\)
Vector Projection Calculator
Find the scalar and vector projections of \(\vec{u}\text{ onto }\vec{v}\text{, if
}\vec{u}=(5, 6, -3)\text{ and }\vec{v}=(1, 4, 5) \).
Show Answer
First, we can determine the scalar projection of \(\vec{u}\text{ onto }\vec{v}\):
\(\vert\text{proj}\left(\vec{u}\text{ on
}\vec{v}\right)\vert=\dfrac{\vec{u}\cdot\vec{v}}{\vert\vec{v}\vert}\)
\(=\dfrac{5(1)+6(4)-3(5)}{\sqrt{1^2+4^2+5^2}}\)
\(=\dfrac{14}{\sqrt{42}}\)
\(=\dfrac{14\sqrt{42}}{42}=\dfrac{1}{3}\sqrt{42}
\)
Therefore, we can determine the scalar projection of \(\vec{u}\text{ onto }\vec{v}\) is \(\dfrac{14}{\sqrt{42}}\).
Next, we can determine the vector projection of \(\vec{u}\text{ onto }\vec{v}\):
\(\text{proj}\)\(\left(\vec{u}\text{ on
}\vec{v}\right)=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\)
\(=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\)
\(=\dfrac{14}{42}(1, 4,
5)\)
\(=\left(\dfrac{1}{3}, \dfrac{4}{3}, \dfrac{5}{3}\right) \)
Therefore, we can determine the vector projection of \(\vec{u}\text{ onto }\vec{v}\) is \(\left(\dfrac{1}{3}, \dfrac{4}{3}, \dfrac{5}{3}\right) \).
Find the scalar and vector projections of \(\vec{v}\text{ onto }\vec{u} \).
Show Answer
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 \(=\dfrac{14}{\sqrt{70}}\).
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 \(\vec{u}\text{ and }\vec{v} \) are non-zero vectors, but \(\text{proj}\vec{u}\text{ onto }\vec{v}=\vec{0} \), what conclusion can be drawn?
Show Answer
If \(\text{proj} \left(\vec{u}\text{ on }\vec{v}\right)=\vec{0} \)
\(\left(\vert\vec{u}\vert\cos{θ}\right)\hat{v}=\vec{0} \)
\(\vert\vec{u}\vert{≠0}\text{ }\vert\hat{v}\vert{≠0 }\text{ }∴\cos{θ}=0 \)
\(θ=90° \)
If \(\vec{u}\text{ and }\vec{v} \) are non-zero vectors, does it follow that \(\text{proj}\vec{u}\text{ onto
}\vec{v}=\vec{0} \)?
Show Answer
Yes. since if \(\vec{u}\perp\vec{v}\text{ then }\vec{v}\perp\vec{u} \)
Find the projection of \(\vec{PQ} \) onto each of the coordinate axes, where \(P \) is the
point \((2, 3, 5)\) and \(Q \) is the point \((-1, 2, 5)\).
Show Answer
\(\vec{PQ}=\left(-1-2, 2-3, 5-5\right) \)
\(\vec{PQ}=(-3, -1, 0) \)
\(Proj \left(\vec{PQ} \text{ on } \hat{i}\right)=\dfrac{\vec{PQ}\cdot\hat{i}}{\hat{i}\cdot\hat{i}}\hat{i}=\dfrac{(-3)(1)+0+0
}{1}(1, 0, 0)=(3, 0, 0) \)
Proj\(\left(\vec{PQ} \text{ on } \hat{j}\right)=(0, -1, 0) \)
Proj\(\left(\vec{PQ} \text{ on } \hat{k}\right)=(0, 0, 0) \)