6.0 Projection of a Vector on a Line

  Vectors

Suppose any vector $\overrightarrow {AB} $ makes an angle $\theta $ with the direction line $l$ in the anticlockwise direction.Then projection of vector $\overrightarrow {AB} $ on line $l$ is given by $$\left| {\overrightarrow {AB} } \right|\cos \theta $$ and direction is same as that of the line depending on the value of $\cos \theta $ being positive or negative. This projection is called projection vector and magnitude is simply called projection of the vector ${\overrightarrow {AB} }$ on the line $l$.

Properties

Property 1: Projection of vector $\overrightarrow a $ on line $l$ is given as $\overrightarrow a .\widehat p$ where $\widehat p$ is the unit vector along the line $l$.

Property 2: Projection of vector $\overrightarrow a $ on $\overrightarrow b $ is given as $\overrightarrow a .\widehat b$ which is $$\overrightarrow a .\widehat b = \overrightarrow a .(\frac{{\overrightarrow b }}{{\left| {\overrightarrow b } \right|}})$$ or it can bewritten as $\frac{1}{{\left| {\overrightarrow b } \right|}}(\overrightarrow a .\overrightarrow b )$

Property 3: For parallel vectors i.e $\theta = 0$ the projection of vector ${\overrightarrow {AB} }$ is the vector itself ${\overrightarrow {AB} }$ and if vectors are antiparallel i.e $\theta = \pi $ , projection of vector ${\overrightarrow {AB} }$ is ${\overrightarrow {BA} }$.

Property 4: If $\theta = \frac{\pi }{2}or\frac{{3\pi }}{2}$, the projection of ${\overrightarrow {AB} }$ is $0$.

Note:

If $\alpha$, $\beta$ and $\lambda $ are the direction angles of the vector $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ makes with $x$, $y$ and $z$ axes respectively, then the direction cosines of this vector are given as $$\begin{equation} \begin{aligned} Cos\alpha = frac{{\overrightarrow a .\widehat i}}{{\left| {\overrightarrow a } \right|\left| {\widehat i} \right|}} = \frac{{{a_1}}}{{\left| {\overrightarrow a } \right|}} \\ Cos\beta = \frac{{{a_2}}}{{\left| {\overrightarrow a } \right|}} \\ Cos\lambda = \frac{{{a_3}}}{{\left| {\overrightarrow a } \right|}} \\\end{aligned} \end{equation} $$

So $\left| {\overrightarrow a } \right|Cos\alpha$ , $\left| {\overrightarrow a } \right|Cos\beta$ and $\left| {\overrightarrow a } \right|Cos\lambda $ are the projections of vector $\overrightarrow a $ along $OX$, $OY$ and $OZ$ respectively. So, the scalar components ${a_1}$, ${a_2}$ and ${a_3}$ are the projections of vector $\overrightarrow a $ along $x$, $y$ and $z$ axis respectively. Further, if $\overrightarrow a $ is a unit vector, then vector can be expressed in terms of direction cosines as follows:

$$\overrightarrow a = Cos\alpha \widehat i + Cos\beta \widehat j + Cos\lambda \widehat k$$

Question 11. If $\overrightarrow a = 5\widehat i - 2\widehat j + 3\widehat k$ and $\overrightarrow a = \widehat i - x\widehat j + 3\widehat k$ are perpendicular vectors, then find the value of $x$.

Solution: For perpendicular vectors $$\overrightarrow a .\overrightarrow b =0 $$

Therefore, $$\begin{equation} \begin{aligned} \overrightarrow a .\overrightarrow b = (5\widehat i - 2\widehat j + 3\widehat k).(\widehat i - x\widehat j + 3\widehat k) \\ 0 = 5 + 2x + 9 \\ 14 + 2x = 0 \\ 2x = - 14 \\ x = - 7 \\\end{aligned} \end{equation} $$

Question 12. Find the angle between the vectors $\overrightarrow a = (\widehat i - 2\widehat j + 3\widehat k)$ and $\overrightarrow b = (2\widehat i - 8\widehat j + 3\widehat k)$.

Solution: We know that \[\begin{gathered} cos\theta = \frac{{(\widehat i - 2\widehat j + 3\widehat k).(2\widehat i - 8\widehat j + 3\widehat k)}}{{\sqrt {1 + 4 + 9} \sqrt {4 + 64 + 9} }} \hspace{1em} \\ cos\theta = \frac{{2 + 16 + 9}}{{\sqrt {14} \sqrt {77} }} \hspace{1em} \\ cos\theta = \frac{{27}}{{7\sqrt {22} }} \hspace{1em} \\ \end{gathered} \]

Question 13. Find the direction cosines of the vector $\overrightarrow a = (\widehat i - 6\widehat j + 4\widehat k)$.

Solution : Direction cosines are given as $$\begin{equation} \begin{aligned} cos\alpha = \frac{1}{{\sqrt {1 + 36 + 16} }} \\ cos\alpha = \frac{1}{{\sqrt {53} }} \\ cos\beta = \frac{{ - 6}}{{\sqrt {53} }} \\ cos\lambda = \frac{4}{{\sqrt {53} }} \\\end{aligned} \end{equation} $$\[\theta = {\cos ^{ - 1}}\frac{{27}}{{7\sqrt {22} }}\]