# Examples On Vector Dot Product Set-1

Go back to  'Vectors and 3-D Geometry'

Example – 16

Find the component of a vector $$\vec b$$ perpendicular to the vector  $$\vec a$$.

Solutions: We need to find $$\vec r$$, the component of  $$\vec b$$perpendicular to $$\vec a$$

We have

\begin{align} &\qquad \overrightarrow {PQ} = \left( {\frac{{\vec b \cdot \vec a}}{{{{\left| {\vec a} \right|}^2}}}} \right)\vec a \hfill \\\\ &\;\Rightarrow \quad \vec r = \vec b - \overrightarrow {PQ} \hfill \\\\&\quad\qquad\;= \vec b - \left( {\frac{{\vec a \cdot \vec b}}{{{{\left| {\vec a} \right|}^2}}}} \right)\vec a \hfill \\ \end{align}

Example – 17

Let $$\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$$ be three mutually perpendicular vectors of equal magnitude. Prove that the vector $$\vec a + \vec b + \vec c$$ is equally inclined to each of the three vectors.

Solutions: Let $${\theta _a}$$ represent the angle between $$\vec a$$ and  $$\vec a + \vec b + \vec c$$ . We have,

\begin{align}&\cos {\theta _a} = \frac{{\vec a \cdot (\vec a + \vec b + \vec c)}}{{\left| {\vec a} \right|\;\left| {\vec a + \vec b + \vec c} \right|}} \hfill \\\\&\qquad\;= \frac{{\vec a \cdot \vec a + \vec a \cdot \vec b + \vec a \cdot \vec c}}{{\left| {\vec a} \right|\;\sqrt {{{\left| {\vec a + \vec b + \vec c} \right|}^2}} }}\qquad\qquad...\left( 1 \right) \hfill \\ \end{align}

Let the magnitude of $$\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$$ be $$\lambda$$. Also since the three vectors are mutually perpendicular, we have $$\vec a \cdot \vec b = \vec a \cdot \vec c = 0$$. Now,

\begin{align}& {\left| {\vec a + \vec b + \vec c} \right|^2} = (\vec a + \vec b + \vec c) \cdot (\vec a + \vec b + \vec c) \hfill \\\\ &\qquad\qquad\quad= \vec a \cdot \vec a + \vec b \cdot \vec b + \vec c \cdot \vec c \hfill \\\\&\qquad\qquad\quad= 3{\lambda ^2} \hfill \\ \end{align}

Using these facts in (1), we have

$\cos {\theta _a} = \frac{{{\lambda ^2}}}{{\lambda \cdot \sqrt 3 \,\lambda }} = \frac{1}{{\sqrt 3 }}$

It is easy to see that $$\cos {\theta _b}$$ and  $$\cos {\theta _c}$$ will have the same value. Thus, $$\vec a + \vec b + \vec c$$ is equally inclined to $$\vec a,\;\;\vec b\;\;{\text{and}}\;\;\vec c$$ .

Example – 18

Find the angle between the two diagonals of a cube.

Solutions:

We now have,

\begin{align}& \overrightarrow {OP} \equiv a\hat i + a\hat j + a\hat k \Rightarrow \left| {\overrightarrow {OP} } \right| = \sqrt 3 a \hfill \\\\&\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \hfill \\\\&\quad\;\;\equiv - a\hat i + a\hat j + a\hat k \Rightarrow \left| {\overrightarrow {AB} } \right| = \sqrt 3 \,a \hfill \\ \end{align}

Let $$\theta$$ denote the angle between OP and AB. Thus,

\begin{align}&\qquad \cos \theta = \frac{{\overrightarrow {OP} \cdot \overrightarrow {AB} }}{{\left| {\overrightarrow {OP} } \right|\;\left| {\overrightarrow {AB} } \right|}} \hfill \\\\&\qquad\qquad = \frac{{(a\hat i + a\hat j + a\hat k) \cdot ( - a\hat i + a\hat j + a\hat k)}}{{(\sqrt 3 a)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\sqrt 3 \,a)}} \hfill \\\\&\qquad \qquad= \frac{{ - {a^2} + {a^2} + {a^2}}}{{3\;{a^2}}} \hfill \\\\&\qquad\qquad = \frac{1}{3} \hfill \\\\ & \quad\Rightarrow \quad\theta = {\cos ^{ - 1}}\frac{1}{3} \hfill \\ \end{align}

This is the angle between any two diagonals of (any) cube.

Vectors
grade 11 | Questions Set 1
Vectors