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.

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