Example – 1

Let ABC and PQR be two triangles in a plane. Assume that the perpendiculars from the points A, B, C to the sides QR, RP, PQ respectively are concurrent. Using vector methods, prove that the perpendiculars from P, Q, R to BC, CA, AB respectively are also concurrent.


We have,

\[\begin{align}&\qquad\;OA \bot QR,\,\,\,\,\,OB \bot RP,\,\,\,\,\,OC \bot PQ \hfill \\\\& \Rightarrow \quad  \vec a \cdot \left( {\vec r - \vec q} \right) = \vec b \cdot \left( {\vec p - \vec r} \right) = \vec c \cdot \left( {\vec q - \vec p} \right) = 0\qquad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align} \]

We now proceed as follows. We draw the perpendiculars from P, Q to BC, CA respectively (not shown in the figure) and assume that these perpendiculars meet in \(Z\left( {\vec z} \right).\)

If we find  \(Z\left( {\vec z} \right).\) and show that RZ is perpendicular to AB, the concurrency will be established.

Since, by assumption,

\[\begin{align}&\qquad\quad\overrightarrow {PZ}  \bot \overrightarrow {BC} \,\,and\,\,\overrightarrow {QZ}  \bot \overrightarrow {CA}  \hfill \\\\&\Rightarrow \quad  \left( {\vec p - \vec z} \right) \cdot \left( {\vec c - \vec b} \right) = 0\,\,and\,\,\left( {\vec q - \vec z} \right) \cdot \left( {\vec a - \vec c} \right) = 0 \hfill \\\\
  & \Rightarrow \quad  \left\{ {\vec p \cdot \left( {\vec c - \vec b} \right) + \vec q \cdot \left( {\vec a - \vec c} \right)} \right\} = \vec z \cdot \left\{ {\left( {\vec c - \vec b} \right) + \left( {\vec a - \vec c} \right)} \right\} \hfill \\ \end{align} \]

The left hand side can be modified using (1), and thus we obtain,

\[\begin{align}& \left\{ { - \vec p \cdot \vec b + \vec q \cdot \vec a + \left( {\vec p - \vec q} \right) \cdot \vec c} \right\} = \vec z \cdot \left( {\vec a - \vec b} \right) \hfill \\\\ &\Rightarrow \quad  \vec q \cdot \vec a - \vec p \cdot \vec b = \vec z \cdot \left( {\vec a - \vec b} \right) \hfill \\\\& \Rightarrow \quad  \left( {\vec a - \vec b} \right) \cdot \vec r = \vec z \cdot \left( {\vec a - \vec b} \right)\qquad\qquad\qquad\left( {Again{\text{ }}using{\text{ }}\left( 1 \right)} \right) \hfill \\\\& \Rightarrow \quad \left( {\vec a - \vec b} \right) \cdot \left( {\vec r - \vec z} \right) = 0 \hfill \\\\&\Rightarrow \quad \vec a - \vec b\,\,is{\text{ }}perpendicular{\text{ }}to\,\,\vec r - \vec z. \hfill \\\\& \Rightarrow  \quad AB\,\,is{\text{ }}perpendicular{\text{ }}to\,\,RZ. \hfill \\ \end{align} \]

This establishes the concurrency of the three perpendiculars.

Example – 2

For any two vectors \(\vec u\) and \(\vec v\), prove that

\[\left( {1 + {{\left| {\vec u} \right|}^2}} \right)\left( {1 + {{\left| {\vec v} \right|}^2}} \right) = {\left( {1 - \vec u \cdot \vec v} \right)^2} + {\left| {\vec u + \vec v + \left( {\vec u \times \vec v} \right)} \right|^2}\]

Solution: It should be apparent that we should start with expanding the right hand side. To expand the second term in the RHS, we use the relation \({\left| {\vec a + \vec b} \right|^2} = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + 2\left( {\vec a \cdot \vec b} \right)\)


\[\begin{align}&RHS = 1 + {\left( {\vec u \cdot \vec v} \right)^2} - 2\vec u \cdot \vec v + \left\{ {\vec u + \vec v + \left( {\vec u \times \vec v} \right)} \right\} \cdot \left\{ {\vec u + \vec v + \left( {\vec u \times \vec v} \right)} \right\} \hfill \\\\&\qquad\;= 1 + {\left( {\vec u \cdot \vec v} \right)^2} - 2\vec u \cdot \vec v + {\left| {\vec u} \right|^2} + {\left| {\vec v} \right|^2} + 2\vec u \cdot \vec v + {\left| {\vec u \times \vec v} \right|^2} \hfill \\\\&\qquad\;   = 1 + {\left| {\vec u} \right|^2} + {\left| {\vec v} \right|^2} + \underline {{{\left( {\vec u \cdot \vec v} \right)}^2} + {{\left( {\vec u \times \vec v} \right)}^2}}  \hfill \\ \end{align} \]

Assuming the angle between  \(\vec u\,\,and\,\,\vec v\) to be \(\theta ,\) the underlined expression can be written simply as \({\left| {\vec u} \right|^2}{\left| {\vec v} \right|^2}{\cos ^2}\theta  + {\left| {\vec u} \right|^2}{\left| {\vec v} \right|^2}{\sin ^2}\theta  = {\left| {\vec u} \right|^2}{\left| {\vec v} \right|^2}.\) Thus,

\[\begin{align}& RHS = 1 + {\left| {\vec u} \right|^2} + {\left| {\vec v} \right|^2} + {\left| {\vec u} \right|^2}{\left| {\vec v} \right|^2} \hfill \\\\& \qquad\;= \left( {1 + {{\left| {\vec u} \right|}^2}} \right)\left( {1 + {{\left| {\vec v} \right|}^2}} \right) \hfill \\ \end{align} \]

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