Examples On Vector Dot Product Set-2

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Example – 19

Let AD, BE and CF be the medians in \(\Delta ABC\) . Prove that

\[\overrightarrow {BC}  \cdot \overrightarrow {AD}  + \overrightarrow {CA}  \cdot \overrightarrow {BE}  + \overrightarrow {AB}  \cdot \overrightarrow {CF}  = 0\]

Solutions:

Since D is the mid-point of BC, we have

\[\begin{align}&\quad\qquad\quad \;  D \equiv \left( {\frac{{\vec b + \vec c}}{2}} \right) \hfill \\\\&
   \Rightarrow  \qquad \overrightarrow {AD}  = ({\text{Position}}\;\,{\text{vector}}\;\,{\text{of}}\;\vec D) - ({\text{Position}}\;\,{\text{vector}}\;\,{\text{of}}\;\vec A) \hfill \\\\&\quad\qquad\qquad    = \frac{{\vec b + \vec c}}{2} - \vec a \hfill \\\\& \quad\qquad\qquad  = \frac{1}{2}(\vec b + \vec c - 2\vec a) \hfill \\ \end{align} \]

Thus,

\[\begin{align} &\overrightarrow {BC}  \cdot \overrightarrow {AD}  = \frac{1}{2}(\vec c - \vec b) \cdot (\vec b + \vec c - 2\vec a) \hfill \\\\ &\qquad\qquad= \frac{1}{2}\left( {{{\left| {\vec c} \right|}^2} - {{\left| {\vec b} \right|}^2} + 2\vec a \cdot (\vec b - \vec c)} \right)\quad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align} \]

Similarly,

\[\begin{align}&\overrightarrow {CA}  \cdot \overrightarrow {BE}  = \frac{1}{2}\left( {{{\left| {\vec a} \right|}^2} - {{\left| {\vec c} \right|}^2} + 2\vec b \cdot (\vec c - \vec a)} \right)\qquad\qquad...\left( 2 \right) \hfill \\\\& \overrightarrow {AB}  \cdot \overrightarrow {CF}  = \frac{1}{2}\left( {{{\left| {\vec b} \right|}^2} - {{\left| {\vec a} \right|}^2} + 2\vec c \cdot (\vec a - \vec b)} \right)\qquad\qquad...\left( 3 \right) \hfill \\ \end{align} \]

It is now immediately apparent that the right hand sides (1), (2) and (3) sum to zero. Thus, the stated assertion is true.

Example – 20

Prove that the altitudes in a triangle are concurrent.

Solution: Assume the three vertices of the triangle to be A, B and C.

Assume H to be the origin  \(\vec 0\) and A, B, C to have the position vectors  \(\vec a,\vec b,\vec c,\) .

Since  \(AH \bot BC,\) we have

\[\begin{align}& \qquad\;\vec a \cdot (\vec b - \vec c) = 0 \hfill \\\\&  \Rightarrow \quad   \vec a \cdot \vec b = \vec a \cdot \vec c\qquad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align} \]

Similarly, since \(BH \bot AC\) ,

\[\begin{align} &\qquad\;\; \vec b \cdot (\vec c - \vec a) = 0 \hfill \\\\ &  \Rightarrow \quad   \vec a \cdot \vec b = \vec b \cdot \vec c\qquad\qquad\qquad...\left( 2 \right) \hfill \\ \end{align} \]

From (1) and (2), we have

\[\begin{align}& \qquad\;\;\vec a \cdot \vec c = \vec b \cdot \vec c \hfill \\\\& \Rightarrow \quad  (\vec a - \vec b) \cdot \vec c = 0 \hfill \\\\&\Rightarrow \quad CH \bot AB \hfill \\ \end{align} \]

Thus, the altitude through C passes through H, implying that the three altitudes are concurrent.

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grade 11 | Answers Set 1
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grade 11 | Questions Set 2
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