H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I)

Foot of Perpendicular is a hot topic for H2 Prelims and A Levels. It comes out almost every year.

There are two versions of Foot of Perpendicular, from point to line, and from point to plane. However, the two are highly similar, and the following article will teach how to understand and remember them.

H2: Vectors (Foot of perpendicular)

From point (B) to Line ( l)

(Picture)

h2-vectors-tuition

Equation (I):

Where does F lie? F lies on the line  l.

\overrightarrow{\mathit{OF}}=\mathbf{a}+\lambda  \mathbf{m}

Equation (II):

Perpendicular:

\overrightarrow{\mathit{BF}}\cdot \mathbf{m}=0

(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}})\cdot  \mathbf{m}=0

Final Step

Substitute Equation (I) into Equation (II) and solve for  \lambda .

Example:

[CJC 2010 P1Q7iii]

Relative to the origin O , the points A , B and C  have position vectors  \left(\begin{matrix}1\\2\\1\end{matrix}\right) , \left(\begin{matrix}2\\1\\3\end{matrix}\right) and \left(\begin{matrix}-1\\2\\3\end{matrix}\right) Find the shortest distance from  C to \mathit{AB} . Hence or otherwise, find the area of triangle \mathit{ABC} .

[Note: There is a 2nd method to this question. (cross product method)]

Solution:

Let the foot of perpendicular from C to AB be F.

Equation (I):

\overrightarrow{\mathit{OF}}=\overrightarrow{\mathit{OA}}+\lambda  \overrightarrow{\mathit{AB}}=\left(\begin{matrix}1+\lambda \\2-\lambda  \\1+2\lambda \end{matrix}\right)

Equation (II):

(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}})\cdot  \overrightarrow{\mathit{AB}}=0

\left(\begin{matrix}2+\lambda \\-\lambda \\-2+2\lambda  \end{matrix}\right)\cdot  \left(\begin{matrix}1\\-1\\2\end{matrix}\right)=0

\lambda =\frac{1}{3}

\overrightarrow{\mathit{CF}}=\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}}=\left(\begin{matrix}2\frac{1}{3}\\-{\frac{1}{3}}\\-1\frac{1}{3}\end{matrix}\right)

\left|{\overrightarrow{{\mathit{CF}}}}\right|=\sqrt{\frac{22}{3}}

Area of  \Delta  \mathit{ABC}=\frac{1}{2}\left|{\overrightarrow{\mathit{AB}}}\right|\left|{\overrightarrow{\mathit{CF}}}\right|=\sqrt{11}

For the next part, please read our article on Foot of Perpendicular (from point to plane).

H2 Maths Tuition

If you are looking for Maths Tuition, contact Mr Wu at:

SMS: 98348087

Email: mathtuition88@gmail.com

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged , , , . Bookmark the permalink.

One Response to H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I)

  1. Pingback: Foot of perpendicular from point to plane | Singapore Maths Tuition

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s