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)

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

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4 thoughts on “H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I)”

1. Jessie says:

How do find the foot of perpendicular from a point C to the line AB, after having found the length of projection of AC tonto the line AB?

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1. You can try finding the length of projection in terms of lambda (see examples of lambda in the post). Then solve for lambda by equating the two expressions of length of projection.
Once you have lambda, substitute it back to the vector OF for finding the foot of perpendicular.

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