H2 Maths Tuition: Foot of Perpendicular (from point to plane) (Part II)

This is a continuation from H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I).

Foot of Perpendicular (from point to plane)

From point (B) to Plane ( $p$ )

Equation (I):

Where does F lie?

F lies on the plane $p$ .

$\overrightarrow{\mathit{OF}}\cdot \mathbf{n}=d$

Equation (II):

Perpendicular

$\overrightarrow{\mathit{BF}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}=k\mathbf{n}+\overrightarrow{OB}$

Final Step

Substitute Equation (II) into Equation (I) and solve for k.

Example

[VJC 2010 P1Q8i]

The planes $\Pi _{1}$ and $\Pi _{2}$ have equations $\mathbf{r\cdot(i+j-k)}=6$ and $\mathbf{r\cdot(2i-4j+k)}=-12$ respectively. The point $A$ has position vector $\mathbf{{9i-7j+5k}}$ .

(i) Find the position vector of the foot of perpendicular from $A$ to $\Pi _{2}$ .

Solution

Let the foot of perpendicular be F.

Equation (I)

$\overrightarrow{\mathit{OF}}\cdot \left(\begin{matrix}2\\-4\\1\end{matrix}\right)=-12$

Equation (II)

$\overrightarrow{\mathit{OF}}=k\left(\begin{matrix}2\\-4\\1\end{matrix}\right)+\left(\begin{matrix}9\\-7\\5\end{matrix}\right)=\left(\begin{matrix}2k+9\\-4k-7\\k+5\end{matrix}\right)$

Subst. (II) into (I)

$2(2k+9)-4(-4k-7)+(k+5)=-12$

Solve for k, $k=-3$ .

$\overrightarrow{\mathit{OF}}=\left(\begin{matrix}3\\5\\2\end{matrix}\right)$

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One Response to H2 Maths Tuition: Foot of Perpendicular (from point to plane) (Part II)

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