Distance from a Point to a Hyperplane: Exercises

Problem 1

Find the distance from the point \(Q\) to the given hyperplane in \(\RNrSpc{2}\).

\(Q(5,3)\), and \(\Pi\) the solutions of the equation \(2x-y=3\)
- Hint Solution Answer
\(Q(-3,-1)\), and \(\Pi\) the solutions of the equation \(x = 3y\)
- Solution Answer

Problem 2

Find the distance from the point \(Q\) to the given hyperplane of \(\RNrSpc{3}\).

\(Q(3,1,-4)\) and \(\Pi\) the solutions of the equation \(2x-y-z=1\).
- Solution Answer
\(Q(0,5,2)\) and \(\Pi\) the solutions of the equation \(y=5\).
- Answer
\(Q(1,-3,-2)\) and \(\Pi\) the solutions of the equation \(y=-z\).
- Answer

Problem 3

In \(\RNrSpc{2}\), find the point \(R\) on the line \(L\) of solutions of the equation \(3x + 2y = -2\) which is closest to the point \(Q(2,2)\).

Hint Solution Answer

Problem 4

In \(\RNrSpc{3}\), find the point \(R\) on the plane \(H\) of solutions of the equation \(4x + 3z = 3\) which is closest to the point \(Q(1,1,1)\).

Answer

Problem 5

Verify that the given two lines in \(\RNrSpc{2}\) are parallel. Then find the distance between them.

\[ \begin{array}{rrcrcl} L_1: &\quad x & - & 3y & = & 0 \\ L_2: &\quad2x & - & 6y & = & 1 \end{array} \]

Hint Solution Answer

Problem 6

Verify that the given two lines in \(\RNrSpc{3}\) are parallel. Then find the distance between them.

\[ \begin{array}{rrcrcrccl} H_1: &\quad x & + & y & - & 2z & = & 0 \\ H_2: &\quad3x & + & 3y & - & 6z & = & 1 \end{array} \]

Hint Hint Hint Hint Hint Hint Answer

Solution for Problem 1 a

Recall that the distance from \(Q\) to \(\Pi\) is

\(\Dstnc{Q}{\Pi}\)

\(=\)

\(\dfrac{ \Abs{ \DotPr{ (\Vect{q} - \Vect{p}) }{ \Vect{a} } } }{ \Norm{ \Vect{a} } }\)

where

\(\Vect{q}\) is the position vector of the point \(Q\). We know that \(\Vect{q}= (5,3)\)
\(\Vect{p}\) is the position vector of a point on \(\Pi\). We are free to use any point we want. \(\Vect{p}\DefEq (0,-3)\) is a convenient choice.
\(\Vect{a}\) is a normal vector to \(\Pi\). The equation for the given line directly offers \(\Vect{a} \DefEq (2,-1)\) as a possible choice.

With this setup we obtain

\(\Dstnc{Q}{\Pi}\)	\(=\)	\(\dfrac{ \Abs{ \DotPr{ ((5,3)-(0,-3)) }{ (2,-1) } } }{ \Abs{ (2,-1) } }\)
\(\)	\(=\)	\(\dfrac{ \Abs{ 4 } }{ \Abs{ \sqrt{5} } }\)
\(\)	\(\)	\(\dfrac{ 4 }{ \sqrt{5} }\)

So the distance from \(Q\) to \(\Pi\) is \(4/\sqrt{5}\).

Solution for Problem 1 b

There are two ways in which we can arrive at an answer to this problem:

Approach 1: the lucky check Substituting the coordinates of \(Q\) into the equation for \(\Pi\) shows

\(-3\)

\(=\)

\(3\cdot (-1)\)

So \(Q\) lies on \(\Pi\). Therefore the distance from \(Q\) to \(\Pi\) is \(0\).

Approach 2: we just use the distance formula In this case we find

\(\Vect{q}=(-3,-1)\) is the position vector of the point \(Q\).
\(\Vect{p}\DefEq(3,1)\) is a point on \(\Pi\).
\(\Vect{a}\DefEq(-1,3)\) is a normal vector for \(\Pi\); found by rewriting the equation for \(\Pi\) as
\(-x + 3y\) \(=\) \(0\)

Substituting these data into the distance formula gives

\(\Dstnc{Q}{\Pi}\)	\(=\)	\(\dfrac{ \Abs{ \DotPr{ ((-3,-1)-(3,1)) }{ (-1,3) } } }{ \Abs{ (-1,3) } }\)
\(\)	\(=\)	\(\dfrac{ \Abs{ \DotPr{ (-6,-2) }{ (-1,3) } } }{ \Abs{ (-1,3) } }\)
\(\)	\(=\)	\(\dfrac{ \Abs{ 0 } }{ \Abs{ \sqrt{10} } }\)
\(\)	\(\)	\(0\)

This confirms the result obtained with ‘lucky check’ approach described above.

Solution for Problem 5

The two lines are seen to be parallel if and only if their normal vectors are parallel. As normal vectors we find

\(\Vect{a}\)	\(\DefEq\)	\((1,-3)\)
\(\Vect{b}\)	\(\DefEq\)	\((2,-6) = 2\cdot (1,-3)\)

This means that the lines have parallel normal vectors. So the lines are parallel. To find the distance between the two lines, we pick a point \(Q\) on \(L_1\), and compute the distance from \(Q\) to \(L_2\): \(Q(3,1)\) with position vector \(\Vect{q}=(3,1)\) lies on \(L_1\). A point on \(L_2\) is \(P(\tfrac{1}{2},0)\). Now the formula for the distance from \(Q\) to \(L_2\) yields

\(\Dstnc{q}{L_2}\)	\(=\)	\(\dfrac{ \Abs{ \DotPr{ ((3,1) - (\tfrac{1}{2},0)) }{ (1,-3) } } }{ \Abs{ (1,-3) } }\)
\(\)	\(=\)	\(\dfrac{ \Abs{ -\tfrac{1}{2} } }{ \sqrt{10} }\)
\(\)	\(=\)	\(\dfrac{ 1 }{ 2\sqrt{10} }\)

\(\Dstnc{Q}{\Pi}\)	\(=\)	\(\dfrac{ \Abs{ \DotPr{ ((3,1,-4)-(0,0,-1)) }{ (2,-1,-1) } } }{ \Norm{ (2,-1,-1) } }\)
\(\)	\(=\)	\(\dfrac{ \Abs{ 8 } }{ \Abs{ \sqrt{6} } }\)
\(\)	\(\)	\(\dfrac{8}{\sqrt{6}}\)

\(\Vect{r}\)	\(=\)	\((2,2) - \dfrac{ \DotPr{ \left((2,2) - (0,-1)\right) }{ (3,2) } }{ \DotPr{ (3,2) }{ (3,2) }} \cdot (3,2)\)
\(\)	\(=\)	\((2,2)\ -\ \dfrac{12}{13}\cdot (3,2)\)
\(\)	\(=\)	\(\dfrac{1}{13}(-10,2)\)