Question

To find: The parametric equations for the line through the point \((0,1,2)\), and perpendicular to the line \(x = 1 + t,y = 1 - t,z = 2t\), and intersects the line \(x = 1 + t,y = 1 - t,z = 2t\).

Short answer

The line through the point's\((0,1,2)\) parametric equations, perpendicular to the line \(x = 1 + t,y = 1 - t,z = 2t\)and intersects the line \(x = 1 + t,y = 1 - t,z = 2t\)are\(x = - 3t,y = 1 + t,z = 2 + 2t\).

Step-by-step solution

Concept of parametric Equations and Formula’s for intersection planes & angles.

The components of the vector equation are the parametric equations of the line, which have the form\({\bf{x}}{\rm{ }} = {\rm{ }}{\bf{x}}{\rm{ }} + {\rm{ }}{\bf{at}},{\rm{ }}{\bf{y}}{\rm{ }} = {\rm{ }}{\bf{y0}}{\rm{ }} + {\rm{ }}{\bf{bt}},\)and\({\bf{z}}{\rm{ }} = {\rm{ }}{\bf{z0}}{\rm{ }} + {\rm{ }}{\bf{ct}}\). The direction numbers of the line are the components\(a,b\)and\(c\).

Formulas

A line passing through a point\(\left( {{x_0},{y_0},{z_0}} \right)\)and parallel to the direction vector\(\langle a,b,c\rangle \)has a parametric equation.

\(x = {x_0} + at,y = {y_0} + bt,z = {z_0} + ct\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(1)\)

Let’s take the expression to find direction vector of required line (v).

\(v = b - \left( {\frac{{a - b}}{{|a{|^2}}}} \right)a\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......(2)\)

Substituting the points.

Here,

\(a\)is the direction vector of perpendicular line \(x = 1 + t,y = 1 - t,z = 2t\)and

\(b\)is the vector joined from the point on the required line to the point on the perpendicular line.

Let’s take the expression of perpendicular line.

\(x = 1 + t,y = 1 - t,z = 2t\)

We can rewrite the expression as follows.

\(x = 1 + t,y = 1 + ( - 1t),z = 0 + 2t\,\,\,\,\,\,\,\,\,\,\,\,\,......(3)\)

Consider the value of scalar parameter \(t\) as \(0\) and obtain the point on the perpendicular line.

We can substitute \(0\) for \(t\) in equation (3) and obtain the point on the perpendicular line.

\(\begin{array}{l}x = 1 + 0,y = 1 + ( - 1)(0),z = 0 + 2(0)\\x = 1,y = 1,z = 0\end{array}\)

The point on the perpendicular line is \((1,1,0)\).

Compare equation (\(3\)) with equation (\(1\)) and obtain the direction vector of perpendicular line (\(a\)).

\(\langle 1, - 1,2\rangle = i - j + 2k\)

Let’s take the expression to find direction vector joined from the point \(P\left( {{x_1},{y_1},{z_1}} \right)\) to \(Q\left( {{x_2},{y_2},{z_2}} \right)\).

\(v = \left\langle {\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,.....(4)\)

The vector \(b\) is the direction vector joined from the points \((1,1,0)\) and \((0,1,2)\).

Direction vector (v) for required line with the help of formula.

Calculation of direction vector b:

We can substitute \(1\) for \({x_1},1\)for\({y_1},0\)for\({z_1},0\)for\({x_2},1\)for\({y_2}\), and \(2\) for \({z_2}\) in equation (4),

\(\begin{array}{l}b = \langle (0 - 1),(1 - 1),(2 - 0)\rangle \\b = \langle - 1,0,2\rangle \\b = - i + 0j + 2k\end{array}\)

Direction vector for required equation

We can substitute \(\left( { - i + 0{\rm{ }}j + 2{\rm{ }}k} \right)\)for \(b\) and \((i - j + 2k)\) for \(a\) in equation (\(2\)),

\(\begin{array}{c}v = ( - i + 0j + 2k) - \left( {\frac{{(i - j + 2k) \cdot ( - i + 0j + 2k)}}{{|(i - j + 2k){|^2}}}} \right)(i - j + 2k)\\v = ( - i + 0j + 2k) - \left( {\begin{array}{*{20}{l}}{(i \cdot ( - i)) + (i - 0j) + (i \cdot 2k) + (( - j) \cdot ( - i)) + (( - j) \cdot - 0j)}\\{ + (( - j) \cdot 2k) + (2k \cdot ( - i)) + (2k \cdot - j) + (2k - 2k)}\\{{{\left( {\sqrt {{1^2} + {{( - 1)}^2} + {2^2}} } \right)}^2}}\end{array}} \right)(i - j + 2k)\\v = ( - i + 0j + 2k) - \left( {\frac{{ - 1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 4}}{{{{(\sqrt 6 )}^2}}}} \right)(i - j + 2k)\\v = ( - i + 0j + 2k) - \left( {\frac{3}{6}} \right)(i - j + 2k)\end{array}\)

Simplify the expression.

\(\begin{array}{c}v = ( - i + 0j + 2k) - \left( {\frac{1}{2}} \right)(i - j + 2k)\\v = - \frac{3}{2}i + \frac{1}{2}j + k\\v = \left\langle { - \frac{3}{2},\frac{1}{2},1} \right\rangle \end{array}\)

The direction vector of required line is\(\langle - 3,1,2\rangle \).

Required line passes through the point\(\langle - 3,1,2\rangle \).

Identify parametric equations through the points.

We can substitute \(0\) for \({x_0},1\)for\({y_0},2\)for\({z_0}, - 3\)for\(a,1\)for\(b\), and \(2\) for \(c\) in equation (\(1\)),

\(\begin{array}{l}x = 0 + ( - 3)t,y = 1 + (1)t,z = 2 + (2)t\\x = - 3t,y = 1 + t,z = 2 + 2t\end{array}\)

Thus, the parametric equations for the line through the point\((0,1,2)\), perpendicular to the line \(x = 1 + t,y = 1 - t,z = 2t\) and intersects the line \(x = 1 + t,y = 1 - t,z = 2t\) are\(x = - 3t,y = 1 + t,z = 2 + 2t\).