Question

In Exercises 21–24, a, b, and c are non-collinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \) and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

22. Let p be a point on the line through a and b. Show that\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

Short answer

It is proved that \(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).

Step-by-step solution

State the equation of line

If a point p lies on the line passing through the points a and b, then the equation of line is\({\bf{p}} = \left( {1 - t} \right){\bf{a}} + t{\bf{b}}\), where\(t\)is aparameter.

Let\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right]\)and\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\)be the points.

Then, the equation of line can be represented as shown below:

\(\begin{array}{c}{\bf{p}} = \left( {1 - t} \right)\left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\left( {1 - t} \right){a_1} + t{b_1}}\\{\left( {1 - t} \right){a_2} + t{b_2}}\end{array}} \right]\end{array}\)

Thus,\({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right]\),\({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right]\), and \({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{\left( {1 - t} \right){a_1} + t{b_1}}\\{\left( {1 - t} \right){a_2} + t{b_2}}\end{array}} \right]\).

Then,\(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right]\)can be written as shown below:

\(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = \det \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{\left( {1 - t} \right){a_1} + t{b_1}}\\{{a_2}}&{{b_2}}&{\left( {1 - t} \right){a_2} + t{b_2}}\\1&1&1\end{array}} \right]\)

Find the determinant of the vectors

Add\( - 1\)times column 1 to column 2 to get column 2. Then, add\( - 1\)times column 1 to column 3 to get column 3 as shown below:

\(\begin{array}{c}\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1} - {a_1}}&{ - t{a_1} + t{b_1}}\\{{a_2}}&{{b_2} - {a_2}}&{ - t{a_2} + t{b_2}}\\1&0&0\end{array}} \right|\\ = 1\left[ {\left( {{b_1} - {a_1}} \right)t\left( {{b_2} - {a_2}} \right) - \left( {{b_2} - {a_2}} \right)t\left( {{b_1} - {a_1}} \right)} \right]\\ = 0\end{array}\)

Thus, \(\det \left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{p}} }\end{array}} \right] = 0\).