Question

Find ‘p’ if the vectors \[\vec i + p\vec j - 3\vec k,2\vec i + \vec j - 4\vec k\]and \[\vec i - \vec j + \vec k\] are coplanar.

Answer 1

Hint: Coplanar generally refers to points and lines which lie on the same geometric plane on a three-dimensional plane. If the scalar triple product of any three vectors is zero then the vectors are said to be coplanar. Before checking whether vectors are coplanar, check whether given vectors are linearly dependent or not. The vectors are linearly dependent when there is at least one non-trivial combination of vectors equal to zero vector, which refers to the variables or terms not equal to zero. Vectors are the line segment having the length and a definite direction, having a beginning point and ending point.

Complete step by step answer:
\[\vec i + p\vec j - 3\vec k = [1{\text{ p - 3]}}\]
\[2\vec i + \vec j - 4\vec k = [2{\text{ 1 - 4]}}\]
\[\vec i - \vec j + \vec k = [1{\text{ - 1 1]}}\]
Here, \[[abc] = \left| {\left[ {\begin{array}{*{20}{c}}
  1&p&{ - 3} \\
  2&1&{ - 4} \\
  1&{ - 1}&1
\end{array}} \right]} \right| = 0\]
Now, expand the given matrix as:
\[
  1\left| {\begin{array}{*{20}{c}}
  1&{ - 4} \\
  { - 1}&1
\end{array}} \right| - p\left| {\begin{array}{*{20}{c}}
  2&{ - 4} \\
  1&1
\end{array}} \right| + \left( { - 3} \right)\left| {\begin{array}{*{20}{c}}
  2&1 \\
  1&{ - 1}
\end{array}} \right| = 0 \\
  1(1 - 4) - p(2 + 4) - 3( - 2 - 1) = 0 \\
  1( - 3) - p(6) - 3( - 3) = 0 \\
   - 3 - 6p + 9 = 0 \\
  6p = 6 \\
  p = 1 \\
 \]
Hence the value of p=1 means that when we put the value of p in the above vector, then all vectors will lie on the same geometric planes that will make them coplanar.

Additional information: Substitute the value $p = 1$ in the given vector,
\[
  1\left| {\begin{array}{*{20}{c}}
  1&{ - 4} \\
  { - 1}&1
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
  2&{ - 4} \\
  1&1
\end{array}} \right| + \left( { - 3} \right)\left| {\begin{array}{*{20}{c}}
  2&1 \\
  1&{ - 1}
\end{array}} \right| = 0 \\
  1(1 - 4) - 1(2 + 4) - 3( - 2 - 1) = 0 \\
  1( - 3) - (6) - 3( - 3) = 0 \\
   - 3 - 6 + 9 = 0 \\
  0 = 0 \\
 \]

We can see that the scalar triple product is zero. Hence, vectors are coplanar which indicates that the determined value of p is correct.

Note: In general, before checking whether the vectors are coplanar, we check whether the 3-vectors are linearly dependent being non-trivial. In the case of n numbers of vectors, we check for not more than two vectors are linearly dependent. The vectors which are parallel on the same plane or lie on the same plane are coplanar vectors.