Question

Find the value of \[p\] if the vectors \[i - 2j + k,2i - 5j + pk\] and \[5i - 9j + 4k\] are coplanar.

Answer 1

Hint: When all the vectors lie in the same plane then we can say that they are coplanar. If the scalar triple product of three vectors is zero, if the three vectors are linearly independent then the vectors are said to be coplanar. The scalar triple product of the vectors \[x = ({a_1},{a_2},{a_3}),y = ({b_1},{b_2},{b_3})\] and \[z = ({c_1},{c_2},{c_3})\] is given by
 \[[x,y,z] = \left| {\begin{array}{*{20}{c}}
  {{c_1}}&{{c_2}}&{{c_3}} \\
  {{a_1}}&{{a_2}}&{{a_3}} \\
  {{b_1}}&{{b_2}}&{{b_3}}
\end{array}} \right|\] .

Complete step by step answer:
The given vectors are \[i - 2j + k,2i - 5j + pk\] and \[5i - 9j + 4k\] .
Our aim is to find the value of \[p\] . Also given that these three vectors are coplanar.
We know that if the vectors are coplanar then their scalar triple product is zero.
Let us first name these vectors, \[x = i - 2j + k,y = 2i - 5j + pk\] and \[z = 5i - 9j + 4k\] .
That is, \[x = (1, - 2,1)\] , \[y = (2, - 5,p)\] and \[z = (5, - 9,4)\] .
Therefore, \[[x,y,z] = 0\] .
 \[ \Rightarrow [x,y,z] = \left| {\begin{array}{*{20}{c}}
  5&{ - 9}&4 \\
  1&{ - 2}&1 \\
  2&{ - 5}&p
\end{array}} \right| = 0\]
Let us find the determinant of this matrix.
 \[\left| {\begin{array}{*{20}{c}}
  5&{ - 9}&4 \\
  1&{ - 2}&1 \\
  2&{ - 5}&p
\end{array}} \right| = 5[( - 2 \times p) - ( - 5 \times 1)] - ( - 9)[(1 \times p) - (2 \times 1)] + 4[(1 \times - 5) - (2 \times - 2)] = 0\]
 \[ \Rightarrow 5[( - 2p) - ( - 5)] - ( - 9)[(p) - (2)] + 4[( - 5) - ( - 4)] = 0\]
Simplifying this we will get,
 \[ \Rightarrow 5[ - 2p + 5] + 9[p - 2] + 4[ - 5 + 4] = 0\]
On further simplification we get,
 \[ \Rightarrow - 10p + 25 + 9p - 18 - 20 + 16 = 0\]
Simplifying this we will get,
 \[ \Rightarrow - p + 3 = 0\]
Thus, we get \[p = 3\] .
Let’s check whether the value of \[p\] is correct or not. Substitute the value of \[p\] in the scalar triple product and check whether we get zero.
 \[ \Rightarrow [x,y,z] = \left| {\begin{array}{*{20}{c}}
  5&{ - 9}&4 \\
  1&{ - 2}&1 \\
  2&{ - 5}&3
\end{array}} \right|\]
Find the determinant of this matrix.
 \[\left| {\begin{array}{*{20}{c}}
  5&{ - 9}&4 \\
  1&{ - 2}&1 \\
  2&{ - 5}&3
\end{array}} \right| = 5[( - 2 \times 3) - ( - 5 \times 1)] - ( - 9)[(1 \times 3) - (2 \times 1)] + 4[(1 \times - 5) - (2 \times - 2)]\]
On simplifying this we get,
 \[ \Rightarrow 5[ - 6 - ( - 5)] + 9[3 - 2] + 4[ - 5 - ( - 4)]\]
On further simplification, we get
 \[ \Rightarrow 5[ - 6 + 5] + 9[1] + 4[ - 5 + 4]\]
After simplifying this further we get,
 \[ \Rightarrow 5[ - 1] + 9 + 4[ - 1]\]
Simplify it further,
 \[
   \Rightarrow - 5 + 9 - 4 \\
   \Rightarrow 0 \\
 \]
Thus, the value of \[p = 3\] is correct.

Note: The scalar triple product of vectors is also known as box product. If \[x,y,z\]are the vectors then the scalar triple product of these vectors is denoted by, \[(x \times y).z\] or\[[x,y,z]\]. The difference between the scalar and the vector: Scalar is nothing but the size or magnitude but the vector will have both the direction and the size.