If \(\vec a = \hat i - \hat j + \hat k\),\(\vec b = \hat i + 2\hat j - \hat k\) and \(\vec c = 3\hat i + p\hat j + 5\hat k\) are coplanar, then p=? [RPET \(1985,86,88,91\)].
A) \( - 6\).
B) \( - 2\)
C) \(2\)
D) \(6\)

Hint: In this question we have to use the concept of co-planarity. Three vectors are said to be coplanar when they all are present in the same plane or we can say that all those vectors which are parallel to the same plane are coplanar. In order to find whether vectors are coplanar or not we have to find a scalar triple product of three vectors. If the value of the scalar triple product is zero then we can say that three given vectors are coplanar.



Formula Used:Scalar triple product of vectors \( = \vec a.\left( {\vec b \times \vec c} \right)\)
Where \(\vec a,\vec b\)and \(\vec c\)are three given vectors.
\(\vec a.\left( {\vec b \times \vec c} \right) = \left| a \right|\left| {\left( {\vec b \times \vec c} \right)} \right|\cos \left( \theta \right)\)
Where is the angle between \(\vec a\)and \(\left( {\vec b \times \vec c} \right)\)

\(\vec a.\left( {\vec b \times \vec c} \right) = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\)
\( = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left\{ {\left( {{b_2}{c_3} - {b_3}{c_2}} \right)\hat i - \left( {{b_1}{c_3} - {b_3}{c_1}} \right)\hat j + \left( {{b_1}{c_2} - {b_2}{c_1}} \right)\hat k} \right\}\)
\( = {a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right)\)




Complete step by step solution:Given: Three vectors a, b and c are coplanar
I.e. \(\vec a.\left( {\vec b \times \vec c} \right) = 0\)

Where,
\(\vec a = \hat i - \hat j + \hat k\)
\(\vec b = \hat i + 2\hat j - \hat k\)
\(\vec c = 3\hat i + p\hat j + 5\hat k\)

\({a_1} = 1,\;{a_{2 = }} - 1,\;{a_3} = 1\)
\({b_1} = 1,\;{b_2} = 2,{b_3} = - 1\)
\({c_1} = 3,\;{c_2} = p,\;{c_3} = 5\)
\(\vec a.\left( {\vec b \times \vec c} \right) = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\)
\( = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left\{ {\left( {{b_2}{c_3} - {b_3}{c_2}} \right)\hat i - \left( {{b_1}{c_3} - {b_3}{c_1}} \right)\hat j + \left( {{b_1}{c_2} - {b_2}{c_1}} \right)\hat k} \right\}\)
\( = {a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right)\)
\(\vec a.\left( {\vec b \times \vec c} \right) = 0\)
\(1\left( {2 \times 5 - \left( { - 1} \right) \times p} \right) + 1\left( {1 \times 5 - \left( { - 1} \right) \times 3} \right) + 1\left( {1 \times p - 2 \times 3} \right) = 0\)
\(\left( {10 + p} \right) + \left( {5 + 3} \right) + \left( {p - 6} \right) = 0\)
\(10 + p + 8 + p - 6 = 0\)
\(2p + 12 = 0\)
\(2p = - 12\)
\(p = - {\bf{6}}\)



Option ‘A’ is correct

Note: Here in this question we have to find the value of p. In order to find the value of p we must know the concept of coplanarity. Scalar triple product formula is used to find the required value. Scalar triple product means product of three vectors i.e. dot product of one of the vectors with cross product of other two vectors.
Scalar triple products are represented as \[\left[ {a{\rm{ }}b{\rm{ }}c{\rm{ }}} \right].\]
The resultant scalar triple product is always scalar. If the scalar triple product is zero then we can say that three vectors are coplanar.
Scalar triple product may be zero, negative and positive.