Question

If the vectors $p\hat i + \hat j + \hat k$,$\hat i + q\hat j + \hat k$ and \hat i + \hat j + r\hat k$$$$\left( {p \ne q \ne r \ne 1} \right) are coplanar, then the value of $pqr - \left( {p + q + r} \right)$ is:
(1) $- 2$
(2) $2$
(3) $0$
(4) $- 1$

Answer 1

First we know a vector is a physical quantity which has both magnitude and direction. Examples are velocity, acceleration, force, etc. Then we have to know that the vectors are coplanar if the scalar triple product of three vectors is zero. Using the formula, find the value of $pqr - \left( {p + q + r} \right)$.

Complete step by step answer:
A scalar is a physical quantity which has only magnitude and does not have direction. Examples are mass, temperature, length. If there are three vectors in a $3d$-space and their scalar triple product is zero, then these three vectors are coplanar. Hence three vectors are coplanar then their scalar product is zero.
The scalar triple product of three vectors $\vec a$, $\vec b$, and $\vec c$ is $\left( {\vec a \times \vec b} \right).\vec c$. It is a scalar product because, just like the dot product, it evaluates to a single number.
If $\vec a = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k$, $\vec b = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k$ and $\vec c = {x_3}\hat i + {y_3}\hat j + {z_3}\hat k$ then

i&j;&k; \\\ {{x_1}}&{{y_1}}&{{z_1}} \\\ {{x_2}}&{{y_2}}&{{z_2}} \end{array}} \right| = ({y_1}{z_2} - {z_1}{y_2})\hat i - ({x_1}{z_2} - {z_1}{x_2})\hat j + ({x_1}{y_2} - {y_1}{x_2})\hat k$$$$\left( {\vec a \times \vec b} \right).\vec c = {x_3}({y_1}{z_2} - {z_1}{y_2}) - {y_3}({x_1}{z_2} - {z_1}{x_2}) + {z_3}({x_1}{y_2} - {y_1}{x_2}) = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}} \\\ {{x_2}}&{{y_2}}&{{z_2}} \\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right|$$ Hence $$\left( {\vec a \times \vec b} \right).\vec c = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}} \\\ {{x_2}}&{{y_2}}&{{z_2}} \\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right|$$. Given the vectors $$p\hat i + \hat j + \hat k$$,$$\hat i + q\hat j + \hat k$$and $$\hat i + \hat j + r\hat k$$$$\left( {p \ne q \ne r \ne 1} \right)$$ are coplanar. Then $$\left| {\begin{array}{*{20}{c}} p&1&1 \\\ 1&q;&1 \\\ 1&1&r; \end{array}} \right| = 0$$ $$ \Rightarrow p\left( {qr - 1} \right) - \left( {r - 1} \right) + 1 - q = 0$$ $$ \Rightarrow pqr - p - r + 1 + 1 - q = 0$$ $$ \Rightarrow pqr - \left( {p + r + q} \right) = - 2$$ Hence, the Option (1) is correct. **So, the correct answer is “Option A”.** **Note:** Note that Coplanar vectors are the vectors which lie on the same plane, in a three-dimensional space. These are vectors which are parallel to the same plane. Simply the three vectors are also coplanar if the vectors are in three-dimensional space and are linearly independent. If more than two vectors are linearly independent; then all the vectors are coplanar.