Vectors 5 Question 15
15. If vectors $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are coplanar, then show that
$$ \left|\begin{array}{ccc} \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{c}} \\ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} \\ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} \end{array}\right|=\overrightarrow{\mathbf{0}} $$
(1989, 2M)
Show Answer
Solution:
- Given that, $\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}$ are coplanar vectors.
$\therefore$ There exists scalars $x, y, z$ not all zero, such that
$$ x \overrightarrow{\mathbf{a}}+y \overrightarrow{\mathbf{b}}+z \overrightarrow{\mathbf{c}}=0 $$
Taking dot with $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ respectively, we get
$$ \begin{aligned} & x(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}})+y(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})+z(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}})=0 \\ & \text { and } \quad x(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})+y(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}})+z(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}})=0 \end{aligned} $$
Since, Eqs. (i), (ii) and (iii) represent homogeneous equations with $(x, y, z) \neq(0,0,0)$.
$\Rightarrow$ Non-trivial solutions
$\therefore \quad \Delta=0 \Rightarrow\left|\begin{array}{ccc}\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{c}} \ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} \ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}\end{array}\right|=\overrightarrow{\mathbf{0}}$
