Question

If $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ are non-zero, non-coplanar vectors, prove that the following vectors are coplanar:
(i) $5\overrightarrow{a}+6\overrightarrow{b}+7\overrightarrow{c,}7\overrightarrow{a}-8\overrightarrow{b}+9\overrightarrow{c}\mathrm{and}3\overrightarrow{a}+20\overrightarrow{b}+5\overrightarrow{c}$

(ii) $\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c},-3\overrightarrow{b}+5\overrightarrow{c}\mathrm{and}-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$

Answer 1

(i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let
$$\begin{gathered} 5\overrightarrow{a}+6\overrightarrow{b}+7\overrightarrow{c}=x\left(7\overrightarrow{a}-8\overrightarrow{b}+9\stackrel{\rightharpoonup }{c}\right)+y\left(3\overrightarrow{a}+20\overrightarrow{b}+5\overrightarrow{c}\right). \\ =\overrightarrow{a}\left(7x+3y\right)+\overrightarrow{b}\left(-8x+20y\right)+\overrightarrow{c}\left(9x+5y\right). \end{gathered}$$
$\Rightarrow 7x+3y=5,-8x+20y=6\mathrm{and}9x+5y=7.$
Solving first two of these equations, we get $x=\frac{1}{2},y=\frac{1}{2}$. Clearly, these values of x and y satisfies the third equation.
Hence, the given vectors are coplanar.

(ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let
$$\begin{gathered} \overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}=x(-3\overrightarrow{b}+5\overrightarrow{c})+y(-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}). \\ =\overrightarrow{a}(-2y)+\overrightarrow{b}(-3x+3y)+\overrightarrow{c}(5x-4y). \end{gathered}$$
$\Rightarrow -2y=1,-3x+3y=-2\mathrm{and}5x-4y=3$
Solving first two of these equations, we get $x=\frac{1}{6},y=-\frac{1}{2}$.
These values of x and y does not satisfy the third equation.
Hence, the given vectors are not coplanar.