Question

Show that the vectors $\overrightarrow{a,}\overrightarrow{b,}\overrightarrow{c}$given by $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=2\hat{i}+\hat{j}+3\hat{k}\mathrm{and}\overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}$ are non-coplanar. Express vector $\overrightarrow{d}=2\hat{i}-\hat{j}-3\hat{k}$ as a linear combination of the vectors $\overrightarrow{a,}\overrightarrow{b}\mathrm{and}\overrightarrow{c}.$

Answer 1

Let the given vectors $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=2\hat{i}+\hat{j}+3\hat{k}$ and $\overrightarrow{c}=\hat{i}+\dot{\hat{j}+\dot{\hat{k}}}$ are coplanar. Then one of the vector is expressible as a linear combination of the other two. Let,
$$\begin{gathered} \hat{i}+2\hat{j}+3\hat{k}=x\left(2\hat{i}+\hat{j}+3\hat{k}\right)+y\left(\hat{i}+\hat{j}+\hat{k}\right). \\ =\hat{i}\left(2x+y\right)+\hat{j}\left(x+y\right)+\hat{k}\left(3x+y\right). \end{gathered}$$
$\Rightarrow 2x+y=1,x+y=2,3x+y=3.$
On solving the first two equations we get $x=-1,y=3$. Clearly the values of $x,y$ does not satisfy the third equation.
Hence the given vectors are non-coplanar.
Now, $\overrightarrow{d}=2\hat{i}-\hat{j}-3\hat{k}$ which can be expressed as
$2\hat{i}-\hat{j}-3\hat{k}=x(\hat{i}+2\hat{j}+3\hat{k})+y(2\hat{i}+\hat{j}+3\hat{k})+z(\hat{i}+\hat{j}+\hat{k}).$
= $\hat{i}(x+2y+z)+\hat{j}(2x+y+z)+\hat{k}(3x+3y+z)$.

$$\begin{gathered} \Rightarrow x+2y+z=2,2x+y+z=-1,3x+3y+z=-3. \\ \Rightarrow x=-\frac{8}{3},y=\frac{1}{3},z=4 \end{gathered}$$
Hence $\overrightarrow{d}$ is expressible as the linear combination of $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$.