Check whether vectors a=2i-3j , b=i+2j , c=i-j are coplanar

Check whether vectors $\vec{a}=2\vec{i}-3\vec{j}$, $\vec{b}=\vec{i}+2\vec{j}$, $\vec{c}=\vec{i}-\vec{j}$ are coplanar.

Step 1: Criterion

By Problem 21, three vectors are coplanar if and only if one of them is a linear combination of the other two. We check whether $\vec{c} = \alpha\vec{a}+\beta\vec{b}$ has a solution.

Step 2: Set up the system

\[ \vec{i}-\vec{j} = \alpha(2\vec{i}-3\vec{j}) + \beta(\vec{i}+2\vec{j}) = (2\alpha+\beta)\vec{i} + (-3\alpha+2\beta)\vec{j} \]

\[ \begin{cases} 2\alpha + \beta = 1 \\ -3\alpha + 2\beta = -1 \end{cases} \]

Step 3: Solve the system

From the first equation $\beta = 1 - 2\alpha$. Substituting into the second: \[-3\alpha + 2(1-2\alpha) = -1 \;\Longrightarrow\; -7\alpha = -3 \;\Longrightarrow\; \alpha = \frac{3}{7}, \qquad \beta = 1 - \frac{6}{7} = \frac{1}{7}.\]

Step 4: Verification

\[ \frac{3}{7}(2\vec{i}-3\vec{j}) + \frac{1}{7}(\vec{i}+2\vec{j}) = \frac{1}{7}(7\vec{i}-7\vec{j}) = \vec{i}-\vec{j} = \vec{c} \;\checkmark \]

Step 5: Conclusion

\[\vec{c} = \frac{3}{7}\,\vec{a} + \frac{1}{7}\,\vec{b}\]

The vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar. This is also clear directly: none of them has a $\vec{k}$ component, so all three lie in the $xy$-plane.

Step 6: Determinant check

Any vectors of the form $x\vec i+y\vec j$ lie in one plane, so the real content of the question is whether $\vec a$ and $\vec b$ form a basis of that plane: \[\det\begin{pmatrix} 2 & 1 \\ -3 & 2 \end{pmatrix} = (2)(2)-(-3)(1) = 7 \neq 0.\]

Since the determinant is non-zero, $\vec{a}$ and $\vec{b}$ are non-collinear, and every vector of the plane (in particular $\vec{c}$) is uniquely expressible through them. $\checkmark$

Step 7: Graphical construction