Three consecutive vertices of parallelogram ABCD : A(-3,2,λ)…

Three consecutive vertices of parallelogram $ABCD$: $A(-3,2,\lambda)$, $B(3,-3,1)$, $C(5,\lambda,2)$. $1^\circ$ Find the fourth vertex $D$. $2^\circ$ Find $\lambda$ such that $|\overrightarrow{AD}|=\sqrt{14}$. $3^\circ$ For the larger value of $\lambda$ found in $2^\circ$, find the linear dependence among $\overrightarrow{AD}$, $\overrightarrow{BD}$ and $\overrightarrow{AC}$, and decompose $\overrightarrow{AC}$ in terms of $\overrightarrow{AD}$ and $\overrightarrow{BD}$.

Step 1: Part $1^\circ$: the fourth vertex

\[ \overrightarrow{BC} = (5-3,\ \lambda+3,\ 2-1) = (2,\ \lambda+3,\ 1) \]

\[ D = A + \overrightarrow{BC} = (-3+2,\ 2+\lambda+3,\ \lambda+1) \]

\[D(-1,\;\lambda+5,\;\lambda+1)\]

Check: $\overrightarrow{AB} = (6,-5,1-\lambda) = \overrightarrow{DC}$ $\checkmark$

Step 2: Part $2^\circ$: finding $\lambda$

\[ \overrightarrow{AD} = (2,\ \lambda+3,\ 1), \qquad |\overrightarrow{AD}|^2 = (\lambda+3)^2 + 5 = 14 \]

\[ (\lambda+3)^2 = 9 \;\Longrightarrow\; \lambda + 3 = \pm 3 \]

\[\lambda_1 = 0, \qquad \lambda_2 = -6\]

The larger value is $\lambda = 0$, giving $A(-3,2,0)$, $B(3,-3,1)$, $C(5,0,2)$, $D(-1,5,1)$. (For $\lambda=-6$: $D(-1,-1,-5)$, $\overrightarrow{AD}=2\vec i-3\vec j+\vec k$, also of length $\sqrt{14}$.)

Step 3: Part $3^\circ$: linear dependence ($\lambda = 0$)

\[ \overrightarrow{AD} = (2,3,1), \qquad \overrightarrow{BD} = (-4,8,0), \qquad \overrightarrow{AC} = (8,-2,2) \]

We seek $\alpha,\beta,\gamma$, not all zero, with $\alpha\overrightarrow{AD}+\beta\overrightarrow{BD}+\gamma\overrightarrow{AC}=\vec 0$: \[\begin{cases} 2\alpha - 4\beta + 8\gamma = 0 & (i)\\ 3\alpha + 8\beta - 2\gamma = 0 & (ii)\\ \alpha + 2\gamma = 0 & (iii) \end{cases}\]

From $(iii)$: $\alpha = -2\gamma$. Then $(i)$: $-4\gamma - 4\beta + 8\gamma = 0$, so $\beta = \gamma$. Check $(ii)$: $-6\gamma+8\gamma-2\gamma = 0$ $\checkmark$. With $\gamma = 1$: $\alpha=-2,\ \beta=1$.

\[ -2\,\overrightarrow{AD} + \overrightarrow{BD} + \overrightarrow{AC} = \vec{0} \]

\[\overrightarrow{AC} = 2\,\overrightarrow{AD} - \overrightarrow{BD}\]

Verification

\[ 2(2,3,1) - (-4,8,0) = (8,-2,2) = \overrightarrow{AC} \;\checkmark \]

Why it must hold

All three vectors lie in the plane of the parallelogram, so they are linearly dependent. Indeed $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$ and $\overrightarrow{AB} = \overrightarrow{AD} - \overrightarrow{BD}$, hence $\overrightarrow{AC} = 2\overrightarrow{AD}-\overrightarrow{BD}$ in every parallelogram.