A(2,6,2), B(-4,0,), C(2,3,-1), D(4,5,0), ||5, are vertices of quadrilateral ABCD.

Question

$A(2,6,2)$, $B(-4,0,\lambda)$, $C(2,3,-1)$, $D(4,5,0)$, $|\lambda|\le5$, are vertices of quadrilateral $ABCD$. If its area is 18 sq. units, then $5-6\lambda$ is:

Answer: 11

Accepted Answer

Area $=\dfrac12|\vec{AC}\times\vec{BD}| = 18$.

$\vec{AC}=(0,-3,-3)$, $\vec{BD}=(8,5,-\lambda)$:
$$\vec{AC}\times\vec{BD} = \begin{vmatrix}\hat i&\hat j&\hat k\0&-3&-3\8&5&-\lambda\end{vmatrix} = (3\lambda+15)\hat i+24\hat j-24\hat k$$

$$|\vec{AC}\times\vec{BD}|=36 \implies (3\lambda+15)^2+576+576=1296 \implies (3\lambda+15)^2=144$$
$$3\lambda+15=\pm12 \implies \lambda=-1\text{ or }-9$$

Since $|\lambda|\le5$: $\lambda=-1$. So $5-6\lambda = 5+6 = 11$.

Solutions · 1