Let $E - ABCD$ be a pyramid on square base $ABCD$ where $A$ is the origin and $B$ and $D$ are lying on positive $x$-axis and $y$-axis respectively. If $E$ is $(0, 2, 3)$ and $\overrightarrow{DE} \cdot (\hat{i} + \hat{j}) = 0$, then:
image of the point $D$ in the plane $ABE$ is $\left(0, \dfrac{-10}{13}, \dfrac{24}{13}\right)$
image of the point $D$ in the plane $ABE$ is $\left(0, \dfrac{-6}{13}, \dfrac{30}{13}\right)$
volume of the tetrahedron $ABDE$ is 2 cubic units
perpendicular distance of the point $D$ from the plane $ABE$ is $\dfrac{9}{\sqrt{13}}$
Step-by-Step Solution
Step 1: Set up coordinates. $A$ is origin $(0,0,0)$. $B$ is on positive $x$-axis: $B = (a, 0, 0)$. $D$ is on positive $y$-axis: $D = (0, d, 0)$. Since $ABCD$ is a square base, $AB = AD$, so $a = d$. Let $B = (a,0,0)$ and $D = (0,a,0)$.
Step 2: Use the condition $\overrightarrow{DE} \cdot (\hat{i}+\hat{j}) = 0$. $\overrightarrow{DE} = E - D = (0-0, 2-a, 3-0) = (0, 2-a, 3)$.
$$\overrightarrow{DE}\cdot(\hat{i}+\hat{j}) = 0 + (2-a) = 0 \Rightarrow a = 2$$
So $D = (0,2,0)$, $B = (2,0,0)$.
Step 3: Find the equation of plane $ABE$. Points: $A=(0,0,0)$, $B=(2,0,0)$, $E=(0,2,3)$.
$$\overrightarrow{AB} = (2,0,0),\quad \overrightarrow{AE} = (0,2,3)$$
$$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AE} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&0&0\\0&2&3\end{vmatrix} = \hat{i}(0\cdot3-0\cdot2)-\hat{j}(2\cdot3-0\cdot0)+\hat{k}(2\cdot2-0\cdot0)$$
$$= (0,-6,4)$$
Plane $ABE$: $0(x-0) - 6(y-0) + 4(z-0) = 0 \Rightarrow -6y + 4z = 0 \Rightarrow 3y - 2z = 0$
Step 4: Find perpendicular distance of $D=(0,2,0)$ from plane $3y-2z=0$:
$$d = \frac{|3(2)-2(0)|}{\sqrt{9+4}} = \frac{6}{\sqrt{13}}$$
Option 4 claims $\frac{9}{\sqrt{13}}$ — this is FALSE.
Step 5: Find image of $D=(0,2,0)$ in plane $3y-2z=0$. The foot of perpendicular from $D$ to plane:
$$\frac{x-0}{0} = \frac{y-2}{3} = \frac{z-0}{-2} = -\frac{3(2)-2(0)}{9+4} = -\frac{6}{13}$$
Foot: $x=0$, $y = 2 + 3\cdot(-6/13) = 2 - 18/13 = 8/13$, $z = 0 + (-2)(-6/13) = 12/13$.
Image: $x=0$, $y = 2(8/13)-2 = 16/13-26/13 = -10/13$, $z = 2(12/13)-0 = 24/13$.
Image $= \left(0, -\frac{10}{13}, \frac{24}{13}\right)$. Option 1 is TRUE.
Step 6: Compute volume of tetrahedron $ABDE$. $A=(0,0,0)$, $B=(2,0,0)$, $D=(0,2,0)$, $E=(0,2,3)$.
$$V = \frac{1}{6}|\overrightarrow{AB}\cdot(\overrightarrow{AD}\times\overrightarrow{AE})|$$
$$\overrightarrow{AB}=(2,0,0),\ \overrightarrow{AD}=(0,2,0),\ \overrightarrow{AE}=(0,2,3)$$
$$\overrightarrow{AD}\times\overrightarrow{AE} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&2&0\\0&2&3\end{vmatrix} = (6-0, 0-0, 0-0) = (6,0,0)$$
$$V = \frac{1}{6}|(2,0,0)\cdot(6,0,0)| = \frac{1}{6}|12| = 2$$
Option 3 is TRUE.
Correct Answer: 1, 3