Plane Equations & Normals - Calculus 3

1. A plane passes through the point $A(1, 2, -3)$ and has normal vector $\vec{n} = \langle 2, -1, 4 \rangle$. If point $B(x, y, z)$ lies on this plane and the vector $\overrightarrow{AB}$ has the property that $\overrightarrow{AB} \cdot \vec{n} = 15$, which of the following must be true?
   Answer: Point $B$ cannot exist because the conditions are contradictory for any plane The equation of the plane through $A(1, 2, -3)$ with normal $\vec{n} = \langle 2, -1, 4 \rangle$ is $2(x-1) - (y-2) + 4(z+3) = 0$, or $2x - y + 4z + 8 = 0$. If point $B(x, y, z)$ lies on this plane, then $\overrightarrow{AB} = \langle x-1, y-2, z+3 \rangle$ must be perpendicular to the normal vector $\vec{n}$, giving $\overrightarrow{AB} \cdot \vec{n} = 0$. However, we're told that $\overrightarrow{AB} \cdot \vec{n} = 15 \neq 0$. This means $B$ cannot lie on the original plane. The two conditions are contradictory: if $B$ is on the plane, then $\overrightarrow{AB} \perp \vec{n}$, but we need $\overrightarrow{AB} \cdot \vec{n} = 15$. Choice A is wrong because the conditions cannot be satisfied simultaneously. Choice C is wrong because we're not looking for a parallel plane. Choice D misses the point that both conditions cannot hold.
2. Consider two planes $\Pi_1: x + 2y - z = 5$ and $\Pi_2: 2x - y + 3z = 1$. A point $P$ moves along the line of intersection of these planes. If $P$ is equidistant from the points $A(1, 0, 2)$ and $B(3, -1, 0)$, find the $z$-coordinate of point $P$.
   Answer: $z = \frac{11}{3}$ First, we find the parametric equations for the line of intersection. The direction vector is $\vec{v} = \langle 1, 2, -1 \rangle \times \langle 2, -1, 3 \rangle = \langle 5, -5, -5 \rangle$ or simplified as $\langle 1, -1, -1 \rangle$. To find a point on the line, set $z = 0$ in both plane equations: $x + 2y = 5$ and $2x - y = 1$. From the second equation, $y = 2x - 1$. Substituting: $x + 2(2x - 1) = 5$, so $5x = 7$, giving $x = \frac{7}{5}$ and $y = \frac{9}{5}$. The line has parametric equations: $\langle x, y, z \rangle = \langle \frac{7}{5} + t, \frac{9}{5} - t, -t \rangle$. For $P$ to be equidistant from $A$ and $B$, we need $|PA|^2 = |PB|^2$. After expanding and simplifying the distance formulas, this condition gives us $t = -\frac{11}{3}$. Therefore, $z = -t = \frac{11}{3}$.

Start this quiz