First slide

Planes in 3D

Question

Given four points,4(2, 1,0), B(1,0, 1),C(3,0, 1)and D(0,0,2).Point D lies on a line L orthogonal to the plane determined by the points A, B and C.

Moderate

Question

The equation of the Plane ABC is

A

x+y+z+3=0
B

y+z-1=0
C

x+z-1=0
D

2y+z-1=0

Solution

$\begin{array}{l} |\begin{array}{llc} x - 2 & y - 1 & z \\ 1 - 2 & 0 - 1 & 1 - 0 \\ 3 - 2 & 0 - 1 & 1 - 0 \end{array}| = 0 \\ (x - 2) [(- 1) - (- 1)] - (y - 1) [(- 1) - 1] + z [1 + 1] = 0 \end{array}$
$\begin{array}{l} or 2 (y - 1) + 2 z = 0 \\ or y + z - 1 = 0 \end{array}$
The vector normal to the plane is $\vec{r} = 0 \hat{i} + \hat{j} + \hat{k}$ The equation of the line through (0, 0, 2) and parallel to $\vec{n} is \vec{r} = 2 \hat{k} + λ (\hat{j} + \hat{k})$
The perpendicular distance of D(0, 0, 2) from plane $ABC is |\frac{2 - 1}{\sqrt{1^{2} + 1^{2}}}| = \frac{1}{\sqrt{2}} .$

Question

The equation of the line I is

A

$\vec{r} = 2 \hat{k} + λ (\hat{i} + \hat{k})$
B

$\vec{r} = 2 \hat{k} + λ (2 \hat{j} + \hat{k})$
C

$\vec{r} = 2 \hat{k} + λ (\hat{j} + \hat{k})$
D

None

Solution

$\begin{array}{l} |\begin{array}{llc} x - 2 & y - 1 & z \\ 1 - 2 & 0 - 1 & 1 - 0 \\ 3 - 2 & 0 - 1 & 1 - 0 \end{array}| = 0 \\ (x - 2) [(- 1) - (- 1)] - (y - 1) [(- 1) - 1] + z [1 + 1] = 0 \end{array}$
$\begin{array}{l} or 2 (y - 1) + 2 z = 0 \\ or y + z - 1 = 0 \end{array}$
The vector normal to the plane is $\vec{r} = 0 \hat{i} + \hat{j} + \hat{k}$ The equation of the line through (0, 0, 2) and parallel to $\vec{n} is \vec{r} = 2 \hat{k} + λ (\hat{j} + \hat{k})$
The perpendicular distance of D(0, 0, 2) from plane $ABC is |\frac{2 - 1}{\sqrt{1^{2} + 1^{2}}}| = \frac{1}{\sqrt{2}} .$

Question

The perpendicular distance of D from the plane ABC is

Solution

$\begin{array}{l} |\begin{array}{llc} x - 2 & y - 1 & z \\ 1 - 2 & 0 - 1 & 1 - 0 \\ 3 - 2 & 0 - 1 & 1 - 0 \end{array}| = 0 \\ (x - 2) [(- 1) - (- 1)] - (y - 1) [(- 1) - 1] + z [1 + 1] = 0 \end{array}$
$\begin{array}{l} or 2 (y - 1) + 2 z = 0 \\ or y + z - 1 = 0 \end{array}$
The vector normal to the plane is $\vec{r} = 0 \hat{i} + \hat{j} + \hat{k}$ The equation of the line through (0, 0, 2) and parallel to $\vec{n} is \vec{r} = 2 \hat{k} + λ (\hat{j} + \hat{k})$
The perpendicular distance of D(0, 0, 2) from plane $ABC is |\frac{2 - 1}{\sqrt{1^{2} + 1^{2}}}| = \frac{1}{\sqrt{2}} .$

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