Consider a pyramid OPQRS located in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) with O as origin, and OP

Q: Wait, why can't I just look at the diagram and guess the angle between $OQ$ and $OS$ is $\frac{\pi}{3}$?

Visual estimation is a trap in 3D geometry! Our eyes are easily deceived by perspective. Always rely on the dot product: $\cos \theta = \frac{\vec{OQ} \cdot \vec{OS}}{|\vec{OQ}| |\vec{OS}|}$. The math never lies, and in this case, it proves the angle is not $\frac{\pi}{3}$.

Q: Why do we use the cross product to find the plane equation? Can't we just use three points?

You absolutely can! You could use the determinant form. However, the cross product $\vec{n} = \vec{v_1} \times \vec{v_2}$ is often faster and less prone to arithmetic errors. It's a powerful tool to keep in your arsenal for any plane problem.

Q: Is there a trick to the distance from $P$ to the plane $x-y=0$?

Yes! Since the plane is $x-y=0$, the distance is simply $\frac{|x_P - y_P|}{\sqrt{1^2 + (-1)^2}}$. Since $P$ is $(3,0,0)$, it's just $\frac{|3-0|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$. Recognizing the simplicity of the plane equation saves you precious time in the exam hall.

Question

Consider a pyramid $OPQRS$ located in the first octant ($x \ge 0, y \ge 0, z \ge 0$) with $O$ as origin, and $OP$ and $OR$ along the x-axis and the y-axis, respectively. The base $OPQR$ of the pyramid is a square with $OP = 3$. The point $S$ is directly above the mid-point, $T$ of diagonal $OQ$ such that $TS = 3$. Then

Options:

(A) the acute angle between $OQ$ and $OS$ is $\frac{\pi}{3}$

(B) the equation of the plane containing the triangle $OQS$ is $x - y = 0$

(C) the length of the perpendicular from $P$ to the plane containing the triangle $OQS$ is $\frac{3}{\sqrt{2}}$

(D) the perpendicular distance from $O$ to the straight line containing $RS$ is $\sqrt{\frac{15}{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Foundation

We begin by establishing the coordinates of the square base $OPQR$ in the $xy$-plane. Given $O$ is at $(0,0,0)$ and $P$ is at $(3,0,0)$, the square's symmetry dictates the positions of the remaining vertices:
$R = (0,3,0)$
$Q = (3,3,0)$

The apex $S$ is located directly above the midpoint $T$ of the diagonal $OQ$. The midpoint $T$ is calculated as:
$$T = \left( \frac{0+3}{2}, \frac{0+3}{2}, 0 ight) = (1.5, 1.5, 0)$$

Given the height $TS = 3$, the coordinates of the apex $S$ are $(1.5, 1.5, 3)$.

### Calculating the Angle between $OQ$ and $OS$

To find the angle $	heta$ between vectors $\vec{OQ}$ and $\vec{OS}$, we define:
$\vec{OQ} = 3\hat{i} + 3\hat{j}$
$\vec{OS} = 1.5\hat{i} + 1.5\hat{j} + 3\hat{k}$

Using the dot product formula $\cos 	heta = \frac{\vec{OQ} \cdot \vec{OS}}{|\vec{OQ}| |\vec{OS}|}$, we compute the components:
$\vec{OQ} \cdot \vec{OS} = (3)(1.5) + (3)(1.5) + (0)(3) = 4.5 + 4.5 = 9$
$|\vec{OQ}| = \sqrt{3^2 + 3^2} = 3\sqrt{2}$
$|\vec{OS}| = \sqrt{1.5^2 + 1.5^2 + 3^2} = \sqrt{2.25 + 2.25 + 9} = \sqrt{13.5} = \frac{3\sqrt{6}}{2}$

Substituting these into the formula:
$$\cos 	heta = \frac{9}{(3\sqrt{2})(\frac{3\sqrt{6}}{2})} = \frac{9}{\frac{9\sqrt{12}}{2}} = \frac{2}{\sqrt{12}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

### The Plane Containing $	riangle OQS$

To find the equation of the plane containing $	riangle OQS$, we determine the normal vector $\vec{n}$ via the cross product $\vec{OQ} 	imes \vec{OS}$:
$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 3 & 0 \ 1.5 & 1.5 & 3 \end{vmatrix} = 9\hat{i} - 9\hat{j} + 0\hat{k}$$

Normalizing this vector, we obtain the plane equation $x - y = 0$. The distance $d$ from point $P(3,0,0)$ to this plane is:
$$d = \frac{|3 - 0|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$$

### Distance from Origin to Line $RS$

The line $RS$ passes through $R(0,3,0)$ with direction vector $\vec{d} = S - R = (1.5, -1.5, 3)$. For simplicity, we scale this to $\vec{d} = (1, -1, 2)$.

The distance $L$ from the origin $O(0,0,0)$ to the line is given by:
$$L = \frac{|\vec{OR} 	imes \vec{d}|}{|\vec{d}|}$$

Calculating the cross product $\vec{OR} 	imes \vec{d}$:
$$\vec{OR} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 3 & 0 \ 1 & -1 & 2 \end{vmatrix} = 6\hat{i} - 0\hat{j} - 3\hat{k}$$

The magnitude of the cross product is $\sqrt{6^2 + (-3)^2} = \sqrt{45}$, and the magnitude of $\vec{d}$ is $\sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}$. The final distance is:
$$L = \sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}}$$

Select Answer:

Visualized Solution

Coordinate Setup & Base

Coordinates of $Q$ and $T$

Coordinates of Apex $S$

Vectors $O Q$ and $O S$

Angle Between $O Q$ and $O S$

Normal to Plane $O QS$

Equation of Plane $O QS$

Distance from $P$ to Plane $O QS$

Direction of Line $R S$

Distance from $O$ to Line $R S$

Final Calculation for Option 4

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Foundation

Calculating the Angle between $O Q$ and $O S$

The Plane Containing $△ O QS$

Distance from Origin to Line $R S$

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Select Answer:

Visualized Solution

Coordinate Setup & Base

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Coordinates of Apex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OS

Angle Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OS

Normal to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQS

Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQS

Distance from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQS

Direction of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS

Distance from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O to Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS