+0

# Cube question help.

+1
186
4

Let ABCDEFGH be a cube of side length 5, as shown. Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1. The plane through C, P, and Q intersects line DH at R. Find DR.

One of hints says "Let the plane through C, P, and Q intersect the extension of line DA at S. What can you say about the diagram? Which two-dimensional figures can you work with?" but I dont know what it means.

Guest Mar 27, 2017

#1
+4777
+4

Here I wrote in the info given in the problem and I drew point S according to the directions in the hint.

Let's just look at triangle DRS by itself:

If only we knew what SA was, we could find RD!

Well, let's look at triangle DCS by itself:

Ah, now we can find SA!

$$\frac{5}{5+SA}=\frac{2}{SA} \\~\\ (5)(SA)=10+(2)(SA) \\~\\ SA=\frac{10}{3}$$

Now looking back at triangle DRS:

$$\frac{DR}{5+\frac{10}{3}}=\frac{1}{\frac{10}{3}} \\~\\ DR*\frac{3}{25}=\frac{3}{10} \\~\\ DR=\frac{3}{10}*\frac{25}{3}=\frac{5}{2}$$

hectictar  Mar 27, 2017
Sort:

#1
+4777
+4

Here I wrote in the info given in the problem and I drew point S according to the directions in the hint.

Let's just look at triangle DRS by itself:

If only we knew what SA was, we could find RD!

Well, let's look at triangle DCS by itself:

Ah, now we can find SA!

$$\frac{5}{5+SA}=\frac{2}{SA} \\~\\ (5)(SA)=10+(2)(SA) \\~\\ SA=\frac{10}{3}$$

Now looking back at triangle DRS:

$$\frac{DR}{5+\frac{10}{3}}=\frac{1}{\frac{10}{3}} \\~\\ DR*\frac{3}{25}=\frac{3}{10} \\~\\ DR=\frac{3}{10}*\frac{25}{3}=\frac{5}{2}$$

hectictar  Mar 27, 2017
#2
+77125
+2

Let  B  = (0,0, 0)      Let C  = (0, 5, 0)     Let  P = ( 3, 0 ,0)    Let D  = (5,5, 0)    Let Q  = ( 5,,0 , 1)

Equation of CP     y = (-5/3)x + 5  →   [ y - 5] / (-5/3)   = x

Equation of AD  =   x  = 5

Y Intersection  of   CP, AD      [ y - 5 ] / (-5/3)   = 5  →   y - 5  = 5 (-5/3)   →  y =  -25/3 + 5   =   -10/3

So "S"   =   (5, -10/3, 0 )

So AS  = 10/3

Now, the plane apex intersects AD at S.....thus, SRD forms a triangle  such that  triangle SRD  ≈ SQA

And  QA/AS   ≈ RD / DS

1 / ( 10/3)   =  RD/ (5 + 10/3)

3/10  = RD / ( 25/3)

(3/10) (25/3)   = RD   = 25 / 10   = 2.5

Check  ...distance from S to R  ≈  8.7  = the hypotenuse of SRA

And  DR^2 + SD^2  =   sqrt [ (5 + 10/3)^2  + 2.5^2 ] ≈  8.7=  SR

CPhill  Mar 27, 2017
#3
+77125
+1

Hey.....hectictar  and I got the same answer....albeit, with slightly different approaches........his is a little better, IMHO, but two great minds couldn't be wrong.....LOL!!!!!

CPhill  Mar 27, 2017
#4
+18626
+3

Let ABCDEFGH be a cube of side length 5, as shown.

Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1.

The plane through C, P, and Q intersects line DH at R. Find DR.

$$Let\ \vec{C} = \langle 0,0,0 \rangle\\ Let\ \vec{P} = \langle 5,3,0 \rangle\\ Let\ \vec{Q} = \langle 5,5,1 \rangle\\ Let\ DR = z \\ Let\ \vec{R} = \langle 0,5,z \rangle$$

The plane through C, P, and Q:

$$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{C} + s \cdot(\vec{P}-\vec{C}) + t \cdot(\vec{Q}-\vec{C}) \quad & | \quad \vec{x} = \vec{R} \quad \vec{C} = \vec{0} \\ \vec{R} &=& \vec{0} + s \cdot(\vec{P}-\vec{0}) + t \cdot(\vec{Q}-\vec{0}) \\ \vec{R} &=& s \cdot \vec{P} + t \cdot \vec{Q} \quad & | \quad \vec{P} =\begin{pmatrix}5 \\ 3 \\ 0\end{pmatrix} \quad \vec{Q}=\begin{pmatrix}5 \\ 5 \\ 1\end{pmatrix} \quad \vec{R}=\begin{pmatrix}0 \\ 5 \\ z\end{pmatrix} \\ \begin{pmatrix}0 \\ 5 \\ z\end{pmatrix} &=& s \cdot \begin{pmatrix}5 \\ 3 \\ 0\end{pmatrix} + t \cdot \begin{pmatrix}5 \\ 5 \\ 1\end{pmatrix} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1) & 0&=& 5s+5t \\ & 0&=& s+ t \\ & s&=& -t \\\\ (2) & 5&=& 3s+5t \\ & 5&=& 3(-t) + 5t \\ & 5&=& 2t \\ & t&=& \frac{5}{2} \\\\ (3) & z&=& 0\cdot s + t \\ & z&=& t \\ & z&=& \frac{5}{2} \\ \hline \end{array}$$

DR = $$\frac{5}{2}$$

heureka  Mar 28, 2017

### 19 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details