0
$\begingroup$

The coordinates of $A$ and $B$ are $(3, -2, 4)$ and $(6, 0, 3)$ respectively.

The line $l_1$ has equation $$r=\begin{bmatrix} 3\\ -2\\ 4\\ \end{bmatrix} +\lambda \begin{bmatrix} 2\\ -1\\ 3\\ \end{bmatrix} $$

The point $D$ lies on $l_1$ where $\lambda=2$. The line $l_2$ passes through $D$ and is parallel to $AB$.

The diagram shows a symmetrical trapezium $ABCD$, with angle $DAB$ equal to angle $ABC$.

enter image description here

The point $C$ lies on the line $l_2$. The length of $AD$ is equal to the length $BC$.

Find the coordinates of $C$.

"$D$ lies on $l_1$ where $\lambda=2$" therefore,

\begin{align} D & = \begin{bmatrix} 3\\ -2\\ 4\\ \end{bmatrix} +2 \begin{bmatrix} 2\\ -1\\ 3\\ \end{bmatrix} \\ \Rightarrow D & = \begin{bmatrix} 7\\ -4\\ 10\\ \end{bmatrix} \\ \end{align}

"$l_2$ is parallel to $AB$"

\begin{align} \vec{AB} & = \begin{bmatrix} 6\\ 0\\ 3\\ \end{bmatrix} - \begin{bmatrix} 3\\ -2\\ 4\\ \end{bmatrix} \\ \Rightarrow \vec{AB} & = \begin{bmatrix} 3\\ 2\\ -1\\ \end{bmatrix} \\ \end{align}

"The line $l_2$ passes through $D$" therefore the line $l_2$ has equation,

$$r=\begin{bmatrix} 7\\ -4\\ 10\\ \end{bmatrix} +\mu \begin{bmatrix} 3\\ 2\\ -1\\ \end{bmatrix} $$

"$C$ lies on the line $l_2$" therefore,

$$C=\begin{bmatrix} 7+3\mu\\ -4+2\mu\\ 10-\mu\\ \end{bmatrix} $$

One way I tried to find the coordinates of $C$ was using the information that "the length of $AD$ is equal to the length $BC$",

\begin{align} \vec{AD} & = \begin{bmatrix} 7\\ -4\\ 10\\ \end{bmatrix} - \begin{bmatrix} 3\\ -2\\ 4\\ \end{bmatrix} = \begin{bmatrix} 4\\ -2\\ 6\\ \end{bmatrix} \\ \Rightarrow AD & = \sqrt{(4)^2+(-2)^2+(6)^2} = \sqrt{56} \\ \end{align}

\begin{align} \vec{BC} & = \begin{bmatrix} 7+3\mu\\ -4+2\mu\\ 10-\mu\\ \end{bmatrix} - \begin{bmatrix} 6\\ 0\\ 3\\ \end{bmatrix} = \begin{bmatrix} 1+3\mu\\ -4+2\mu\\ 7-\mu\\ \end{bmatrix} \\ \Rightarrow BC & = \sqrt{(1+3\mu)^2+(-4+2\mu)^2+(7-\mu)^2} \\ \end{align}

\begin{align} \sqrt{(1+3\mu)^2+(-4+2\mu)^2+(7-\mu)^2} & = \sqrt{56} \\ (7\mu-5)(\mu-1) & = 0 \\ \end{align}

I do not know which value of $\mu$ gives $C$

So then I thought I would use the information "with angle $DAB$ equal to angle $ABC$" to find $C$,

\begin{align} DAB \Rightarrow \cos\theta & = \frac{(3\cdot2)+(2\cdot-1)+(-1\cdot3)}{\sqrt{(3)^2+(2)^2+(-1)^2}\cdot\sqrt{(2)^2+(-1)^2+(3)^2}} \\ & = \frac{1}{14} \\ \end{align}

\begin{align} ABC \Rightarrow \cos\theta & = \frac{(3\cdot(1+3\mu))+(2\cdot(-4+2\mu))+(-1\cdot(7-\mu))}{\sqrt{(3)^2+(2)^2+(-1)^2}\cdot\sqrt{(4)^2+(-2)^2+(6)^2}} \\ & = \frac{-12+14\mu}{28} \\ \end{align}

\begin{align} \frac{-12+14\mu}{28} & = \frac{1}{14} \\ \mu & = 1 \\ \end{align}

Answer in the Mark Scheme $$\mu=\frac{5}{7}$$

$\endgroup$

2 Answers 2

1
$\begingroup$

Somewhat along the lines you chose, there are some details you don't need to work out. We are told that $ \ \vec{AD} \ = \ \langle 4, \ -2 \ , \ 6 \rangle \ $ and you have worked out that $ \ \vec{AB} \ = \ \langle 3, \ 2, \ -1 \rangle \ $ . So the cosine of the included angle is $$ \ \frac{\vec{AD} \ \cdot \ \vec{AB}}{\vert \ \vec{AD} \ \vert \ \vert \ \vec{AB} \ \vert} \ = \ \frac{2}{\vert \ \vec{AD} \ \vert \ \vert \ \vec{AB} \ \vert} \ \ . $$

You found $ \ D \ ( 7, \ -4 , \ 10) \ $ , so $ \ C \ $ is somewhere along line $ \ l_2 \ $ at $ \ ( 7 + 3 \ \kappa , \ -4 + 2 \ \kappa , \ 10 - \kappa ) \ $ . Hence,

$$ \vec{BC} \ = \ \langle 7 + 3 \ \kappa - 6 , \ -4 + 2 \ \kappa - 0 , \ 10 - \kappa - 3 \rangle \ = \ \langle 1 + 3 \ \kappa , \ -4 + 2 \ \kappa , \ 7 - \kappa \rangle \ \ . $$

The included angle between $ \ \vec{BA} \ $ and $ \ \vec{BC} \ $ is the same as the one above, and we are told (although we can find easily enough for this symmetrical trapezium that it is true) that $ \ \vert \ \vec{BC} \ \vert \ = \ \vert \ \vec{AD} \ \vert \ $ . So the cosine of this included angle is

$$ \ \frac{\vec{BC} \ \cdot \ \vec{BA}}{\vert \ \vec{BC} \ \vert \ \vert \ \vec{AB} \ \vert} \ \ = \ \frac{\vec{AD} \ \cdot \ \vec{AB}}{\vert \ \vec{AD} \ \vert \ \vert \ \vec{AB} \ \vert} \ = \ \frac{2}{\vert \ \vec{AD} \ \vert \ \vert \ \vec{AB} \ \vert} \ \ , $$

from which we can infer that
$$ \vec{BC} \ \cdot \ \vec{BA} \ = \ 2 \ = \ \langle -3, \ -2, \ 1 \rangle \ \cdot \ \langle 1 + 3 \ \kappa , \ -4 + 2 \ \kappa , \ 7 - \kappa \rangle $$

$$ \Rightarrow \ \ -3 \ - \ 9 \ \kappa \ + \ 8 \ - 4 \ \kappa \ + \ 7 \ - \kappa \ = \ 2 \ \ \Rightarrow \ \ \kappa \ = \ \frac{-10}{-14} \ = \ \frac{5}{7} \ \ . $$

This places $ \ C \ $ at

$$ \ ( 7 + 3 \ \left[\frac{5}{7} \right] , \ -4 + 2 \ \left[\frac{5}{7} \right] , \ 10 - \left[\frac{5}{7} \right] ) \ = \ ( \frac{64}{7} , \ -\frac{18}{7} , \ \frac{65}{7} ) \ \ . $$

EDIT -- I think I've tracked down the point where your calculation goes astray near the end: when you compute $ \ \cos \theta \ $ for $ \ \angle ABC \ $ , you need to reverse the orientation of $ \ \vec{AB} \ $ to use $ \ \vec{BA} \ $ emanating from vertex $ \ B \ $ . So a sign error is introduced there. (Another interpretation is that what you have found in your last calculation is the cosine of the supplementary angle, which gives $ \ - \cos \theta \ $ . )

$\endgroup$
0
$\begingroup$

You are working a little bit too hard.

lets translate A to the origin.

This translates $D$ to $D' = (4,-2,6)$ and $B$ to $B' = (3,2,-1)$

$C' = D'+B' - 2(D' proj B')$

The projection of $D'$ onto $B' = ||D'|| \cos\theta \frac B{||B'||} = \dfrac {D'\cdot B'}{B'\cdot B'} B' = \frac 27(3,2,1)$

$C = D + \frac 57 B' = (7,-4,10) + (\frac{15}{7}, \frac{10}{7},\frac{5}{7})$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .