Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 02 (d)

MATH152 April 2013

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Question B 02 (d)
Let $A=(0,1,2)$, $B=(1,0,3)$ and $C=(1,1,4)$. Let $\Delta$ be the triangle whose vertices are A,B, and C. Find the point D on the plane P from part (c) such that A,B,C and D form a rectangle.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First, let $\displaystyle D=(x,y,z)$. Compute the vectors ${\vec {AD}}$ and ${\vec {CD}}$.

Hint 2

Compute the dot products with all four of the vectors from the hint and part (b) corresponding to the right angles. These are ${\begin{aligned}{\vec {AD}}\cdot {\vec {CD}}\\{\vec {CD}}\cdot {\vec {CB}}\\{\vec {AD}}\cdot {\vec {AB}}\\{\vec {CB}}\cdot {\vec {AB}}\end{aligned}}$ Notice the last one was done already in part (b). Use this information and the plane from part (c) to get a system of equations that one can solve.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $\displaystyle D=(x,y,z)$. First we compute the vectors as stated in hint one. These are given by ${\begin{aligned}{\vec {AD}}&=(x,y-1,z-2)\\{\vec {CD}}&=(x-1,y-1,z-4)\\{\vec {AB}}&=(1,-1,1)\\{\vec {CB}}&=(0,-1,-1)\end{aligned}}$ Taking dot products of these values as stated in hint 2 we have ${\begin{aligned}0={\vec {AD}}\cdot {\vec {CD}}&=x(x-1)+(y-1)^{2}+(z-2)(z-4)\\0={\vec {CD}}\cdot {\vec {CB}}&=(1-y)+(4-z)=-y-z+5\\0={\vec {AD}}\cdot {\vec {AB}}&=x-(y-1)+(z-2)=x-y+z-1\\0={\vec {CB}}\cdot {\vec {AB}}&=(0)(1)+(-1)(-1)+(1)(-1)\end{aligned}}$ It turns out we will not use the first equation. Rearranging the middle two equations above gives ${\begin{aligned}5&=y+z\\1-x&=-y+z\end{aligned}}$ Lastly, recall that the plane P is given by $\displaystyle 4x+2y-2z=-2$. Using this with the two equations above gives $\displaystyle -2=4x+2y-2z=4x-2(-y+z)=4x-2(1-x)=6x-2$ and solving yields that $x=0$. Using this, the system of two equations above reduces to ${\begin{aligned}5&=y+z\\1&=-y+z\end{aligned}}$ Summing these yields $\displaystyle 6=2z$ and so $\displaystyle z=3$ and substituting this above gives $\displaystyle y=2$. Thus the solution is $\displaystyle D=(x,y,z)=(0,2,3)$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From part (b) we have that the triangle ABC is a right-angled triangle and so if we reflect along the hypotenuse connecting AC we will have a rectangle. We can achieve this by adding the vector BC to the vector A. We have $\displaystyle {}BC=<1,1,4>-<1,0,3>=<0,1,1>$ and so $\displaystyle {}D=A+BC=<0,1,2>+<0,1,1>=<0,2,3>.$ Alternatively we could add the vector BA to the vector C: $\displaystyle {}BA=<0,1,2>-<1,0,3>=<-1,1,-1>$ and $\displaystyle {}D=C+BA=<1,1,4>+<-1,1,-1>=<0,2,3>.$ This point is on the plane because the vectors BA and BC span the plane and this point is written as a linear combination of these vectors.

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