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The question goes as follows.

Consider $3$ planes $$1)2x+y+z=3$$ $$2)x-y+2z=4$$ $$3)x+y=2$$ such that they don't intersect in a single line and form a prism.Another plane $4)$ is made through some point $P$ on line of intersection of $2),3)i.e.L_1L_2$ such that it is made perpendicular to line $L_1L_2$ intersecting line $L_3L_4$ in $Q$ and line $L_5L_6$ in $R$ such that $\triangle PQR$ is made.

Here is a figure.

enter image description here

The question is to find which of the following option(s) are correct if $L$ denotes area of triangle PQR

$a)\lfloor \frac 1L \rfloor \lt 11$

$b)\lfloor \frac 1L +1 \rfloor \geq 12$

$c)(\lfloor L+1 \rfloor)^{-1} \lt 3$

$d) \lfloor \frac 2L -1 \rfloor \in [18, 20]$

I tried by finding out the line of intersection of the planes $2$ and $3$.And since the plane of the triangle is perpendicular to line $ L_1L_2 $ so the $ DR$ of plane is given by the DR of the line. Using this i could find out the DR of ths plane of triangle and by taking $x=0$ in the line of intersection of planes i could find out the equation of the plane of the triangle. But i could not find out the range of values that $ L$ can take. Any ideas? Thanks.
Navin
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  • Since $P$ is arbitrary, we can assume that the three lines are parallel (you can check this by computing the three cross products of the normals). I’d rotate the whole mess so that the three planes are perpendicular to a coordinate plane and reduce it to a two-dimensional problem. Perhaps there’s some clever way of doing this without computing $L$ explicitly, but nothing immediately comes to mind. – amd Apr 15 '17 at 00:31

2 Answers2

1

$$P=\left(\frac53,\frac13,\frac43\right)$$ $$Q=\left(1,-\frac{1}3,\frac43\right)$$ $$R=\left(1,1,0\right)$$ $$L=\frac{4\sqrt{3}}9$$

san
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  • @Harsh Kumar: Thanks for the edits. – san Apr 19 '17 at 16:59
  • Thanks for your answe. But that is one possible value of $L$ could you please suggest what will the range of values of $L$ . Thanks. – Navin Apr 20 '17 at 00:52
  • There are different possible values for $P$, $Q$ and $R$. But the value of $L$ is always the same. – san Apr 20 '17 at 01:03
  • It is the area of the ortogonal cross section of a triangular prisma, so it is always the same. – san Apr 20 '17 at 01:07
  • So a) and c) are both correct. – san Apr 20 '17 at 01:10
  • How did you find out the values of P ,Q,R – user471651 Jan 26 '18 at 13:11
  • I first took any point R on the line which is the intersection of the planes 1) and 3). Then I determined teh intersection of the plane 4) given by ((x,y,z)-R).(-1,1,1)=0 with the planes 2) and 3), giving P, and then the intersection of 4), 1) and 2), giving Q. – san Jan 29 '18 at 22:59
0

Take the intersection of all planes with x, y plane, i.e z=0, Now notice that the third equation is x+y=2 that means the third plane is parallel to z axis. We take value of L in (x, y) plane for compare with options .we have three lines in x, y plane:

2x+y=3

x-y=4

x+y=2

these lines intersect at A(7/3, -5/3), B(3, -1), C(1, 1). We find:

AB=a=(2^1.5)/3,

BC=b=2^1.5,

AC =c= (5^0.5)4/3

Suppose

c = AB + BC + AC = (2/3)2^0.5 + 2(2^0.5) +4/3(5^0.5)

and p=c/2

Using formula L = (p(P-a)(p-b)(p-c))^0.5 we find L = 4/3. Therefore option

3 i.e. (L+1)^-1 < 3 is correct.

sirous
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  • The plane $z=0$ is not perpendicular to $L_1L_2$: $$(x,y,z)=(1/3)(5,1,4)+t(-1,1,1)$$ – san Apr 20 '17 at 20:42
  • True. the triangle on plane PQR is the projection of triangle on z=0 plane that is its area is less than 4/3. so the third option is correct anyway. – sirous Apr 21 '17 at 05:53
  • The third option is true for any $L\ge 0$. – san Apr 21 '17 at 11:02