In R^3, let L be a straight lines passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P_1:x + 2y - z + 1 = 0 and P_2:2x - y + z - 1 = 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane P_1. Which of the following points lie (s) on M ?

( 0, - 5 \over 6, - 2 \over 3 ), ( - 1 \over 6, - 1 \over 3,1 \over 6 )

( 0, - 5 \over 6, - 2 \over 3 )

( - 1 \over 6, - 1 \over 3,1 \over 6 )

3D Geometry PYQ — JEE Advanced 2015 · Shift 1 · Q34 with Solution

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Correct answer: B, A

Condition for points on the line $L$

Let a general point on the line $L$ through the origin be $Q=(x,y,z).$

Its perpendicular distances from the planes $P_1: x+2y-z+1=0$ and $P_2: 2x-y+z-1=0$ are $d_1=\frac{|x+2y-z+1|}{\sqrt{1^2+2^2+(-1)^2}}=\frac{|x+2y-z+1|}{\sqrt6},$ $d_2=\frac{|2x-y+z-1|}{\sqrt{2^2+(-1)^2+1^2}}=\frac{|2x-y+z-1|}{\sqrt6}.$

Given that for every point on $L$ , these distances are equal, we get $|x+2y-z+1|=|2x-y+z-1|.$

So points of $L$ must satisfy one of the two equations:

$x+2y-z+1=2x-y+z-1$ or $x+2y-z+1=-(2x-y+z-1).$

This gives:

Case 1: $-x+3y-2z+2=0$
Case 2: $3x+y=0$

Since $L$ passes through the origin, every point on $L$ must satisfy the relevant equation at $(0,0,0)$ .

Check origin:

In Case 1: $-0+0-0+2=2\neq 0$ , so this cannot contain the origin.
In Case 2: $3(0)+0=0$ , valid.

Hence $L$ lies in the plane $3x+y=0.$

Since $L$ is a line through the origin and all its points satisfy equal distance from the two planes, we also need the equality to hold identically along that line. Substituting $y=-3x$ into the absolute-value condition:

$|x+2(-3x)-z+1|=|2x-(-3x)+z-1|$ $|-5x-z+1|=|5x+z-1|.$

But note that $-5x-z+1=-(5x+z-1),$ so this is automatically true for every point with $y=-3x$ .

Thus the entire locus of such points is the plane $3x+y=0$ , and any line through origin contained in this plane works. Therefore the feet of perpendiculars from points of $L$ to $P_1$ will lie on the orthogonal projection of such a line onto $P_1$ .

To identify the possible locus $M$ , let us derive the set of feet from all points on the plane $3x+y=0$ ; since every valid line is inside this plane, the feet must lie in the intersection of $P_1$ with the plane obtained by projection along normal to $P_1$ .

Foot of perpendicular from a point $(x,y,z)$ to plane $P_1$

Plane $P_1$ has normal vector $\mathbf n_1=(1,2,-1).$

For point $Q=(x,y,z)$ , the foot $F$ on $P_1$ is

=Q-\frac{x+2y-z+1}{6}(1,2,-1).$$ So if $F=(X,Y,Z)$, $$X=x-\frac{x+2y-z+1}{6},$$ $$Y=y-\frac{2(x+2y-z+1)}{6},$$ $$Z=z+\frac{x+2y-z+1}{6}.$$ Also $F$ must satisfy $P_1$: $$X+2Y-Z+1=0.$$ Now $Q$ lies on $3x+y=0$, i.e. $$y=-3x.$$ Let $$s=x+2y-z+1=x-6x-z+1=-5x-z+1.$$ Then $$X=x-\frac{s}{6},\quad Y=-3x-\frac{s}{3},\quad Z=z+\frac{s}{6}.$$ Instead of parametrizing further, observe a cleaner geometric fact: The set of feet from points in the plane $3x+y=0$ onto $P_1$ is the intersection of $P_1$ with the plane generated by projecting $3x+y=0$ along the normal $(1,2,-1)$. So we eliminate $(x,y,z)$ from $$X=x-\frac{t}{6},\quad Y=y-\frac{2t}{6},\quad Z=z+\frac{t}{6},$$ with $y=-3x$. Thus $$x=X+\frac{t}{6},\quad y=Y+\frac{t}{3},\quad z=Z-\frac{t}{6}.$$ Using $y=-3x$, $$Y+\frac{t}{3}=-3\left(X+\frac{t}{6}\right)$$ $$Y+\frac{t}{3}=-3X-\frac{t}{2}$$ $$Y+3X+\frac{5t}{6}=0.$$ So $$t=-\frac{6}{5}(3X+Y).$$ But since $F=(X,Y,Z)$ lies on $P_1$, $$X+2Y-Z+1=0.$$ To test the options, it is easier to check whether each point lies on $P_1$ and whether it can arise from projection of some point satisfying $3x+y=0$. For a foot $F$, the original point is on the normal through $F$: $$Q=F+\lambda(1,2,-1).$$ For this $Q$ to belong to the plane $3x+y=0$, $$3(X+\lambda)+(Y+2\lambda)=0$$ $$3X+Y+5\lambda=0.$$ So for any $F$, such a $\lambda$ always exists: $$\lambda=-\frac{3X+Y}{5}.$$ Hence the only additional condition is that $F$ lies on plane $P_1$. Therefore, $$M\subset P_1,$$ and for every point of $P_1$, there is some point in the plane $3x+y=0$ projecting to it. Since $L$ is a line, $M$ is a line inside $P_1$, but among the given points, the relevant necessary condition is membership in $P_1$. Let us check options. --- 3. **Check each option in $P_1: x+2y-z+1=0$** ### Option A: $\left(0,-\frac56,-\frac23\right)$ $$x+2y-z+1=0+2\left(-\frac56\right)-\left(-\frac23\right)+1$$ $$=-\frac{10}{6}+\frac{4}{6}+1=-1+1=0.$$ So A lies on $P_1$. ### Option B: $\left(-\frac16,-\frac13,\frac16\right)$ $$x+2y-z+1=-\frac16+2\left(-\frac13\right)-\frac16+1$$ $$=-\frac16-\frac46-\frac16+1=-1+1=0.$$ So B lies on $P_1$. ### Option C: $\left(-\frac56,0,\frac16\right)$ $$x+2y-z+1=-\frac56+0-\frac16+1=0.$$ So C lies on $P_1$. ### Option D: $\left(-\frac13,0,\frac23\right)$ $$x+2y-z+1=-\frac13+0-\frac23+1=0.$$ So D also lies on $P_1$. So all options lie on $P_1$. We now determine the actual line $M$. --- 4. **Find the actual line $M$** Take direction vector of $L$ as $\mathbf v=(a,b,c)$ with origin on it. Since every point $t(a,b,c)$ satisfies equal distances, $$|t(a+2b-c)+1|=|t(2a-b+c)-1|\quad \forall t.$$ For this to hold for all $t$, we must have $$a+2b-c=-(2a-b+c),$$ which gives $$3a+b=0.$$ Thus $b=-3a$, and direction vector is $$\mathbf v=(a,-3a,c)=a(1,-3,0)+c(0,0,1).$$ Since $L$ is a single line, choose general direction $$\mathbf v=(1,-3,m).$$ A point on $L$ is $$Q(t)=(t,-3t,mt).$$ Its foot on $P_1$ is $$F(t)=Q(t)-\frac{x+2y-z+1}{6}(1,2,-1).$$ Here $$x+2y-z+1=t-6t-mt+1=1-(5+m)t.$$ Hence $$F(t)=\left(t-\frac{1-(5+m)t}{6}, -3t-\frac{1-(5+m)t}{3}, mt+\frac{1-(5+m)t}{6}\right).$$ This is a line depending on $m$. Now among given points, we seek those that lie on **every such possible** $M$? Actually the problem asks for points lying on $M$, implying $M$ is uniquely determined. That happens when we use the special line obtained as intersection of the two bisector planes and passing through origin. Since only one bisector plane passes through origin, and the line through origin satisfying the condition for all its points is the intersection of that bisector plane with the plane through origin perpendicular to both normals, giving direction $$\mathbf v=\mathbf n_1\times \mathbf n_2=(1,2,-1)\times(2,-1,1)=(1,-3,-5).$$ Also this satisfies $3x+y=0$. So take $$L: (x,y,z)=t(1,-3,-5).$$ Then $$Q(t)=(t,-3t,-5t).$$ Now $$x+2y-z+1=t-6t+5t+1=1.$$ So every point on $L$ is at constant distance $\frac1{\sqrt6}$ from both planes. Hence the foot on $P_1$ is $$F(t)=Q(t)-\frac{1}{6}(1,2,-1) =\left(t-\frac16,-3t-\frac13,-5t+\frac16\right).$$ Thus $M$ is the line $$M:\left(X,Y,Z\right)=\left(-\frac16,-\frac13,\frac16\right)+t(1,-3,-5).$$ --- 5. **Check options on this line** ### A: $\left(0,-\frac56,-\frac23\right)$ Compare with base point $\left(-\frac16,-\frac13,\frac16\right)$: $$\Delta=\left(\frac16,-\frac12,-\frac56\right)=\frac16(1,-3,-5).$$ So A lies on $M$. ### B: $\left(-\frac16,-\frac13,\frac16\right)$ This is the base point itself ($t=0$), so B lies on $M$. ### C: $\left(-\frac56,0,\frac16\right)$ Difference from base point: $$\left(-\frac23,\frac13,0\right),$$ not proportional to $(1,-3,-5)$. So C does not lie on $M$. ### D: $\left(-\frac13,0,\frac23\right)$ Difference: $$\left(-\frac16,\frac13,\frac12\right),$$ not proportional to $(1,-3,-5)$. So D does not lie on $M$. --- 6. **Final answer** The points lying on $M$ are $$\boxed{A,\ B}.$$

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View written solutionFree

Correct answer: B, A

Condition for points on the line $L$

Let a general point on the line $L$ through the origin be $Q=(x,y,z).$

Given that for every point on $L$ , these distances are equal, we get $|x+2y-z+1|=|2x-y+z-1|.$

So points of $L$ must satisfy one of the two equations:

$x+2y-z+1=2x-y+z-1$ or $x+2y-z+1=-(2x-y+z-1).$

This gives:

Case 1: $-x+3y-2z+2=0$
Case 2: $3x+y=0$

Since $L$ passes through the origin, every point on $L$ must satisfy the relevant equation at $(0,0,0)$ .

Check origin:

In Case 1: $-0+0-0+2=2\neq 0$ , so this cannot contain the origin.
In Case 2: $3(0)+0=0$ , valid.

Hence $L$ lies in the plane $3x+y=0.$

$|x+2(-3x)-z+1|=|2x-(-3x)+z-1|$ $|-5x-z+1|=|5x+z-1|.$

But note that $-5x-z+1=-(5x+z-1),$ so this is automatically true for every point with $y=-3x$ .

Foot of perpendicular from a point $(x,y,z)$ to plane $P_1$

Plane $P_1$ has normal vector $\mathbf n_1=(1,2,-1).$

For point $Q=(x,y,z)$ , the foot $F$ on $P_1$ is

=Q-\frac{x+2y-z+1}{6}(1,2,-1).$$ So if $F=(X,Y,Z)$, $$X=x-\frac{x+2y-z+1}{6},$$ $$Y=y-\frac{2(x+2y-z+1)}{6},$$ $$Z=z+\frac{x+2y-z+1}{6}.$$ Also $F$ must satisfy $P_1$: $$X+2Y-Z+1=0.$$ Now $Q$ lies on $3x+y=0$, i.e. $$y=-3x.$$ Let $$s=x+2y-z+1=x-6x-z+1=-5x-z+1.$$ Then $$X=x-\frac{s}{6},\quad Y=-3x-\frac{s}{3},\quad Z=z+\frac{s}{6}.$$ Instead of parametrizing further, observe a cleaner geometric fact: The set of feet from points in the plane $3x+y=0$ onto $P_1$ is the intersection of $P_1$ with the plane generated by projecting $3x+y=0$ along the normal $(1,2,-1)$. So we eliminate $(x,y,z)$ from $$X=x-\frac{t}{6},\quad Y=y-\frac{2t}{6},\quad Z=z+\frac{t}{6},$$ with $y=-3x$. Thus $$x=X+\frac{t}{6},\quad y=Y+\frac{t}{3},\quad z=Z-\frac{t}{6}.$$ Using $y=-3x$, $$Y+\frac{t}{3}=-3\left(X+\frac{t}{6}\right)$$ $$Y+\frac{t}{3}=-3X-\frac{t}{2}$$ $$Y+3X+\frac{5t}{6}=0.$$ So $$t=-\frac{6}{5}(3X+Y).$$ But since $F=(X,Y,Z)$ lies on $P_1$, $$X+2Y-Z+1=0.$$ To test the options, it is easier to check whether each point lies on $P_1$ and whether it can arise from projection of some point satisfying $3x+y=0$. For a foot $F$, the original point is on the normal through $F$: $$Q=F+\lambda(1,2,-1).$$ For this $Q$ to belong to the plane $3x+y=0$, $$3(X+\lambda)+(Y+2\lambda)=0$$ $$3X+Y+5\lambda=0.$$ So for any $F$, such a $\lambda$ always exists: $$\lambda=-\frac{3X+Y}{5}.$$ Hence the only additional condition is that $F$ lies on plane $P_1$. Therefore, $$M\subset P_1,$$ and for every point of $P_1$, there is some point in the plane $3x+y=0$ projecting to it. Since $L$ is a line, $M$ is a line inside $P_1$, but among the given points, the relevant necessary condition is membership in $P_1$. Let us check options. --- 3. **Check each option in $P_1: x+2y-z+1=0$** ### Option A: $\left(0,-\frac56,-\frac23\right)$ $$x+2y-z+1=0+2\left(-\frac56\right)-\left(-\frac23\right)+1$$ $$=-\frac{10}{6}+\frac{4}{6}+1=-1+1=0.$$ So A lies on $P_1$. ### Option B: $\left(-\frac16,-\frac13,\frac16\right)$ $$x+2y-z+1=-\frac16+2\left(-\frac13\right)-\frac16+1$$ $$=-\frac16-\frac46-\frac16+1=-1+1=0.$$ So B lies on $P_1$. ### Option C: $\left(-\frac56,0,\frac16\right)$ $$x+2y-z+1=-\frac56+0-\frac16+1=0.$$ So C lies on $P_1$. ### Option D: $\left(-\frac13,0,\frac23\right)$ $$x+2y-z+1=-\frac13+0-\frac23+1=0.$$ So D also lies on $P_1$. So all options lie on $P_1$. We now determine the actual line $M$. --- 4. **Find the actual line $M$** Take direction vector of $L$ as $\mathbf v=(a,b,c)$ with origin on it. Since every point $t(a,b,c)$ satisfies equal distances, $$|t(a+2b-c)+1|=|t(2a-b+c)-1|\quad \forall t.$$ For this to hold for all $t$, we must have $$a+2b-c=-(2a-b+c),$$ which gives $$3a+b=0.$$ Thus $b=-3a$, and direction vector is $$\mathbf v=(a,-3a,c)=a(1,-3,0)+c(0,0,1).$$ Since $L$ is a single line, choose general direction $$\mathbf v=(1,-3,m).$$ A point on $L$ is $$Q(t)=(t,-3t,mt).$$ Its foot on $P_1$ is $$F(t)=Q(t)-\frac{x+2y-z+1}{6}(1,2,-1).$$ Here $$x+2y-z+1=t-6t-mt+1=1-(5+m)t.$$ Hence $$F(t)=\left(t-\frac{1-(5+m)t}{6}, -3t-\frac{1-(5+m)t}{3}, mt+\frac{1-(5+m)t}{6}\right).$$ This is a line depending on $m$. Now among given points, we seek those that lie on **every such possible** $M$? Actually the problem asks for points lying on $M$, implying $M$ is uniquely determined. That happens when we use the special line obtained as intersection of the two bisector planes and passing through origin. Since only one bisector plane passes through origin, and the line through origin satisfying the condition for all its points is the intersection of that bisector plane with the plane through origin perpendicular to both normals, giving direction $$\mathbf v=\mathbf n_1\times \mathbf n_2=(1,2,-1)\times(2,-1,1)=(1,-3,-5).$$ Also this satisfies $3x+y=0$. So take $$L: (x,y,z)=t(1,-3,-5).$$ Then $$Q(t)=(t,-3t,-5t).$$ Now $$x+2y-z+1=t-6t+5t+1=1.$$ So every point on $L$ is at constant distance $\frac1{\sqrt6}$ from both planes. Hence the foot on $P_1$ is $$F(t)=Q(t)-\frac{1}{6}(1,2,-1) =\left(t-\frac16,-3t-\frac13,-5t+\frac16\right).$$ Thus $M$ is the line $$M:\left(X,Y,Z\right)=\left(-\frac16,-\frac13,\frac16\right)+t(1,-3,-5).$$ --- 5. **Check options on this line** ### A: $\left(0,-\frac56,-\frac23\right)$ Compare with base point $\left(-\frac16,-\frac13,\frac16\right)$: $$\Delta=\left(\frac16,-\frac12,-\frac56\right)=\frac16(1,-3,-5).$$ So A lies on $M$. ### B: $\left(-\frac16,-\frac13,\frac16\right)$ This is the base point itself ($t=0$), so B lies on $M$. ### C: $\left(-\frac56,0,\frac16\right)$ Difference from base point: $$\left(-\frac23,\frac13,0\right),$$ not proportional to $(1,-3,-5)$. So C does not lie on $M$. ### D: $\left(-\frac13,0,\frac23\right)$ Difference: $$\left(-\frac16,\frac13,\frac12\right),$$ not proportional to $(1,-3,-5)$. So D does not lie on $M$. --- 6. **Final answer** The points lying on $M$ are $$\boxed{A,\ B}.$$

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