Question 11 - Jee advanced Math 2022 P1 Questions with Solutions

Let \(P_{1}\) and \(P_{2}\) be two planes given by 

\(P_{1} : 10x + 15y + 12z - 60 = 0\).

\(P_{2} : -2x + 5y + 4z - 20 = 0\).

Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on \(P_{1}\) and \(P_{2}\)?

A) \(\frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5}\)

B) \(\frac{x - 6}{-5} = \frac{y}{2} = \frac{z}{3}\)

C) \(\frac{x}{-2} = \frac{y - 4}{5} = \frac{z}{4}\)

D) \(\frac{x}{1} = \frac{y - 4}{-2} = \frac{z}{3}\)

Sol : 

The two faces of a tetrahedron lie on the planes \(P_{1}\) and \(P_{2}\). Let L be the line of intersection of the planes. L is one of the six edges of the tetrahedron.

In the figure above, the other five edges of the tetrahedron are \(l_{1}\), \(l_{2}\), \(l_{3}\), \(l_{4}\) and \(l_{5}\), out of which, \(l_{1}\), \(l_{2}\), \(l_{3}\) and \(l_{4}\), all four of them completely lie in either plane \(P_{1}\) or \(P_{2}\) …AND… they each intersect the line L at a single point(A and B in the figure). 

So the lines in the given options must satisfy both these conditions if they are any one of these four edges \(l_{1}\), \(l_{2}\), \(l_{3}\) and \(l_{4}\) of the tetrahedron. 

If an option is line \(l_{5}\) then it must not fully lie in either of the planes but it must intersect them at one point each(C and D in the figure above). Also, it must not intersect the line of intersection L.  

And finally, if L is itself present in any of the options then that for sure is one of our answers. 

Let’s first find the equation of the line of intersection L.

\(P_{1} - 3P_{2} \implies\)

\(10x + 15y + 12z - 60 - (-6x + 15y + 12z - 60) = 0 - 0\)

\(16x = 0\)

\(\implies x = 0\)

\(\implies\) the x-coordinate of all points on the line of intersection is 0.

\(\implies 5y + 4z - 20 = 0\)

For any value of \(z = 0\)(say)

\(\implies 5y = 20\)

\(\implies y = 4\)

\(\implies (0, 4, 0)\) is a point on L.

From the equations of \(P_{1}\) and \(P_{2}\),

\(10\hat{i} + 15\hat{j} + 12\hat{k}\) and \(-2\hat{i} + 5\hat{j} + 4\hat{k}\) are vectors perpendicular to \(P_{1}\) and \(P_{2}\). 

\(\implies\) their cross product is parallel to the line of intersection L.

Therefore, the equation of L is:

\(\frac{x - 0}{0} = \frac{y - 4}{-64} = \frac{z - 0}{80}\).

Let’s check the options now.

A) \(\frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5}\)

The x-coordinate of every point on this line is 1. So this clearly does not intersect the line of intersection L of the planes where x-coordinate of every point is 0.

Also, from the equation, \((1, 1, 1)\) is a point on this line. 

\(10(1) + 15(1) + 12(1) - 60 = -23 \neq 0\) and

\(-2(1) + 5(1) + 4(1) - 20 = -13 \neq 0\)

So, it does not fully lie in either \(P_{1}\) or \(P_{2}\)(otherwise all points on this line will also have been on the plane).

So it could be the edge \(l_{5}\) provided that it intersects both the planes in one point each.

Let \(\frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5} = b\)

\(\implies x = 1; y = 1; z = 5b + 1\)

Substituting in equations of \(P_{1}\) and \(P_{2}\),

\(10(1) + 15(1) + 12(5b + 1) - 60 = 0\) and

\(-2(1) + 5(1) + 4(5b + 1) - 20 = 0\)

\(\implies b = \frac{23}{60}\) and

\(b = \frac{13}{20}\)

So the line cuts both \(P_{1}\) and \(P_{2}\) at points \((1, 1, 5\frac{23}{60} + 1)\) and \((1, 1, 5\frac{13}{20} + 1)\)each. So this line definitely can be an edge of the tetrahedron.

B) \(\frac{x - 6}{-5} = \frac{y}{2} = \frac{z}{3}\)

Let \(\frac{x - 6}{-5} = \frac{y}{2} = \frac{z}{3} = b\)

\(\implies x = -5b + 6; y = 2b; z = 3b\)

Substituting in the equation of L,

\(\frac{-5b + 6}{0} = \frac{2b - 4}{-64} = \frac{3b - 0}{80}\)

\(\implies b = \frac{6}{5}\) and \(10b - 20 = -12b\)

\(\implies b = \frac{6}{5}\) and \(b = \frac{20}{22} = \frac{10}{11}\)

Two different values of \(b\) implies it is not intersecting L.

From the equation of the line, (6, 0, 0) is a point on this line which clearly satisfies the equation of \(P_{1}\). 

\((10(6) + 15(0) + 12(0) - 60 = 0 \implies 0 = 0)\)

For any b = 1(say), 

\(\implies x = -5(1)+ 6; y = 2(1); z = 3(1) \implies (1, 2, 3)\) be some other point on the line.

\(10(1) + 15(2) + 12(3) - 60 = 0 \implies 16 = 0\) which is not true.

\(\implies\) the line intersects the plane \(P_{1}\) at one point \((6, 0, 0)\).

From the equation of \(P_{2}\),

\(-2(-5b + 6) + 5(2b) + 4(3b) - 20 = 0\)

\(\implies 32b = 32\)

\(\implies b = 1\)

\(\implies\) the line intersects the plane \(P_{2}\) at the point \((1, 2, 3)\). 

Therefore this also could be the edge \(l_{5}\) of the tetrahedron. 

C) \(\frac{x}{-2} = \frac{y - 4}{5} = \frac{z}{4}\)

Let \(\frac{x}{-2} = \frac{y - 4}{5} = \frac{z}{4} = b\)

\(\implies x = -2b; y = 5b + 4; z = 4b\)

From the equation, \((0, 4, 0)\) is a point on the line, and from equation of L, we see that this point is also on L.

So this line intersects L at this point. This point will also satisfy equations of both the planes. 

For any b = 1(say), 

\(\implies x = -2(1); y = 5(1) + 4; z = 4(1) \implies (-2, 9, 4)\) be some other point on the line.

\(10(-2) + 15(9) + 12(4) - 60 = 0 \implies 103 = 0\) which is not true.

\(-2(-2) + 5(9) + 4(4) - 20 = 0 \implies 45 = 0\) which is not true.

So the given line does not fully lie in either of the planes.

Hence this line cannot be an edge of the tetrahedron.

D) \(\frac{x}{1} = \frac{y - 4}{-2} = \frac{z}{3}\)

Let \(\frac{x}{1} = \frac{y - 4}{-2} = \frac{z}{3} = b\)

\(\implies x = b; y = -2b + 4; z = 3b\)

From the equation, \((0, 4, 0)\) is a point on the line, and from equation of L, we see that this point is also on L.

So this line intersects L at \((0, 4, 0)\). This point will also satisfy equations of both the planes. 

For any b = 1(say), 

\(\implies x = 1; y = -2(1) + 4; z = 3(1) \implies (1, 2, 3)\) be some other point on the line.

\(10(1) + 15(2) + 12(3) - 60 = 0 \implies 16 = 0\) which is not true.

\(-2(1) + 5(2) + 4(3) - 20 = 0 \implies 0 = 0\) which is true.

So the given line is in plane \(P_{2}\).

This can be an edge of the tetrahedron.

\(\implies\) Options A, B and D are TRUE.