Question Video: Determining Which Point Lies on the Line of Intersection of Two Planes | Nagwa Question Video: Determining Which Point Lies on the Line of Intersection of Two Planes | Nagwa

# Question Video: Determining Which Point Lies on the Line of Intersection of Two Planes Mathematics • Third Year of Secondary School

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Which of the following points lies on the line of intersection of the two planes β9π₯ + 16π¦ β 6π§ β 11 = 0 and β13π₯ + 2π¦ + 4π§ + 1 = 0? [A] (1, 2, β2) [B] (1, 2, 2) [C] (1, β2, 2) [D] (1, 3, 2) [E] (2, 3, 5).

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### Video Transcript

Which of the following points lies on the line of intersection of the two planes negative nine π₯ plus 16π¦ minus six π§ minus 11 is equal to zero and minus 13π₯ plus two π¦ plus four π§ plus one is equal to zero? (A) One, two, negative two; (B) one, two, two; (C) one, negative two, two; (D) one, three, two; or (E) two, three, five.

Weβre given the equations for two planes in general form, that is, in the form ππ₯ plus ππ¦ plus ππ§ plus π is equal to zero, where π, π, π, and π are constants. Weβre told that these two planes intersect over a straight line. And since there are infinitely many points on a straight line, there must be infinitely many points of intersection of the two planes. And weβre given five potential points of intersection (A), (B), (C), (D), and (E). We can approach this problem in a couple of different ways. We could use the two plane equations to find the equation of the line of intersection. We would then substitute each point into the equation of the line to see which point satisfy the equation of the line and therefore the two plane equations.

However, there is a simpler way to approach this. What we can do instead is simply substitute each of the given points into the two plane equations and see which point or points satisfied them. Our two planes which weβll call π one and π two have equations negative nine π₯ plus 16π¦ minus six π§ minus 11 is equal to zero and negative 13π₯ plus two π¦ plus four π§ plus one is equal to zero, respectively. So letβs begin with point (A). In point (A), we have π₯ is equal to one, π¦ is equal to two, and π§ is negative two. And substituting these values into our plane π one, this gives us negative nine times one plus 16 times two minus six times negative two minus 11. And if point (A) lies in plane one, this should be equal to zero.

However, our left-hand side evaluates to 24, and this is not equal to zero. Hence, point (A) does not lie in plane one. And since it doesnβt lie in plane one, it cannot lie on the line of intersection of the two planes. Therefore, we can discount point (A). Now, letβs try point (B). This has coordinates one, two, two, where π₯ is equal to one, π¦ is equal to two, and π§ is also two. Substituting these values into the equation for plane one, this gives us negative nine times one plus 16 times two minus six times two minus 11. That is negative nine plus 32 minus 12 minus 11. And this does evaluate to zero. So point (B) does sit in plane one.

So now, letβs see if point (B) sits in plane two, and if it does, this means it must lie on the line of intersection. So substituting these values into plane two, this gives us negative 13 times one plus two times two plus four times two plus one. That is negative 13 plus four plus eight plus one. Our left-hand side evaluates to zero, and therefore point (B) lies in plane two. And since point (B) lies in both planes, it must lie on the line of intersection of the two planes.

Now, letβs check point (C). Point (C) has coordinates one, negative two, two. And into the equation for plane one, this gives us negative nine minus 32 minus 12 minus 11. That is negative 64, and this is not equal to zero. So point (C) can be discounted. If it doesnβt lie in plane one, then it cannot lie on the line of intersection of the two planes.

Point (D) has coordinates one, three, and two. And substituting this into plane one gives us negative nine plus 48 minus 12 minus 11. And this evaluates to 16, which is of course not equal to zero. So we can discount point (D). Finally, point (E) has coordinates two, three, and five. Into the equation for plane one, this gives us negative 18 plus 48 minus 30 minus 11. This evaluates to negative 11, which of course is nonzero. And so we can discount point (E).

We see then that only point (B) thatβs with coordinates one, two, two satisfies both plane equations and therefore sits on the line of intersection of the two planes. Our answer is therefore point (B).

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