Question

Consider points A(0,0,0), B(1,1,1) and plane P : x - y + z = 3. Let Q(a,b,c) be a moving point on plane P then identify the correct statement(s)

• A. If QA + QB is minimum then a - c - 10b is equal to 1
• B. If QA + QB is minimum then a - c - 10b is equal to 2
• C. If |QA - QB| is maximum then a - 2b + c is equal to 0
• D. If |QA - QB| is minimum then Q is unique

Answer 1

Answer: (B), (C)

Options (B) and (C) are correct.

Answer 2

We are given:

A = (0,0,0), B = (1,1,1), and the plane P: x – y + z = 3.

A moving point Q = (a, b, c) lies on P (so a – b + c = 3). We need to check the given statements.

1. Minimizing QA + QB

The standard trick is to “reflect” one focus about the plane. Reflect A about P.

• Plane P is written as: x – y + z – 3 = 0

• The normal is n = (1, –1, 1) and |n|² = 3.

• Distance (with sign) from A to P: 0 – 0 + 0 – 3 = –3.

• The reflection formula gives:

  A′ = A – 2*( (A–P distance)/|n|² ) * n
  = (0, 0, 0) – 2*(–3/3)(1, –1, 1)
  = (0,0,0) + 2(1, –1, 1) = (2, –2, 2).

Now, the minimal path from A to Q to B is the same as the straight line from A′ to B.

• Line A′B:

  Parameterize: R(t) = A′ + t(B – A′)
  = (2, –2, 2) + t((1,1,1) – (2, –2, 2))
  = (2, –2, 2) + t(–1, 3, –1).

Find t such that R(t) lies on P. Let Q = R(t) = (2–t, –2+3t, 2–t). Substitute in P:

(2–t) – (–2+3t) + (2–t) = 3
⟹ 2 – t + 2 – 3t + 2 – t = 6 – 5t = 3
⟹ 5t = 3 ⇒ t = 3/5.

Thus, Q = (2 – 3/5, –2 + 9/5, 2 – 3/5) = (7/5, –1/5, 7/5).

For Q = (a, b, c) = (7/5, –1/5, 7/5):
Compute a – c – 10b:
= (7/5) – (7/5) – 10*(–1/5)
= 0 + 10/5 = 2.

Thus, statement (B) is true and (A) is false.

2. Maximizing |QA – QB|

For any two fixed points, by the triangle inequality, we have
|QA – QB| ≤ AB, with equality if and only if Q, A, and B are collinear.

• The line through A and B is: (t, t, t).

• For Q on this line and on P, substitute (t, t, t) into P:

  t – t + t = t = 3 ⟹ t = 3.

So the unique Q is (3, 3, 3).

Now, for Q = (3, 3, 3):
Compute a – 2b + c:
= 3 – 2(3) + 3 = 3 – 6 + 3 = 0.

Thus, statement (C) is true.

3. Minimizing |QA – QB|

The least possible value of |QA – QB| is 0 (i.e. when QA = QB).

• Squaring distances:

  QA² = a² + b² + c²
  QB² = (a–1)² + (b–1)² + (c–1)² = a² + b² + c² – 2(a+b+c) + 3.

Setting QA = QB gives
a² + b² + c² = a² + b² + c² – 2(a+b+c) + 3 ⟹ a+b+c = 3/2.

But Q must also satisfy P: a – b + c = 3.

Solving these two equations:

a + b + c = 3/2 …(1)
a – b + c = 3 …(2)

Adding (1) and (2):
2a + 2c = 9/2 ⟹ a + c = 9/4.
Subtracting (1) from (2):
–2b = 3 – 3/2 = 3/2 ⟹ b = –3/4.

Since a + c = 9/4, the pair (a, c) can take infinitely many values subject to that condition. Thus there is an entire line of Q with QA = QB; so the minimum (zero difference) is attained on a continuum, i.e. Q is not unique.

Thus, statement (D) is false.