Exercise 11.2 answers: Introduction to Three Dimensional Geometry

Open NCERT p. 213Matches NCERT’s answer

(i) (2, 3, 5) and (4, 3, 1)

1. Distance formula: PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].
2. PQ = √[(4 − 2)² + (3 − 3)² + (1 − 5)²] = √(4 + 0 + 16) = √20 = 2√5.

Answer2√5

(ii) (–3, 7, 2) and (2, 4, –1)

1. PQ = √[(2 − (−3))² + (4 − 7)² + (−1 − 2)²] = √(25 + 9 + 9) = √43.

Answer√43

(iii) (–1, 3, – 4) and (1, –3, 4)

1. PQ = √[(1 − (−1))² + (−3 − 3)² + (4 − (−4))²] = √(4 + 36 + 64) = √104 = 2√26.

Answer2√26

(iv) (2, –1, 3) and (–2, 1, 3)

1. PQ = √[(−2 − 2)² + (1 − (−1))² + (3 − 3)²] = √(16 + 4 + 0) = √20 = 2√5.

Answer2√5

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Open NCERT p. 213One way to think about it

1. Call the points A(−2, 3, 5), B(1, 2, 3) and C(7, 0, −1).
2. Find AB = √((1+2)² + (2−3)² + (3−5)²) = √(9+1+4) = √14.
3. Find BC = √((7−1)² + (0−2)² + (−1−3)²) = √(36+4+16) = √56 = 2√14.
4. Find AC = √((7+2)² + (0−3)² + (−1−5)²) = √(81+9+36) = √126 = 3√14.
5. Check: AB + BC = √14 + 2√14 = 3√14 = AC, so the three points lie on one straight line.

In shortYes, A, B, C are collinear because AB + BC = AC (each equal to a multiple of √14), with B between A and C.

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Open NCERT p. 213Checked by computer

(i) … are the vertices of an isosceles triangle

1. Call the points A(0, 7, −10), B(1, 6, −6), C(4, 9, −6).
2. AB = √[1² + (−1)² + 4²] = √18 = 3√2.
3. BC = √[3² + 3² + 0²] = √18 = 3√2.
4. CA = √[(−4)² + (−2)² + (−4)²] = √36 = 6.
5. AB = BC, so two sides are equal and the triangle is isosceles.

Answeryes

(ii) … are the vertices of a right angled triangle

1. Call the points A(0, 7, 10), B(−1, 6, 6), C(−4, 9, 6).
2. AB² = (−1)² + (−1)² + (−4)² = 18.
3. BC² = (−3)² + 3² + 0² = 18.
4. CA² = 4² + (−2)² + 4² = 36.
5. AB² + BC² = 18 + 18 = 36 = CA², so by Pythagoras the angle at B is 90° and the triangle is right angled.

Answeryes

(iii) … are the vertices of a parallelogram

1. Call the points A(−1, 2, 1), B(1, −2, 5), C(4, −7, 8), D(2, −3, 4), in that order.
2. AB = √(4 + 16 + 16) = 6, BC = √(9 + 25 + 9) = √43, CD = √(4 + 16 + 16) = 6, DA = √(9 + 25 + 9) = √43, so opposite sides are equal.
3. In space, equal opposite sides alone do not prove it, because four points need not lie in one plane. So also check that the diagonals bisect each other.
4. Midpoint of AC = ((−1 + 4)/2, (2 − 7)/2, (1 + 8)/2) = (3/2, −5/2, 9/2).
5. Midpoint of BD = ((1 + 2)/2, (−2 − 3)/2, (5 + 4)/2) = (3/2, −5/2, 9/2).
6. The diagonals AC and BD have the same midpoint, so they bisect each other and ABCD is a parallelogram.

Answeryes

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Open NCERT p. 213Matches NCERT’s answer

1. Let P(x, y, z) be a point equidistant from A(1, 2, 3) and B(3, 2, −1), so PA² = PB².
2. PA² = (x−1)² + (y−2)² + (z−3)², PB² = (x−3)² + (y−2)² + (z+1)².
3. Set them equal. The (y−2)² terms are the same on both sides, so they cancel.
4. Expanding the rest and simplifying gives 4x − 8z = 0, i.e. x − 2z = 0.

Answerx − 2z = 0

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Open NCERT p. 213Matches NCERT’s answer

1. Let P be (x, y, z). Then PA = √[(x − 4)² + y² + z²] and PB = √[(x + 4)² + y² + z²], and PA + PB = 10.
2. Write PA = 10 − PB and square both sides: PA² = 100 − 20·PB + PB².
3. PA² − PB² = (x − 4)² − (x + 4)² = −16x, so −16x = 100 − 20·PB, which gives PB = 5 + 4x/5.
4. Square again: (x + 4)² + y² + z² = (5 + 4x/5)², i.e. x² + 8x + 16 + y² + z² = 25 + 8x + 16x²/25.
5. The 8x terms cancel: x² − 16x²/25 + y² + z² = 9, i.e. 9x²/25 + y² + z² = 9.
6. Multiply by 25: 9x² + 25y² + 25z² = 225.

Answer9x² + 25y² + 25z² − 225 = 0

Watch this explained “Conditions, equations, and a gap”, 13:00 into Applying the right-triangle rule twice to get out of the plane