Exercise 11.1 answers: Three Dimensional Geometry

Open NCERT p. 381Matches NCERT’s answer

1. The direction cosines are just the cosines of the three given angles, taken in order x, y, z.
2. cos 90° = 0.
3. cos 135° = −cos 45° = −1/√2 (135° is in the second quadrant, so cosine is negative).
4. cos 45° = 1/√2.
5. So the direction cosines are l = 0, m = −1/√2, n = 1/√2.
6. Check: l² + m² + n² = 0 + 1/2 + 1/2 = 1. Correct.

Answer0, −1/√2, 1/√2

Watch this explained “A minus sign is not a mistake”, 11:36 into Three angles with the axes, and why their cosines square up to one

Open NCERT p. 381Checked by computerAnswers can differ: one example

1. Equal angles means equal cosines, so let l = m = n = c.
2. Use the identity l² + m² + n² = 1: c² + c² + c² = 1.
3. 3c² = 1, so c² = 1/3.
4. Take the square root of both sides now, not before: c = ±1/√3.
5. So the direction cosines are 1/√3, 1/√3, 1/√3 (or all three negative).

Answer±1/√3, ±1/√3, ±1/√3 (all three the same sign)

Watch this explained “Equal angles with all three”, 12:56 into Three angles with the axes, and why their cosines square up to one

Open NCERT p. 381Matches NCERT’s answer

1. To turn direction ratios a, b, c into direction cosines, divide each by √(a² + b² + c²).
2. Here a = −18, b = 12, c = −4.
3. a² + b² + c² = 324 + 144 + 16 = 484, and √484 = 22.
4. Divide each ratio by 22: −18/22, 12/22, −4/22.
5. Simplify: −9/11, 6/11, −2/11.
6. Check: 81/121 + 36/121 + 4/121 = 121/121 = 1. Correct.

Answer−9/11, 6/11, −2/11

Watch this explained “Two worked recoveries”, 6:45 into Direction ratios as any proportional triple, and normalising back to cosines

Open NCERT p. 381One way to think about it

1. Call the points A(2, 3, 4), B(–1, –2, 1), C(5, 8, 7).
2. Find the direction ratios of AB: (−1−2, −2−3, 1−4) = (−3, −5, −3).
3. Find the direction ratios of BC: (5−(−1), 8−(−2), 7−1) = (6, 10, 6).
4. Check if BC's ratios are a multiple of AB's ratios: 6 = (−2)×(−3), 10 = (−2)×(−5), 6 = (−2)×(−3).
5. Yes, every entry matches with the same multiplier −2, so AB is parallel to BC.
6. Point B lies on both AB and BC (it is the shared end point).
7. A line parallel to BC through the shared point B must be the same line as BC, so A, B, C all lie on one line.
8. Therefore A, B, C are collinear.

In shortA, B, C are collinear, since AB and BC have proportional direction ratios and share the point B.

Watch this explained “The same verdict, from any pair”, 11:27 into Direction ratios as any proportional triple, and normalising back to cosines

Open NCERT p. 381Matches NCERT’s answer

1. Call the vertices A(3, 5, –4), B(–1, 1, 2), C(–5, –5, –2).
2. Side AB: gaps are (–1–3, 1–5, 2–(–4)) = (–4, –4, 6). Length = √(16+16+36) = √68 = 2√17.
3. Direction cosines of AB: –4/(2√17), –4/(2√17), 6/(2√17), i.e. –2/√17, –2/√17, 3/√17.
4. Side BC: gaps are (–5–(–1), –5–1, –2–2) = (–4, –6, –4). Length = √(16+36+16) = √68 = 2√17.
5. Direction cosines of BC: –4/(2√17), –6/(2√17), –4/(2√17), i.e. –2/√17, –3/√17, –2/√17.
6. Side CA: gaps are (3–(–5), 5–(–5), –4–(–2)) = (8, 10, –2). Length = √(64+100+4) = √168 = 2√42.
7. Direction cosines of CA: 8/(2√42), 10/(2√42), –2/(2√42), i.e. 4/√42, 5/√42, –1/√42.
8. Along the way, AB and BC came out the same length, 2√17 — so the triangle is isosceles, though the question never asked that.

AnswerAB: −2/√17, −2/√17, 3/√17; BC: −2/√17, −3/√17, −2/√17; CA: 4/√42, 5/√42, −1/√42

Watch this explained “Two worked cases”, 15:38 into Three angles with the axes, and why their cosines square up to one