Questions: Three Dimensional Geometry
Introduction to Three Dimensional Geometry22 min readLesson 4 of 5
Q1. Name the octant in which the point (1, 2, 3) lies.
Answer: All three coordinates are positive, so the point lies in the first octant. The magnitudes do not matter for the octant, only the signs.
Hinglish: Teeno coordinates positive hain, isliye point pehle octant me hai.
Q2. Name the octant containing the point (minus 3, 1, 2).
Answer: The signs are minus, plus, plus. Matching with the sign table, x negative with y and z positive is the second octant.
Hinglish: Signs (−, +, +) hain, jo second octant ka pattern hai.
Q3. Find the distance between the points (1, 2, 3) and (4, 6, 3).
Answer: The z-coordinates are equal, so the distance is the square root of (4 minus 1) squared plus (6 minus 2) squared plus 0, which is the square root of 9 plus 16, equal to 5.
Hinglish: z same hai, isliye √(3² + 4²) = 5. Distance 5 units.
Q4. Find the distance of the point (6, 6, 7) from the origin.
Answer: The distance is the square root of 36 plus 36 plus 49, which is the square root of 121, equal to 11.
Hinglish: √(6² + 6² + 7²) = √121 = 11 units.
Q5. Find the mid-point of the segment joining (2, 4, 6) and (0, minus 2, minus 5).
Answer: Average each coordinate: x is 1, y is 1 and z is 1 by 2. The mid-point is (1, 1, 1 by 2).
Hinglish: Har coordinate ka average: ((2+0)/2, (4−2)/2, (6−5)/2) = (1, 1, 1/2).
Q6. Find the centroid of the triangle with vertices (0, 4, 1), (2, 3, minus 1) and (4, 5, 0).
Answer: The centroid coordinates are the averages: x is 2, y is 4 and z is 0. So the centroid is (2, 4, 0).
Hinglish: Centroid = ((0+2+4)/3, (4+3+5)/3, (1−1+0)/3) = (2, 4, 0).
Q7. Show that the points (0, 4, 1), (2, 3, minus 1) and (4, 5, 0) form a right-angled triangle.
Answer: The squared lengths are AB squared equal to 4 plus 1 plus 4 equal to 9, BC squared equal to 4 plus 4 plus 1 equal to 9 and AC squared equal to 16 plus 1 plus 1 equal to 18. Since 9 plus 9 equals 18, the triangle is right-angled at B.
Hinglish: AB² = 9, BC² = 9, AC² = 18 aur 9 + 9 = 18, isliye right angle B par hai.
Q8. Find the coordinates of the point on the x-axis that is equidistant from (1, 2, 3) and (4, 6, 3).
Answer: A point on the x-axis has the form (x, 0, 0). Equating the squared distances gives (x minus 1) squared plus 4 plus 9 equal to (x minus 4) squared plus 36 plus 9. Expanding and cancelling gives 6x equal to 51, so x is 8.5 and the point is (8.5, 0, 0).
Hinglish: Point (x, 0, 0) maano aur dono distances barabar rakho: solve karne par x = 8.5, point (8.5, 0, 0).
Q9. Show that the points (1, minus 1, 3), (2, minus 4, 5) and (5, minus 13, 11) are collinear.
Answer: The squared distances are AB squared equal to 1 plus 9 plus 4 equal to 14, BC squared equal to 9 plus 81 plus 36 equal to 126 and AC squared equal to 16 plus 144 plus 64 equal to 224. The distances are root 14, 3 root 14 and 4 root 14, and the largest equals the sum of the other two, so the points are collinear with B between A and C.
Hinglish: Distances √14, 3√14 aur 4√14 hain; sabse badi baaki do ka sum hai, isliye points collinear hain.
Q10. The centroid of a triangle is the origin and two vertices are (2, 4, 6) and (0, minus 2, minus 5). Find the third vertex.
Answer: The sum of the three vertices must be zero because the centroid is the origin. So the third vertex is the negative of the sum of the given two, which is (minus 2, minus 2, minus 1).
Hinglish: Centroid origin hai to vertices ka sum zero hoga; third vertex = −[(2,4,6)+(0,−2,−5)] = (−2,−2,−1).
Before writing the final answer to any distance question, count the terms: a space distance must combine three squared differences, so an answer built from two terms is automatically wrong. This count is faster than re-deriving the formula.
When a question asks for a ratio or a trisection point, find both coordinates and then check that the midpoint of the two trisection points is the midpoint of the segment. That single check catches almost every arithmetic slip in section questions.
A working habit for the questions page: draw a small sketch of the axes and mark the given points roughly before substituting anything. The sketch shows which coordinate must change, and it catches a wrong point on an axis or plane almost immediately.
A final habit: after every distance computation, write the three squared differences in a row and add them only once. Most errors in this chapter come from squaring one difference and then adding it twice, which is invisible in a calculator answer but shows in the working.
Two more checks before writing the final answer. First, substitute the computed point back into the condition that produced it, such as equal distances or a given centroid. Second, confirm the units and the number of coordinates: a space answer always carries three numbers, and an answer with two numbers cannot be right.
Common mistakes on this chapter
The z-coordinate is the one most often dropped in the distance formula. The sign pattern for octants is often confused with the name of the planes, so write the three signs before naming the octant. In centroid questions, forgetting that the sum of the coordinates is three times the centroid gives a wrong third vertex. In collinearity questions, using only two distances gives no information, because two distances always exist; the test needs all three and the additive relation.
Isosceles and right-angled questions need the three squared lengths before any conclusion. Keep the squares rather than the roots, because comparing squares avoids surds and reduces errors.
Quiz
The quiz below has five multiple choice questions on octants, distance, midpoint and centroid. Attempt it after the worked questions above, then move to the previous year page.