This MCQ module is based on: Distance Between Two Points in 3D and Exercises
TOPIC 35 OF 45
Distance Between Two Points in 3D and Exercises
🎓 Class 11
Mathematics
CBSE
Theory
Ch 11 — Introduction to Three Dimensional Geometry
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Distance Between Two Points in 3D and Exercises
Targeting Class 11 level in Coordinate Geometry, with Advanced difficulty.
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11.4 Distance Between Two Points
In the 2D plane, the distance between two points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) is obtained using the Pythagoras theorem: \[PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\] We now extend this idea to three dimensions. Given two points \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\), we derive a closed-form formula for \(PQ\) using the Pythagoras theorem applied twice.
Derivation of the 3D Distance Formula
Construction
Through \(P\) and \(Q\), draw planes parallel to the coordinate planes, so that a rectangular parallelepiped (cuboid) is formed whose opposite vertices are \(P\) and \(Q\). The edges of this cuboid are parallel to the coordinate axes and have lengths \(|x_2-x_1|\), \(|y_2-y_1|\), \(|z_2-z_1|\).
Fig 11.3: Rectangular parallelepiped with \(P\) and \(Q\) at opposite vertices; \(R\) is the foot of the perpendicular from \(Q\) to the plane through \(P\) parallel to the XY-plane.
Step 1. Let \(R\) be the vertex of the cuboid such that \(PR\) is a diagonal of the "base" rectangle (the rectangle through \(P\) whose plane is parallel to the XY-plane) and \(RQ\) is the vertical edge. Then: \[PR^2=(x_2-x_1)^2+(y_2-y_1)^2 \qquad (\text{Pythagoras in the base plane}).\]
Step 2. Since \(RQ\) is perpendicular to the base plane, triangle \(PRQ\) is right-angled at \(R\): \[PQ^2=PR^2+RQ^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2.\]
Distance Formula in Three Dimensions
The distance? between \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) is:
\[\boxed{\;PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\;}\]
Distance from the Origin
Taking \(P=O(0,0,0)\) and \(Q(x,y,z)\), the formula collapses to: \[OQ=\sqrt{x^2+y^2+z^2}.\]
Example 4
Find the distance between the points \(P(1,-3,4)\) and \(Q(-4,1,2)\).
\(PQ=\sqrt{(-4-1)^2+(1-(-3))^2+(2-4)^2}=\sqrt{25+16+4}=\sqrt{45}=3\sqrt{5}\) units.
Example 5
Show that the points \(P(-2,3,5)\), \(Q(1,2,3)\), and \(R(7,0,-1)\) are collinear.
\(PQ=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}=\sqrt{9+1+4}=\sqrt{14}\).
\(QR=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{36+4+16}=\sqrt{56}=2\sqrt{14}\).
\(PR=\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{81+9+36}=\sqrt{126}=3\sqrt{14}\).
Since \(PQ+QR = \sqrt{14}+2\sqrt{14}=3\sqrt{14}=PR\), the three points are collinear.
\(QR=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{36+4+16}=\sqrt{56}=2\sqrt{14}\).
\(PR=\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{81+9+36}=\sqrt{126}=3\sqrt{14}\).
Since \(PQ+QR = \sqrt{14}+2\sqrt{14}=3\sqrt{14}=PR\), the three points are collinear.
Example 6
Are the points \(A(3,6,9)\), \(B(10,20,30)\) and \(C(25,-41,5)\) the vertices of a right-angled triangle?
\(AB^2=(10-3)^2+(20-6)^2+(30-9)^2=49+196+441=686\).
\(BC^2=(25-10)^2+(-41-20)^2+(5-30)^2=225+3721+625=4571\).
\(CA^2=(3-25)^2+(6+41)^2+(9-5)^2=484+2209+16=2709\).
Check: \(AB^2+CA^2=686+2709=3395\neq 4571=BC^2\); \(AB^2+BC^2=686+4571=5257\neq 2709=CA^2\); \(BC^2+CA^2=4571+2709=7280\neq 686=AB^2\). No, the triangle is not right-angled.
\(BC^2=(25-10)^2+(-41-20)^2+(5-30)^2=225+3721+625=4571\).
\(CA^2=(3-25)^2+(6+41)^2+(9-5)^2=484+2209+16=2709\).
Check: \(AB^2+CA^2=686+2709=3395\neq 4571=BC^2\); \(AB^2+BC^2=686+4571=5257\neq 2709=CA^2\); \(BC^2+CA^2=4571+2709=7280\neq 686=AB^2\). No, the triangle is not right-angled.
Example 7
Find the equation of the set of points \(P(x,y,z)\) which are equidistant from the points \(A(1,2,3)\) and \(B(3,2,-1)\).
Condition: \(PA=PB\), i.e., \(PA^2=PB^2\).
\((x-1)^2+(y-2)^2+(z-3)^2=(x-3)^2+(y-2)^2+(z+1)^2\).
Expand: \(x^2-2x+1+z^2-6z+9=x^2-6x+9+z^2+2z+1\).
Simplify: \(-2x-6z+10=-6x+2z+10\) ⟹ \(4x-8z=0\) ⟹ \(\boxed{x-2z=0}\). This is the equation of a plane — the perpendicular bisector of \(AB\).
Exercise 11.2 (Selected Problems)
Q1. Find the distance between the following pairs of points: (i) \((2,3,5)\) and \((4,3,1)\); (ii) \((-3,7,2)\) and \((2,4,-1)\); (iii) \((-1,3,-4)\) and \((1,-3,4)\); (iv) \((2,-1,3)\) and \((-2,1,3)\).
(i) \(\sqrt{4+0+16}=\sqrt{20}=2\sqrt 5\).
(ii) \(\sqrt{25+9+9}=\sqrt{43}\).
(iii) \(\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}\).
(iv) \(\sqrt{16+4+0}=\sqrt{20}=2\sqrt 5\).
(ii) \(\sqrt{25+9+9}=\sqrt{43}\).
(iii) \(\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}\).
(iv) \(\sqrt{16+4+0}=\sqrt{20}=2\sqrt 5\).
Q2. Show that the points \((-2,3,5)\), \((1,2,3)\) and \((7,0,-1)\) are collinear.
Refer Example 5: the sum of two distances equals the third, proving collinearity.
Q3. Verify whether \(A(0,7,10)\), \(B(-1,6,6)\), \(C(-4,9,6)\) form an isosceles right triangle.
\(AB^2=1+1+16=18\); \(BC^2=9+9+0=18\); \(CA^2=16+4+16=36\).
\(AB=BC=\sqrt{18}\) ⟹ isosceles. Also \(AB^2+BC^2=18+18=36=CA^2\) ⟹ right-angled at \(B\). Hence \(ABC\) is an isosceles right triangle.
\(AB=BC=\sqrt{18}\) ⟹ isosceles. Also \(AB^2+BC^2=18+18=36=CA^2\) ⟹ right-angled at \(B\). Hence \(ABC\) is an isosceles right triangle.
Q4. Find the equation of the set of points equidistant from \((1,2,3)\) and \((3,2,-1)\).
See Example 7: \(x-2z=0\).
Q5. Find the equation of the set of points \(P\), the sum of whose distances from \(A(4,0,0)\) and \(B(-4,0,0)\) is equal to 10.
\(PA+PB=10\). Let \(P(x,y,z)\).
\(\sqrt{(x-4)^2+y^2+z^2}=10-\sqrt{(x+4)^2+y^2+z^2}\).
Square: \((x-4)^2+y^2+z^2=100-20\sqrt{(x+4)^2+y^2+z^2}+(x+4)^2+y^2+z^2\).
Simplify: \(-16x-100=-20\sqrt{(x+4)^2+y^2+z^2}\) ⟹ \(\sqrt{(x+4)^2+y^2+z^2}=\tfrac{100+16x}{20}=5+\tfrac{4x}{5}\).
Square again: \((x+4)^2+y^2+z^2=25+8x+\tfrac{16x^2}{25}\).
Expand: \(x^2+8x+16+y^2+z^2=25+8x+\tfrac{16x^2}{25}\) ⟹ \(\tfrac{9x^2}{25}+y^2+z^2=9\) ⟹ \(\boxed{\tfrac{x^2}{25}+\tfrac{y^2}{9}+\tfrac{z^2}{9}=1}\). An ellipsoid of revolution.
Activity: Measure the Space Diagonal
L3 ApplyMaterials: A shoe-box (cuboid), a flexible measuring tape, a ruler, a calculator.
Predict: Before measuring, estimate the length of the longest straight line that fits inside the box (corner-to-opposite-corner).
- Measure the three edge lengths of the shoe-box: \(a\) (length), \(b\) (breadth), \(c\) (height).
- Compute the space-diagonal length using \(d=\sqrt{a^2+b^2+c^2}\).
- Now place one corner at the origin. Write the coordinates of all 8 corners of the box.
- Use the distance formula to verify the space diagonal between \((0,0,0)\) and \((a,b,c)\).
- Finally, physically measure the diagonal with the tape. Compare with the calculated value.
The physical measurement should closely match \(\sqrt{a^2+b^2+c^2}\). The tiny difference is due to measurement error. This is exactly the 3D distance formula in action — and why television diagonals, cable lengths through rooms, and pack-ability of long rods into boxes are all computed this way.
Competency-Based Questions
Scenario: A warehouse uses 3D coordinates to track packages. Package \(P_1\) is at \((3,4,12)\), \(P_2\) at \((0,0,0)\) (the loading dock), and \(P_3\) at \((6,8,24)\). All distances are in metres.
Q1. What is the distance from the dock \(P_2\) to package \(P_1\)?
L3 ApplyAnswer: (a) 13 m. \(\sqrt{9+16+144}=\sqrt{169}=13\).
Q2. Analyse: are the three packages collinear? Show your working.
L4 AnalyseAnswer: \(P_2P_1=13\); \(P_1P_3=\sqrt{9+16+144}=13\); \(P_2P_3=\sqrt{36+64+576}=\sqrt{676}=26\). Since \(P_2P_1+P_1P_3=13+13=26=P_2P_3\), they are collinear, with \(P_1\) the mid-point.
Q3. Evaluate: the manager claims the shortest forklift path from the dock to \(P_3\) via \(P_1\) adds no extra distance compared with going directly. Is the manager correct? Justify.
L5 EvaluateAnswer: Since the three points are collinear (Q2), the path \(P_2\to P_1\to P_3\) has total length \(13+13=26\) m, the same as the direct distance \(P_2P_3=26\) m. Hence the manager is correct — there is no detour penalty.
Q4. Create a new package location \(P_4\) such that its distance from the dock is exactly 26 m but it is NOT on the line \(P_2P_1P_3\). Provide coordinates and verify.
L6 CreateSample: \(P_4=(24,0,10)\). Check: \(\sqrt{576+0+100}=\sqrt{676}=26\) ✓. For non-collinearity, the ratio \(24:0:10\) differs from \(3:4:12\) (zero y-component breaks the ratio), so \(P_4\) is off the line. Many valid answers exist (any point on the sphere \(x^2+y^2+z^2=676\) not lying on the line \(x/3=y/4=z/12\)).
Assertion–Reason Questions
Assertion (A): The distance from the origin to \((3,-4,12)\) is 13 units.
Reason (R): Distance from origin in 3D equals \(\sqrt{x^2+y^2+z^2}\).
Reason (R): Distance from origin in 3D equals \(\sqrt{x^2+y^2+z^2}\).
Answer: (a). \(\sqrt{9+16+144}=\sqrt{169}=13\). R is the formula that gives 13, so R explains A.
Assertion (A): Three non-collinear points in 3D always form a triangle.
Reason (R): Any three points in 3D satisfy the triangle inequality.
Reason (R): Any three points in 3D satisfy the triangle inequality.
Answer: (c). A is true. R is false because three collinear points satisfy triangle inequality with equality, yet form no triangle. The correct reason is non-collinearity.
Assertion (A): The locus of points equidistant from two fixed points in 3D is a plane.
Reason (R): The perpendicular bisector of a segment in 3D is a plane.
Reason (R): The perpendicular bisector of a segment in 3D is a plane.
Answer: (a). Both statements are true, and R is the geometric reason behind A, as verified in Example 7.
Frequently Asked Questions
How is the 3D distance formula derived?
Apply the 2D Pythagoras theorem twice - once in the horizontal plane to get the horizontal distance, then combine with the vertical difference to get the full 3D distance.
How do you check if three 3D points are collinear?
Compute the three pairwise distances. If the largest equals the sum of the other two, the points are collinear; otherwise they form a triangle.
What is the distance of P(x, y, z) from the origin?
Distance from origin = root( x squared + y squared + z squared ).
What is the summary of Class 11 Chapter 11?
Chapter 11 introduces 3D coordinate axes, coordinate planes, octants, the distance formula between two points in space and a preview of the section formula.
Where do we use 3D coordinate geometry in real life?
3D coordinates are used in computer graphics, GPS positioning, robotics, aeronautics (aircraft position) and molecular chemistry (atomic positions).
Is 3D geometry important for Class 12 and JEE?
Yes. Class 12 Chapter 11 extends to vectors and planes in space, and 3D geometry contributes significant marks in JEE Main and Advanced.
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