Chapter 11 — INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

Theory Used

In three-dimensional geometry, the distance between two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) is obtained using the extension of the Pythagoras theorem in space:

\[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \]

This formula represents the length of the line segment joining the two points in 3D space.

Solution Roadmap

• Identify coordinates of both points clearly.
• Apply distance formula directly.
• Compute coordinate differences carefully (sign errors are common).
• Square → Add → Simplify square root.

Solution

(i) Distance between \( (2, 3, 5) \) and \( (4, 3, 1) \):

\[ \begin{aligned} d&=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}\\ &=\sqrt{2^2+0^2+(-4)^2}\\ &=\sqrt{4+16}\\ &=2\sqrt{5} \end{aligned} \]

(ii) Distance between \( (-1, 3, -4) \) and \( (1, -3, 4) \):

\[ \begin{aligned} d&=\sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2}\\ &=\sqrt{2^2+(-6)^2+8^2}\\ &=\sqrt{4+36+64}\\ &=2\sqrt{26} \end{aligned} \]

(iii) Distance between \( (-3, 7, 2) \) and \( (2, 4, -1) \):

\[ \begin{aligned} d&=\sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2}\\ &=\sqrt{5^2+(-3)^2+(-3)^2}\\ &=\sqrt{25+9+9}\\ &=\sqrt{43} \end{aligned} \]

(iv) Distance between \( (2, -1, 3) \) and \( (-2, 1, 3) \):

\[ \begin{aligned} d&=\sqrt{(-2-2)^2+(1-(-1))^2+(3-3)^2}\\ &=\sqrt{(-4)^2+2^2+0^2}\\ &=\sqrt{16+4}\\ &=2\sqrt{5} \end{aligned} \]

Exam Significance

• This is a foundational formula used throughout 3D geometry.
• Frequently appears directly in CBSE board exams (1–3 mark questions).
• Acts as a base for vector geometry, section formula, and direction cosines.
• In JEE/NEET, often used inside multi-step problems involving lines and planes.
• Accuracy in sign handling is critical for competitive exams.

Theory Used

Three points in space are said to be collinear if they lie on the same straight line. This can be verified using either of the following conditions:

• Distance Method: If \( AB + BC = AC \), then points are collinear.
• Vector Method: If \( \vec{AB} = \lambda \vec{BC} \), then points are collinear.

In this solution, we use the distance method, which is straightforward and commonly used in board exams.

Solution Roadmap

• Assign coordinates to points A, B, and C.
• Compute distances AB, BC, and AC using 3D distance formula.
• Check whether \( AB + BC = AC \).
• Conclude collinearity based on equality.

Solution

Let the given points be \(A(-2, 3, 5)\), \(B(1, 2, 3)\) and \(C(7, 0, -1)\).

Distance between \(A\) and \(B\):

\[ \begin{aligned} AB&=\sqrt{(1-(-2))^2+(2-3)^2+(3-5)^2}\\ &=\sqrt{3^2+(-1)^2+(-2)^2}\\ &=\sqrt{9+1+4}\\ &=\sqrt{14} \end{aligned} \]

Distance between \(B\) and \(C\):

\[ \begin{aligned} BC&=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\\ &=\sqrt{6^2+(-2)^2+(-4)^2}\\ &=\sqrt{36+4+16}\\ &=2\sqrt{14} \end{aligned} \]

Distance between \(A\) and \(C\):

\[ \begin{aligned} AC&=\sqrt{(7-(-2))^2+(0-3)^2+(-1-5)^2}\\ &=\sqrt{9^2+(-3)^2+(-6)^2}\\ &=\sqrt{81+9+36}\\ &=3\sqrt{14} \end{aligned} \]

Since \( AC = AB + BC \), we get

\[ 3\sqrt{14} = \sqrt{14} + 2\sqrt{14} \]

Hence, point \(B\) lies on the line segment joining \(A\) and \(C\). Therefore, the given points are collinear.

Exam Significance

• Very common 2–3 mark CBSE board question.
• Tests understanding of distance formula and geometric interpretation.
• Vector method may be preferred in JEE for faster verification.
• Forms base for advanced topics like direction ratios and line equations.
• High probability concept in coordinate geometry mixed problems.

Theory Used

• Isosceles Triangle: Two sides are equal.
• Right Angled Triangle: Follows Pythagoras theorem \(a^2 + b^2 = c^2\).
• Parallelogram: Opposite sides are equal OR diagonals bisect each other.

Solution Roadmap

• Use 3D distance formula to compute side lengths.
• Compare lengths (or squares) depending on requirement.
• Apply geometric condition (equal sides / Pythagoras / opposite sides).

Solution

(i) Let \(A(0,7,-10)\), \(B(1,6,-6)\), \(C(4,9,-6)\)

\[ \begin{aligned} AB&=\sqrt{18}, \quad BC=\sqrt{18}, \quad AC=\sqrt{36} \end{aligned} \]

Since \(AB = BC\), the triangle is isosceles.

(ii) Let \(A(0,7,10)\), \(B(-1,6,6)\), \(C(-4,9,6)\)

\[ AB^2=18,\quad BC^2=18,\quad AC^2=36 \]

\[ AB^2 + BC^2 = AC^2 \]

Hence, the triangle is right angled at point \(B\).

(iii) Let \(A(-1,2,1)\), \(B(1,-2,5)\), \(C(4,-7,8)\), \(D(2,-3,4)\)

\[ AB=6,\quad CD=6,\quad BC=\sqrt{43},\quad AD=\sqrt{43} \]

Since opposite sides are equal, \(AB = CD\) and \(BC = AD\), the points form a parallelogram.

Exam Significance

• Very important mixed-concept problem (triangle + quadrilateral).
• Frequently asked in CBSE boards (3–5 marks).
• Right-angle verification is a direct application of Pythagoras (JEE staple).
• Parallelogram verification may also be done using vector method (faster in JEE).
• Builds strong base for vectors, direction ratios, and 3D geometry.

Theory Used

The set (locus) of all points equidistant from two fixed points in space forms a plane. This plane is the perpendicular bisector of the line segment joining the two points.

Algebraically, if a point \(P(x,y,z)\) is equidistant from points \(A\) and \(B\), then:

\[ PA = PB \]

Using distance formula and simplifying gives the required plane equation.

Solution Roadmap

• Assume a general point \(P(x,y,z)\).
• Apply condition \(PA = PB\).
• Square both sides to eliminate roots.
• Simplify and reduce to linear equation.

Solution

Let \(P(x, y, z)\) be a point equidistant from \(A(1,2,3)\) and \(B(3,2,-1)\).

\[ \sqrt{(x-1)^2+(y-2)^2+(z-3)^2} = \sqrt{(x-3)^2+(y-2)^2+(z+1)^2} \]

Squaring both sides:

\[ (x-1)^2+(y-2)^2+(z-3)^2 = (x-3)^2+(y-2)^2+(z+1)^2 \]

Cancelling \((y-2)^2\) and expanding:

\[ \begin{aligned} (x^2-2x+1)-(x^2-6x+9) + (z^2-6z+9)-(z^2+2z+1) &= 0 \\ 4x - 8 - 8z + 8 &= 0 \\ 4x - 8z &= 0 \\ x - 2z &= 0 \end{aligned} \]

Hence, the required equation of the locus is:

\[ x - 2z = 0 \]

Exam Significance

• Standard CBSE board question (3 marks).
• Concept of locus → very important for higher coordinate geometry.
• Frequently used in JEE in plane and vector-based problems.
• Understanding geometric meaning (perpendicular bisector plane) gives strong conceptual advantage.
• Shortcut: symmetric cancellation reduces calculation time significantly.

Theory Used

The locus of a point whose sum of distances from two fixed points is constant represents an ellipsoid in three-dimensional geometry.

Here, A and B are symmetric about the origin on the x-axis, so the resulting surface will be centered at the origin and aligned along coordinate axes.

Solution Roadmap

• Let \(P(x,y,z)\) be a general point.
• Write expressions for \(PA\) and \(PB\).
• Apply condition \(PA + PB = 10\).
• Eliminate radicals carefully by squaring twice.
• Simplify into standard quadratic form.

Solution

Let \(P(x, y, z)\) be any point. Then,

\[ \begin{aligned} PA &=\sqrt{(x-4)^2+y^2+z^2}, \\ PB &=\sqrt{(x+4)^2+y^2+z^2} \end{aligned} \]

\[ \sqrt{(x-4)^2+y^2+z^2} + \sqrt{(x+4)^2+y^2+z^2} = 10 \]

Isolating and squaring:

\[ (x-4)^2+y^2+z^2 = 100 + (x+4)^2+y^2+z^2 - 20\sqrt{(x+4)^2+y^2+z^2} \]

\[ (x-4)^2 - (x+4)^2 = 100 - 20PB \]

\[ \begin{aligned} -16x &= 100 - 20PB \\\Rightarrow PB &= 5 + \frac{4x}{5} \end{aligned} \]

Squaring again:

\[ (x+4)^2 + y^2 + z^2 = \left(5 + \frac{4x}{5}\right)^2 \]

\[ 25(x^2+8x+16+y^2+z^2) = (25+4x)^2 \]

\[ 25x^2+200x+400+25y^2+25z^2 = 625+200x+16x^2 \]

\[ 9x^2 + 25y^2 + 25z^2 = 225 \]

Hence, the required equation is:

\[ 9x^2 + 25y^2 + 25z^2 = 225 \]

Exam Significance

• Classic locus problem → very important for CBSE (3–5 marks).
• Direct extension of ellipse concept from 2D to 3D.
• Frequently appears in JEE in disguised forms (distance constraints).
• Understanding symmetry simplifies problems significantly.
• Double squaring technique is a key algebraic skill for competitive exams.