Conic Sections

In three-dimensional geometry, the distance between two points represents the length of the straight line segment joining them. Since space has three mutually perpendicular directions, the distance must account for displacement along all three axes.

Definition

Let \( A(x_1,y_1,z_1) \) and \( B(x_2,y_2,z_2) \). The distance between them is denoted by \( AB \).

Key Idea

The displacement from A to B can be decomposed into three perpendicular components:

• \(\Delta x = x_2 - x_1\)
• \(\Delta y = y_2 - y_1\)
• \(\Delta z = z_2 - z_1\)

Step-by-Step Derivation (No Shortcut)

Step 1: Projection on \(xy\)-plane

Consider projections of A and B onto the \(xy\)-plane. Distance between projections:

\[ d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]

Step 2: Introduce vertical separation

Now include height difference \( |z_2 - z_1| \). A right triangle is formed in space:

\[ AB^2 = d^2 + (z_2 - z_1)^2 \]

Step 3: Substitute

\[ AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 \]

Final Result

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

Geometric Interpretation

The two points form opposite corners of a rectangular box. The distance \(AB\) is the space diagonal of that box.

Vector Interpretation (Advanced Insight)

Position vectors:

\[ \vec{A} = (x_1,y_1,z_1), \quad \vec{B} = (x_2,y_2,z_2) \]

Displacement vector:

\[ \vec{AB} = (x_2-x_1,\; y_2-y_1,\; z_2-z_1) \]

Distance is magnitude:

\[ AB = |\vec{AB}| \]

This connects coordinate geometry directly to vector algebra.

3D Visualization (Rectangular Box Model)

Reduction Cases (Very Important)

• If \(z_1 = z_2\) → reduces to 2D distance
• If \(y_1 = y_2, z_1 = z_2\) → reduces to 1D distance
• If one point is origin: \[ OP = \sqrt{x^2 + y^2 + z^2} \]

Important Properties

• \(AB \ge 0\)
• \(AB = 0 \Rightarrow A \equiv B\)
• Symmetry: \(AB = BA\)
• Independent of coordinate system orientation

Exam + JEE Insights

• Always use coordinate differences (avoid sign mistakes)
• Recognize rectangular box model instantly
• Distance = magnitude of displacement vector
• Frequently used in section formula & vectors

Concept Bridge

• Section formula (3D)
• Direction ratios
• Vector magnitude
• Equation of line in space

In Fig 11.3, if \(P(2,4,5)\), find the coordinates of point \(F\).

Concept Used

The point \(F\) lies on the plane where the distance measured along the \(y\)-axis is zero. This means:

\[ \text{Point lies on } xz\text{-plane} \Rightarrow y = 0 \]

Step-by-Step Solution

Given point:

\[ P = (2,4,5) \]

Interpretation:

• \(x = 2\) → distance from \(yz\)-plane
• \(y = 4\) → distance from \(xz\)-plane
• \(z = 5\) → distance from \(xy\)-plane

To obtain point \(F\), we drop a perpendicular from \(P\) onto the \(xz\)-plane.

This removes the \(y\)-component but keeps \(x\) and \(z\) unchanged.

Final Answer

\[ \boxed{F = (2,0,5)} \]

Geometric Interpretation

Point \(F\) is the projection of \(P\) on the \(xz\)-plane. This is obtained by drawing a perpendicular from \(P\) to the plane \(y = 0\).

Visualization

General Rule (Very Important)

• Projection on \(xy\)-plane → set \(z = 0\)
• Projection on \(yz\)-plane → set \(x = 0\)
• Projection on \(xz\)-plane → set \(y = 0\)

Exam + JEE Insight

• Projection questions are very common
• Never change all coordinates — only one becomes zero
• Understand plane equations: \(x=0, y=0, z=0\)

Find the octant in which the points \( (-3,1,2) \) and \( (-3,1,-2) \) lie.

Core Concept

The octant of a point depends solely on the signs of its coordinates:

\[ (x,y,z) \Rightarrow (\text{sign of }x,\; \text{sign of }y,\; \text{sign of }z) \]

Systematic Method (Error-Free)

1. Identify sign of each coordinate
2. Write sign triplet \((\pm,\pm,\pm)\)
3. Match with standard octant table

Step 1: Point \( (-3,1,2) \)

Signs:

\[ (-,\; +,\; +) \]

This corresponds to the Second Octant.

Step 2: Point \( (-3,1,-2) \)

Signs:

\[ (-,\; +,\; -) \]

This corresponds to the Sixth Octant.

Final Answer

\[ (-3,1,2) \text{ lies in Second Octant} \] \[ (-3,1,-2) \text{ lies in Sixth Octant} \]

Reference Octant Table

Visualization of Sign Change

Memory Trick

First 4 octants → \(z > 0\) Last 4 octants → \(z < 0\)

Then follow 2D quadrant logic for \(x\) and \(y\).

Exam + JEE Insights

• Do NOT memorize blindly — use sign logic
• Most errors occur due to sign misreading
• Negative sign in front of number must be handled carefully
• Quick check: \(z\) decides upper/lower half

Find the distance between the points \(P(1,-3,4)\) and \(Q(-4,1,2)\).

Given

\[ P = (1,-3,4), \quad Q = (-4,1,2) \]

Formula Used

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

Step-by-Step Calculation

Step 1: Compute coordinate differences

• \(x_2 - x_1 = -4 - 1 = -5\)
• \(y_2 - y_1 = 1 - (-3) = 4\)
• \(z_2 - z_1 = 2 - 4 = -2\)

Step 2: Square the differences

\[ (-5)^2 = 25,\quad 4^2 = 16,\quad (-2)^2 = 4 \]

Step 3: Add

\[ 25 + 16 + 4 = 45 \]

Step 4: Take square root

\[ d = \sqrt{45} = 3\sqrt{5} \]

Final Answer

\[ \boxed{PQ = 3\sqrt{5}} \]

Geometric Interpretation

The points \(P\) and \(Q\) form opposite vertices of a rectangular box. The distance \(PQ\) is the space diagonal of that box.

Vector Interpretation

Displacement vector:

\[ \vec{PQ} = (-5,\;4,\;-2) \]

Distance = magnitude:

\[ |\vec{PQ}| = \sqrt{45} \]

Visualization

Common Mistake (Very Important)

Do NOT write \( (-4 - 1)^2 = 5^2 \) directly without sign check. Always compute difference first → then square.

Exam + JEE Insights

• Distance = magnitude of displacement vector
• Always compute differences carefully (sign errors common)
• Simplify radicals properly: \( \sqrt{45} = 3\sqrt{5} \)
• Visualize rectangular box → helps in derivation questions

Show that the points \(P(-2,3,5)\), \(Q(1,2,3)\), and \(R(7,0,-1)\) are collinear.

Method 1: Distance Approach

If three points are collinear, then:

\[ PQ + QR = PR \]

Step 1: Find \(PQ\)

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

Step 2: Find \(QR\)

\[ \begin{aligned} QR &= \sqrt{(7-1)^2 + (0-2)^2 + (-1-3)^2}\\\\ &= \sqrt{6^2 + (-2)^2 + (-4)^2}\\\\ &= \sqrt{56} = 2\sqrt{14} \end{aligned} \]

Step 3: Find \(PR\)

\[ \begin{aligned} PR &= \sqrt{(7+2)^2 + (0-3)^2 + (-1-5)^2}\\\\ &= \sqrt{9^2 + (-3)^2 + (-6)^2}\\\\ &= \sqrt{126} = 3\sqrt{14} \end{aligned} \]

Step 4: Verify

\[ \begin{aligned} PQ + QR &= \sqrt{14} + 2\sqrt{14} \\\\&= 3\sqrt{14} \\\\&= PR \end{aligned} \]

Hence, the points are collinear.

Method 2: Vector Approach (Stronger & Faster)

Compute direction vectors:

\[ \begin{aligned} \vec{PQ}& = (3,-1,-2), \\\\ \vec{QR} &= (6,-2,-4) \end{aligned} \]

Observe:

\[ \vec{QR} = 2 \cdot \vec{PQ} \]

Since one vector is a scalar multiple of the other, all three points lie on the same straight line.

Ratio Interpretation

\[ PQ : QR = 1 : 2 \]

So point \(Q\) divides segment \(PR\) internally in ratio \(1:2\).

Visualization (Collinear Points)

Key Result

Three points are collinear if:

• \(PQ + QR = PR\)
• OR direction vectors are proportional

Exam + JEE Insights

• Vector method is faster and more reliable
• Distance method is safer for board exams
• Always check proportionality carefully
• Ratio interpretation helps in section formula problems

Are the points \(A(3,6,9)\), \(B(10,20,30)\), and \(C(25,-41,5)\) the vertices of a right-angled triangle?

Method 1: Distance (Pythagoras Test)

For a right-angled triangle:

\[ (\text{Longest side})^2 = (\text{other two sides})^2 \]

Step 1: Compute \(AB^2\)

\[ \begin{aligned} AB^2 &= (10-3)^2 + (20-6)^2 + (30-9)^2\\\\ &= 7^2 + 14^2 + 21^2\\\\ &= 49 + 196 + 441 = 686 \end{aligned} \]

Step 2: Compute \(BC^2\)

\[ \begin{aligned} BC^2 &= (25-10)^2 + (-41-20)^2 + (5-30)^2\\\\ &= 15^2 + (-61)^2 + (-25)^2\\\\ &= 225 + 3721 + 625 = 4571 \end{aligned} \]

Step 3: Compute \(CA^2\)

\[ \begin{aligned} CA^2 &= (25-3)^2 + (-41-6)^2 + (5-9)^2\\\\ &= 22^2 + (-47)^2 + (-4)^2\\\\ &= 484 + 2209 + 16 = 2709 \end{aligned} \]

Step 4: Identify longest side

\(BC^2 = 4571\) is largest

Step 5: Verify Pythagoras

\[ \begin{aligned} AB^2 + CA^2 &= 686 + 2709 \\\\&= 3395 \neq 4571 \end{aligned} \]

Condition not satisfied ⇒ Not a right-angled triangle

Method 2: Vector Dot Product (Fastest)

For a right angle at a vertex, corresponding vectors must be perpendicular:

\[ \vec{AB} \cdot \vec{AC} = 0 \]

At point A:

\[ \begin{aligned} \vec{AB} &= (7,14,21), \\\\ \vec{AC} &= (22,-47,-4) \end{aligned} \] \[ \begin{aligned} \vec{AB} \cdot \vec{AC} &= (7)(22) + (14)(-47) + (21)(-4)\\\\ &= 154 - 658 - 84 \\\\&= -588 \neq 0 \end{aligned} \]

Not perpendicular ⇒ no right angle at A

Similarly check other vertices (optional):

No pair gives zero ⇒ No right angle anywhere

Final Conclusion

\[ \boxed{\text{The points do NOT form a right-angled triangle}} \]

Visualization

Key Result

• Pythagoras holds ⇒ right angle exists
• Dot product = 0 ⇒ vectors perpendicular

Exam + JEE Insights

• Always compare squares (avoid square roots)
• Dot product method is fastest in MCQs
• Check largest side first (reduces work)
• Common mistake: not identifying longest side correctly

Find the equation of the locus of point \(P\) such that \(PA^2 + PB^2 = 2k^2\), where \(A(3,4,5)\), \(B(-1,3,-7)\).

Step 1: Let Variable Point

\[ P(x,y,z) \]

Step 2: Distance Expressions

\[ PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2 \] \[ PB^2 = (x+1)^2 + (y-3)^2 + (z+7)^2 \]

Step 3: Apply Given Condition

\[ (x-3)^2 + (y-4)^2 + (z-5)^2 + (x+1)^2 + (y-3)^2 + (z+7)^2 = 2k^2 \]

Step 4: Expand Carefully

\[ (x-3)^2 + (x+1)^2 = 2x^2 - 4x + 10 \] \[ (y-4)^2 + (y-3)^2 = 2y^2 - 14y + 25 \] \[ (z-5)^2 + (z+7)^2 = 2z^2 + 4z + 74 \]

Step 5: Combine

\[ 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = 2k^2 \] Divide by 2: \[ x^2 + y^2 + z^2 - 2x - 7y + 2z + \frac{109}{2} = k^2 \]

Step 6: Complete Squares

\[ (x^2 - 2x) = (x-1)^2 - 1 \] \[ (y^2 - 7y) = (y-\tfrac{7}{2})^2 - \tfrac{49}{4} \] \[ (z^2 + 2z) = (z+1)^2 - 1 \] Substitute: \[ (x-1)^2 + (y-\tfrac{7}{2})^2 + (z+1)^2 = k^2 - \tfrac{109}{2} + 1 + \tfrac{49}{4} + 1 \] Simplify RHS: \[ \begin{aligned} &= k^2 - \tfrac{109}{2} + \tfrac{53}{4}\\\\ &= k^2 - \tfrac{218}{4} + \tfrac{53}{4}\\\\ &= k^2 - \tfrac{165}{4} \end{aligned} \]

Final Equation (Standard Form)

\[ \boxed{ (x-1)^2 + \left(y-\tfrac{7}{2}\right)^2 + (z+1)^2 = k^2 - \tfrac{165}{4} } \]

Geometric Interpretation

This represents a sphere.

• Centre: \( \left(1,\tfrac{7}{2},-1\right) \)
• Radius: \( \sqrt{k^2 - \tfrac{165}{4}} \)

Important Insight

The centre is the midpoint of A and B:

\[ \left(\frac{3+(-1)}{2}, \frac{4+3}{2}, \frac{5+(-7)}{2}\right) = \left(1,\tfrac{7}{2},-1\right) \]

This is a standard result:

\[ PA^2 + PB^2 = \text{constant} \Rightarrow \text{sphere with centre at midpoint of } AB \]

Visualization (Conceptual)

Exam + JEE Insights

• Always divide by 2 before completing squares
• Recognize midpoint pattern immediately (shortcut)
• This type ALWAYS gives a sphere
• Check radius positivity: \(k^2 > \tfrac{165}{4}\)

Show that \(A(1,2,3)\), \(B(-1,-2,-1)\), \(C(2,3,2)\), \(D(4,7,6)\) form a parallelogram, but not a rectangle.

Method 1: Distance Approach

Compute side lengths:

\[ \begin{aligned} AB &= \sqrt{(-2)^2 + (-4)^2 + (-4)^2} \\\\ &= 6 \end{aligned} \] \[ \begin{aligned} BC &= \sqrt{3^2 + 5^2 + 3^2} \\\\ &= \sqrt{43} \end{aligned} \] \[ \begin{aligned} CD &= \sqrt{2^2 + 4^2 + 4^2} \\\\ &= 6 \end{aligned} \] \[ \begin{aligned} AD &= \sqrt{3^2 + 5^2 + 3^2} \\\\ &= \sqrt{43} \end{aligned} \]

Opposite sides are equal ⇒ parallelogram

Method 2: Vector Approach (Best Method)

Compute direction vectors:

\[ \begin{aligned} \vec{AB} &= (-2,-4,-4), \\\\ \vec{DC} &= (2-4,3-7,2-6)\\\\&=(-2,-4,-4) \end{aligned} \] \[ \begin{aligned} \vec{BC} &= (3,5,3), \\\\ \vec{AD} &= (3,5,3) \end{aligned} \]

Opposite sides are equal and parallel ⇒ parallelogram

Method 3: Diagonal Midpoint Test

Midpoint of \(AC\):

\[ \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) = \left(\tfrac{3}{2}, \tfrac{5}{2}, \tfrac{5}{2}\right) \]

Midpoint of \(BD\):

\[ \left(\frac{-1+4}{2}, \frac{-2+7}{2}, \frac{-1+6}{2}\right) = \left(\tfrac{3}{2}, \tfrac{5}{2}, \tfrac{5}{2}\right) \]

Diagonals bisect each other ⇒ parallelogram

Check for Rectangle

A parallelogram is a rectangle if adjacent sides are perpendicular:

\[ \begin{aligned} \vec{AB} \cdot \vec{BC} &= (-2)(3) + (-4)(5) + (-4)(3)\\\\ &= -6 -20 -12 \\\\&= -38 \neq 0 \end{aligned} \]

Not perpendicular ⇒ Not a rectangle

(Alternatively: diagonals unequal ⇒ not rectangle)

\[ \begin{aligned} AC &= \sqrt{3}, \\\\ BD &= \sqrt{155}, \\\\ AC &\ne BD \end{aligned} \]

Final Conclusion

\[ \boxed{\text{ABCD is a parallelogram but not a rectangle}} \]

Visualization

Key Results

• Opposite sides equal ⇒ parallelogram
• Vectors equal ⇒ strongest test
• Diagonals bisect ⇒ confirmation
• Perpendicular adjacent sides ⇒ rectangle

Exam + JEE Insights

• Vector method is fastest and most reliable
• Dot product = 0 ⇒ right angle
• Diagonal midpoint test is very powerful
• Use multiple checks only if needed

Find the equation of the locus of point \(P\) such that its distances from \(A(3,4,-5)\) and \(B(-2,1,4)\) are equal.

Step 1: Let Variable Point

\[ P(x,y,z) \]

Step 2: Apply Condition

Since \(P\) is equidistant from \(A\) and \(B\):

\[ PA = PB \]

Step 3: Square Both Sides

\[ (x-3)^2 + (y-4)^2 + (z+5)^2 = (x+2)^2 + (y-1)^2 + (z-4)^2 \]

Step 4: Expand Carefully

Left side: \[ x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 + 10z + 25 \] Right side: \[ x^2 + 4x + 4 + y^2 - 2y + 1 + z^2 - 8z + 16 \]

Step 5: Simplify

Cancel common terms \(x^2, y^2, z^2\): \[ -6x - 8y + 10z + 50 = 4x - 2y - 8z + 21 \] Rearranging: \[ -10x - 6y + 18z + 29 = 0 \]

Final Answer

\[ \boxed{10x + 6y - 18z - 29 = 0} \]

Geometric Interpretation

The locus is a plane perpendicular to the line joining \(A\) and \(B\), and passing through the midpoint of \(AB\).

Key Insight (Shortcut Method)

Midpoint of \(AB\):

\[ \begin{aligned} \left(\frac{3+(-2)}{2}, \frac{4+1}{2}, \frac{-5+4}{2}\right) = \left(\tfrac{1}{2}, \tfrac{5}{2}, -\tfrac{1}{2}\right) \end{aligned} \]

Direction vector \(AB\):

\[ \vec{AB} = (-5,-3,9) \]

This acts as the normal vector of the plane.

Using point-normal form:

\[ -5(x-\tfrac{1}{2}) -3(y-\tfrac{5}{2}) + 9(z+\tfrac{1}{2}) = 0 \]

Simplifies to the same equation:

\[ 10x + 6y - 18z - 29 = 0 \]

Visualization

Key Result

Points equidistant from two fixed points form a perpendicular bisector plane.

Exam + JEE Insights

• Always square distances to eliminate roots
• Recognize this standard locus → directly a plane
• Use midpoint + normal vector for faster solution
• Avoid algebra mistakes while expanding

The centroid of triangle \(ABC\) is \(G(1,1,1)\). Given \(A(3,-5,7)\), \(B(-1,7,-6)\), find coordinates of \(C\).

Key Formula (Centroid)

\[ G = \left(\frac{x_1+x_2+x_3}{3},\; \frac{y_1+y_2+y_3}{3},\; \frac{z_1+z_2+z_3}{3}\right) \]

Step 1: Let Unknown Point

\[ C = (x,y,z) \]

Step 2: Apply Centroid Formula

\[ \begin{aligned} \frac{3 + (-1) + x}{3} &= 1 \\\\ \frac{-5 + 7 + y}{3} &= 1 \\\\ \frac{7 + (-6) + z}{3} &= 1 \end{aligned} \]

Step 3: Solve Each Coordinate

x-coordinate:

\[ \begin{aligned} \frac{2 + x}{3} &= 1 \\\\\Rightarrow x &= 1 \end{aligned} \]

y-coordinate:

\[ \begin{aligned} \frac{2 + y}{3} &= 1 \\\\\Rightarrow y &= 1 \end{aligned} \]

z-coordinate:

\[ \begin{aligned} \frac{1 + z}{3} &= 1 \\\\\Rightarrow z &= 2 \end{aligned} \]

Final Answer

\[ \boxed{C = (1,1,2)} \]

Vector Method (Fastest Approach)

Using centroid relation:

\[ \vec{OG} = \frac{\vec{OA} + \vec{OB} + \vec{OC}}{3} \]

Rearranging: \[ \vec{OC} = 3\vec{OG} - \vec{OA} - \vec{OB} \] Substitute: \[ C = 3(1,1,1) - (3,-5,7) - (-1,7,-6) \] \[ \begin{aligned} &= (3,3,3) - (3,-5,7) - (-1,7,-6)\\\\ &= (1,1,2) \end{aligned} \]

Same result obtained quickly.

Geometric Insight

The centroid is the balance point of the triangle and divides medians in the ratio \(2:1\).

Visualization

Key Result

Centroid = average of coordinates of vertices

Exam + JEE Insights

• Direct formula is easiest for board exams
• Vector method is fastest for JEE
• Always solve coordinate-wise independently
• Very common question type (reverse centroid)