OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3 Dimensions Exercise 26 (B)

Question 1. Find the coordinates of the points which divide the join of the points (2, -1, 3) and (4, 3, 1) in the ratio 3: 4 internally.
Answer: Let the points be \( A(2, -1, 3) \) and \( B(4, 3, 1) \). We need to find the coordinates of point R that divides the line segment AB in the ratio 3:4 internally. We use the section formula for internal division. This formula helps us find a point that lies between two other points on a line segment.
\[ R = \left( \frac { 3 \times 4 + 4 \times 2 }{ 3+4 }, \frac { 3 \times 3 + 4 \times (-1) }{ 3+4 }, \frac { 3 \times 1 + 4 \times 3 }{ 3+4 } \right) \] \[ R = \left( \frac { 12 + 8 }{ 7 }, \frac { 9 - 4 }{ 7 }, \frac { 3 + 12 }{ 7 } \right) \] \[ R = \left( \frac { 20 }{ 7 }, \frac { 5 }{ 7 }, \frac { 15 }{ 7 } \right) \]
The coordinates of the point are \( \left( \frac { 20 }{ 7 }, \frac { 5 }{ 7 }, \frac { 15 }{ 7 } \right) \).
In simple words: To find a point that divides a line segment inside, we use a special formula. We multiply the coordinates of each end point by the opposite part of the ratio, then add them up and divide by the sum of the ratio numbers.

🎯 Exam Tip: Remember the section formula for internal division: \( \left( \frac { m x_2 + n x_1 }{ m+n }, \frac { m y_2 + n y_1 }{ m+n }, \frac { m z_2 + n z_1 }{ m+n } \right) \). Always keep track of which point corresponds to \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) and the ratio \( m:n \).

Question 2. Find the coordinates of the points which divide the line joining the points (2, -4, 3), (-4, 5, -6) in the ratio
(i) 1: -4
(ii) 2:1
Answer: Let the given points be \( A(2, -4, 3) \) and \( B(-4, 5, -6) \). We will find the coordinates for each ratio using the section formula. A negative ratio indicates external division.
(i) For ratio 1: -4 (external division, or \( 1:4 \) externally):
Let point P divide the line segment AB in the ratio \( 1:-4 \). Using the section formula:
\[ P = \left( \frac { 1 \times (-4) + (-4) \times 2 }{ 1 + (-4) }, \frac { 1 \times 5 + (-4) \times (-4) }{ 1 + (-4) }, \frac { 1 \times (-6) + (-4) \times 3 }{ 1 + (-4) } \right) \] \[ P = \left( \frac { -4 - 8 }{ -3 }, \frac { 5 + 16 }{ -3 }, \frac { -6 - 12 }{ -3 } \right) \] \[ P = \left( \frac { -12 }{ -3 }, \frac { 21 }{ -3 }, \frac { -18 }{ -3 } \right) \] \[ P = (4, -7, 6) \] The coordinates of point P are \( (4, -7, 6) \).
(ii) For ratio 2:1 (internal division):
Let point Q divide the line segment AB in the ratio \( 2:1 \) internally. Using the section formula:
\[ Q = \left( \frac { 2 \times (-4) + 1 \times 2 }{ 2+1 }, \frac { 2 \times 5 + 1 \times (-4) }{ 2+1 }, \frac { 2 \times (-6) + 1 \times 3 }{ 2+1 } \right) \] \[ Q = \left( \frac { -8 + 2 }{ 3 }, \frac { 10 - 4 }{ 3 }, \frac { -12 + 3 }{ 3 } \right) \] \[ Q = \left( \frac { -6 }{ 3 }, \frac { 6 }{ 3 }, \frac { -9 }{ 3 } \right) \] \[ Q = (-2, 2, -3) \] The coordinates of point Q are \( (-2, 2, -3) \). It's good practice to understand that a negative ratio often implies external division.
In simple words: We find the coordinates of points that split a line in two different ways. For the first way, it's like extending the line because of the negative ratio. For the second way, the point is directly in between the two main points.

🎯 Exam Tip: Pay close attention to the sign of the ratio. A positive ratio means internal division (point is between the two given points), while a negative ratio (like 1:-4) indicates external division, where the point lies outside the line segment. The section formula can handle both.

Question 3. Find the ratio in which the line joining the points (2, 4, 5),(3, 5, -4) is divided by the yz-plane.
Answer: Let the given points be \( A(2, 4, 5) \) and \( B(3, 5, -4) \). Let a point P divide the line segment AB in the ratio \( k:1 \). Using the section formula, the coordinates of P are:
\[ P = \left( \frac { k \times 3 + 1 \times 2 }{ k+1 }, \frac { k \times 5 + 1 \times 4 }{ k+1 }, \frac { k \times (-4) + 1 \times 5 }{ k+1 } \right) \] \[ P = \left( \frac { 3k+2 }{ k+1 }, \frac { 5k+4 }{ k+1 }, \frac { -4k+5 }{ k+1 } \right) \]
Since the line segment AB is divided by the yz-plane, the x-coordinate of point P must be 0. This is a key property of the yz-plane, where all points have an x-value of zero.
So, we set the x-coordinate of P to 0:
\( \frac { 3k+2 }{ k+1 } = 0 \)
\( \implies 3k+2 = 0 \)
\( \implies 3k = -2 \)
\( \implies k = -\frac { 2 }{ 3 } \)
Thus, the required ratio is \( k:1 = -\frac { 2 }{ 3 } : 1 \), which simplifies to \( -2:3 \). The negative sign means the yz-plane divides the line segment externally.
In simple words: We find a point on the line segment using a ratio \( k:1 \). Since this point is on the "yz-plane," its first number (x-coordinate) must be zero. We use this fact to find the ratio \( k \). If \( k \) is negative, it means the plane cuts the line outside the segment.

🎯 Exam Tip: When a plane divides a line segment, one of the coordinates of the division point will be zero. For the yz-plane, \( x=0 \); for the xz-plane, \( y=0 \); and for the xy-plane, \( z=0 \). This is the key to solving such problems quickly.

Question 4. The three points A(0, 0, 0), B(2, -3, 3), C(-2, 3, -3) are collinear. Find in what ratio each point divides the segment joining the other two.
Answer: We need to find the ratio in which each point divides the segment formed by the other two.
**Case 1: Point B divides AC**
Let point B\( (2, -3, 3) \) divide the line segment AC (where \( A(0, 0, 0) \) and \( C(-2, 3, -3) \)) in the ratio \( k:1 \) internally.
Using the section formula, the coordinates of B are: \[ B = \left( \frac { k \times (-2) + 1 \times 0 }{ k+1 }, \frac { k \times 3 + 1 \times 0 }{ k+1 }, \frac { k \times (-3) + 1 \times 0 }{ k+1 } \right) \] \[ B = \left( \frac { -2k }{ k+1 }, \frac { 3k }{ k+1 }, \frac { -3k }{ k+1 } \right) \]
We are given that B is \( (2, -3, 3) \). So, we can equate any coordinate:
\( \frac { -2k }{ k+1 } = 2 \)
\( \implies -2k = 2(k+1) \)
\( \implies -2k = 2k + 2 \)
\( \implies -4k = 2 \)
\( \implies k = -\frac { 2 }{ 4 } = -\frac { 1 }{ 2 } \)
We can verify with the other coordinates:
\( \frac { 3k }{ k+1 } = -3 \)
\( \implies 3k = -3(k+1) \)
\( \implies 3k = -3k - 3 \)
\( \implies 6k = -3 \)
\( \implies k = -\frac { 3 }{ 6 } = -\frac { 1 }{ 2 } \)
\( \frac { -3k }{ k+1 } = 3 \)
\( \implies -3k = 3(k+1) \)
\( \implies -3k = 3k + 3 \)
\( \implies -6k = 3 \)
\( \implies k = -\frac { 3 }{ 6 } = -\frac { 1 }{ 2 } \)
All coordinates give \( k = -\frac { 1 }{ 2 } \). So, the ratio is \( -\frac { 1 }{ 2 } : 1 \), which is \( -1:2 \). This means B divides AC externally in the ratio 1:2.
**Case 2: Point C divides AB**
Let point C\( (-2, 3, -3) \) divide the line segment AB (where \( A(0, 0, 0) \) and \( B(2, -3, 3) \)) in the ratio \( \lambda:1 \) internally.
Using the section formula, the coordinates of C are: \[ C = \left( \frac { \lambda \times 2 + 1 \times 0 }{ \lambda+1 }, \frac { \lambda \times (-3) + 1 \times 0 }{ \lambda+1 }, \frac { \lambda \times 3 + 1 \times 0 }{ \lambda+1 } \right) \] \[ C = \left( \frac { 2\lambda }{ \lambda+1 }, \frac { -3\lambda }{ \lambda+1 }, \frac { 3\lambda }{ \lambda+1 } \right) \]
We are given that C is \( (-2, 3, -3) \). Equating the x-coordinate:
\( \frac { 2\lambda }{ \lambda+1 } = -2 \)
\( \implies 2\lambda = -2(\lambda+1) \)
\( \implies 2\lambda = -2\lambda - 2 \)
\( \implies 4\lambda = -2 \)
\( \implies \lambda = -\frac { 2 }{ 4 } = -\frac { 1 }{ 2 } \)
We can verify with the other coordinates:
\( \frac { -3\lambda }{ \lambda+1 } = 3 \)
\( \implies -3\lambda = 3(\lambda+1) \)
\( \implies -3\lambda = 3\lambda + 3 \)
\( \implies -6\lambda = 3 \)
\( \implies \lambda = -\frac { 3 }{ 6 } = -\frac { 1 }{ 2 } \)
\( \frac { 3\lambda }{ \lambda+1 } = -3 \)
\( \implies 3\lambda = -3(\lambda+1) \)
\( \implies 3\lambda = -3\lambda - 3 \)
\( \implies 6\lambda = -3 \)
\( \implies \lambda = -\frac { 3 }{ 6 } = -\frac { 1 }{ 2 } \)
All coordinates give \( \lambda = -\frac { 1 }{ 2 } \). So, the ratio is \( -\frac { 1 }{ 2 } : 1 \), which is \( -1:2 \). This means C divides AB externally in the ratio 1:2.
**Case 3: Point A divides BC**
Let point A\( (0, 0, 0) \) divide the line segment BC (where \( B(2, -3, 3) \) and \( C(-2, 3, -3) \)) in the ratio \( p:1 \) internally.
Using the section formula, the coordinates of A are: \[ A = \left( \frac { p \times (-2) + 1 \times 2 }{ p+1 }, \frac { p \times 3 + 1 \times (-3) }{ p+1 }, \frac { p \times (-3) + 1 \times 3 }{ p+1 } \right) \] \[ A = \left( \frac { -2p+2 }{ p+1 }, \frac { 3p-3 }{ p+1 }, \frac { -3p+3 }{ p+1 } \right) \]
We are given that A is \( (0, 0, 0) \). Equating the x-coordinate to 0:
\( \frac { -2p+2 }{ p+1 } = 0 \)
\( \implies -2p+2 = 0 \)
\( \implies -2p = -2 \)
\( \implies p = 1 \)
We can verify with the other coordinates:
\( \frac { 3p-3 }{ p+1 } = 0 \)
\( \implies 3p-3 = 0 \)
\( \implies 3p = 3 \)
\( \implies p = 1 \)
\( \frac { -3p+3 }{ p+1 } = 0 \)
\( \implies -3p+3 = 0 \)
\( \implies -3p = -3 \)
\( \implies p = 1 \)
All coordinates give \( p = 1 \). So, the ratio is \( 1:1 \). This means A divides BC internally in the ratio 1:1, meaning A is the midpoint of BC.
In simple words: We check how each point (A, B, or C) splits the line made by the other two points. We found that B divides AC externally by 1:2, C divides AB externally by 1:2, and A divides BC internally by 1:1, meaning A is exactly in the middle of B and C. This confirms the points are indeed on the same straight line.

🎯 Exam Tip: To prove collinearity or find ratios for three given points, assume one point divides the segment formed by the other two in a ratio \( k:1 \). If \( k \) is constant across all coordinates and is positive, it's internal division; if negative, it's external. If \( k=1 \), the point is the midpoint.

Question 5. Find the coordinates of the points which trisect AB given that A(2, 1, -3) and B (5, -8, 3).
Answer: To trisect a line segment means to divide it into three equal parts. Let the given points be \( A(2, 1, -3) \) and \( B(5, -8, 3) \).
If a line segment AB is trisected by points P and Q, then P divides AB in the ratio 1:2, and Q divides AB in the ratio 2:1. This is a common method for finding points that divide a segment into equal parts.
**1. Coordinates of Point P (divides AB in ratio 1:2):**
Using the section formula for internal division: \[ P = \left( \frac { 1 \times 5 + 2 \times 2 }{ 1+2 }, \frac { 1 \times (-8) + 2 \times 1 }{ 1+2 }, \frac { 1 \times 3 + 2 \times (-3) }{ 1+2 } \right) \] \[ P = \left( \frac { 5 + 4 }{ 3 }, \frac { -8 + 2 }{ 3 }, \frac { 3 - 6 }{ 3 } \right) \] \[ P = \left( \frac { 9 }{ 3 }, \frac { -6 }{ 3 }, \frac { -3 }{ 3 } \right) \] \[ P = (3, -2, -1) \]
**2. Coordinates of Point Q (divides AB in ratio 2:1):**
Using the section formula for internal division: \[ Q = \left( \frac { 2 \times 5 + 1 \times 2 }{ 2+1 }, \frac { 2 \times (-8) + 1 \times 1 }{ 2+1 }, \frac { 2 \times 3 + 1 \times (-3) }{ 2+1 } \right) \] \[ Q = \left( \frac { 10 + 2 }{ 3 }, \frac { -16 + 1 }{ 3 }, \frac { 6 - 3 }{ 3 } \right) \] \[ Q = \left( \frac { 12 }{ 3 }, \frac { -15 }{ 3 }, \frac { 3 }{ 3 } \right) \] \[ Q = (4, -5, 1) \]
The coordinates of the points that trisect AB are \( P(3, -2, -1) \) and \( Q(4, -5, 1) \).
In simple words: To find the points that cut a line into three equal pieces, we use a formula twice. First, we find the point that is one-third of the way from the start. Then, we find the point that is two-thirds of the way.

🎯 Exam Tip: When trisecting a line segment, remember that you'll have two points of trisection. The first point divides the segment in a 1:2 ratio, and the second point divides it in a 2:1 ratio. Always draw a small diagram to visualize these ratios.

Question 6. Find the coordinates of the point which is three-fifths of the way from (3, 4, 5) to (-2, -1, 0).
Answer: Let the given points be \( A(3, 4, 5) \) and \( E(-2, -1, 0) \). We need to find the coordinates of a point P that is three-fifths of the way from A to E. This means that P divides the line segment AE in the ratio \( 3:2 \). If it's three-fifths of the way from A, then three parts are from A to P, and two parts are from P to E.
Using the section formula for internal division with ratio \( 3:2 \): \[ P = \left( \frac { 3 \times (-2) + 2 \times 3 }{ 3+2 }, \frac { 3 \times (-1) + 2 \times 4 }{ 3+2 }, \frac { 3 \times 0 + 2 \times 5 }{ 3+2 } \right) \] \[ P = \left( \frac { -6 + 6 }{ 5 }, \frac { -3 + 8 }{ 5 }, \frac { 0 + 10 }{ 5 } \right) \] \[ P = \left( \frac { 0 }{ 5 }, \frac { 5 }{ 5 }, \frac { 10 }{ 5 } \right) \] \[ P = (0, 1, 2) \]
The coordinates of the point are \( (0, 1, 2) \). This point lies on the yz-plane because its x-coordinate is 0.
In simple words: We want to find a point that is 3/5 of the distance from one point to another. We use the section formula, taking the ratio as 3 for the first part and 2 for the remaining part (since 3+2=5).

🎯 Exam Tip: The phrase "three-fifths of the way from A to B" implies an internal division in the ratio 3:2. Always clarify the total number of parts (the denominator of the fraction) to correctly identify the ratio components (numerator for first part, total minus numerator for second part).

Question 7. Show that the point (1, -1, 2) is common to the lines which join (6, -7, 0) to (16, -19, -4) and (0, 3, -6) to (2, -5, 10).
Answer: We need to show that the given point lies on both line segments. To do this, we can find the general coordinates of a point on each line using the section formula in ratio \( k:1 \) (or \( k':1 \)) and then equate them to the given common point.
**1. For the line joining \( A(6, -7, 0) \) and \( B(16, -19, -4) \):**
Let point P divide AB in the ratio \( k:1 \). \[ P = \left( \frac { k \times 16 + 1 \times 6 }{ k+1 }, \frac { k \times (-19) + 1 \times (-7) }{ k+1 }, \frac { k \times (-4) + 1 \times 0 }{ k+1 } \right) \] \[ P = \left( \frac { 16k+6 }{ k+1 }, \frac { -19k-7 }{ k+1 }, \frac { -4k }{ k+1 } \right) \]
**2. For the line joining \( C(0, 3, -6) \) and \( D(2, -5, 10) \):**
Let point Q divide CD in the ratio \( k':1 \). \[ Q = \left( \frac { k' \times 2 + 1 \times 0 }{ k'+1 }, \frac { k' \times (-5) + 1 \times 3 }{ k'+1 }, \frac { k' \times 10 + 1 \times (-6) }{ k'+1 } \right) \] \[ Q = \left( \frac { 2k' }{ k'+1 }, \frac { -5k'+3 }{ k'+1 }, \frac { 10k'-6 }{ k'+1 } \right) \]
If P and Q coincide at the common point \( (1, -1, 2) \), then their coordinates must be equal. We can also equate P and Q, and then solve for k and k' to find if such values exist.
Equating the coordinates of P and Q to \( (1, -1, 2) \):
From P's coordinates and the common point \( (1, -1, 2) \):
\( \frac { 16k+6 }{ k+1 } = 1 \)
\( \implies 16k+6 = k+1 \)
\( \implies 15k = -5 \)
\( \implies k = -\frac { 5 }{ 15 } = -\frac { 1 }{ 3 } \)
Let's check this value of \( k \) with the y and z coordinates:
For y-coordinate: \( \frac { -19k-7 }{ k+1 } = \frac { -19(-\frac { 1 }{ 3 }) - 7 }{ -\frac { 1 }{ 3 } + 1 } = \frac { \frac { 19 }{ 3 } - \frac { 21 }{ 3 } }{ \frac { 2 }{ 3 } } = \frac { -\frac { 2 }{ 3 } }{ \frac { 2 }{ 3 } } = -1 \). This matches.
For z-coordinate: \( \frac { -4k }{ k+1 } = \frac { -4(-\frac { 1 }{ 3 }) }{ -\frac { 1 }{ 3 } + 1 } = \frac { \frac { 4 }{ 3 } }{ \frac { 2 }{ 3 } } = 2 \). This also matches.
So, point \( P \) divides AB externally in the ratio \( 1:3 \), and the point \( (1, -1, 2) \) lies on line AB.
Now, from Q's coordinates and the common point \( (1, -1, 2) \):
\( \frac { 2k' }{ k'+1 } = 1 \)
\( \implies 2k' = k'+1 \)
\( \implies k' = 1 \)
Let's check this value of \( k' \) with the y and z coordinates:
For y-coordinate: \( \frac { -5k'+3 }{ k'+1 } = \frac { -5(1)+3 }{ 1+1 } = \frac { -5+3 }{ 2 } = \frac { -2 }{ 2 } = -1 \). This matches.
For z-coordinate: \( \frac { 10k'-6 }{ k'+1 } = \frac { 10(1)-6 }{ 1+1 } = \frac { 10-6 }{ 2 } = \frac { 4 }{ 2 } = 2 \). This also matches.
So, point \( Q \) divides CD internally in the ratio \( 1:1 \), meaning it's the midpoint of CD, and the point \( (1, -1, 2) \) lies on line CD.
Since \( (1, -1, 2) \) lies on both lines, it is a common point. Finding \( k \) and \( k' \) first and then checking consistency is often an efficient way to demonstrate this.
In simple words: We imagine the common point is on the first line, dividing it in some ratio. We find that ratio. Then we do the same for the second line. If the common point fits both lines with a valid ratio, then it's a point on both. In this case, it means the point (1, -1, 2) sits on both lines.

🎯 Exam Tip: To show a point is common to two lines, assume it divides each line segment in a ratio (\( k:1 \) and \( k':1 \)). Solve for \( k \) and \( k' \) for each line. If consistent values are found for all coordinates, the point is indeed common. Remember that \( k \) and \( k' \) can be negative for external division.

Question 8. Find the lengths of the medians of the triangle whose vertices are A(2, -3, 1), B (-6, 5, 3), C(8, 7, -7).
Answer: A median of a triangle connects a vertex to the midpoint of the opposite side. First, we need to find the midpoints of each side. Let D, E, and F be the midpoints of BC, CA, and AB respectively.
**1. Midpoint D of BC:**
Using the midpoint formula \( \left( \frac { x_1+x_2 }{ 2 }, \frac { y_1+y_2 }{ 2 }, \frac { z_1+z_2 }{ 2 } \right) \) for \( B(-6, 5, 3) \) and \( C(8, 7, -7) \): \[ D = \left( \frac { -6+8 }{ 2 }, \frac { 5+7 }{ 2 }, \frac { 3+(-7) }{ 2 } \right) \] \[ D = \left( \frac { 2 }{ 2 }, \frac { 12 }{ 2 }, \frac { -4 }{ 2 } \right) \] \[ D = (1, 6, -2) \]
**2. Midpoint E of CA:**
For \( C(8, 7, -7) \) and \( A(2, -3, 1) \): \[ E = \left( \frac { 8+2 }{ 2 }, \frac { 7+(-3) }{ 2 }, \frac { -7+1 }{ 2 } \right) \] \[ E = \left( \frac { 10 }{ 2 }, \frac { 4 }{ 2 }, \frac { -6 }{ 2 } \right) \] \[ E = (5, 2, -3) \]
**3. Midpoint F of AB:**
For \( A(2, -3, 1) \) and \( B(-6, 5, 3) \): \[ F = \left( \frac { 2+(-6) }{ 2 }, \frac { -3+5 }{ 2 }, \frac { 1+3 }{ 2 } \right) \] \[ F = \left( \frac { -4 }{ 2 }, \frac { 2 }{ 2 }, \frac { 4 }{ 2 } \right) \] \[ F = (-2, 1, 2) \]
Now, we calculate the length of each median using the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \).
**1. Length of median AD:**
Points are \( A(2, -3, 1) \) and \( D(1, 6, -2) \): \[ AD = \sqrt{(1-2)^2 + (6-(-3))^2 + (-2-1)^2} \] \[ AD = \sqrt{(-1)^2 + (9)^2 + (-3)^2} \] \[ AD = \sqrt{1 + 81 + 9} = \sqrt{91} \]
**2. Length of median BE:**
Points are \( B(-6, 5, 3) \) and \( E(5, 2, -3) \): \[ BE = \sqrt{(5-(-6))^2 + (2-5)^2 + (-3-3)^2} \] \[ BE = \sqrt{(11)^2 + (-3)^2 + (-6)^2} \] \[ BE = \sqrt{121 + 9 + 36} = \sqrt{166} \]
**3. Length of median CF:**
Points are \( C(8, 7, -7) \) and \( F(-2, 1, 2) \): \[ CF = \sqrt{(-2-8)^2 + (1-7)^2 + (2-(-7))^2} \] \[ CF = \sqrt{(-10)^2 + (-6)^2 + (9)^2} \] \[ CF = \sqrt{100 + 36 + 81} = \sqrt{217} \]
The lengths of the medians are \( AD = \sqrt{91} \), \( BE = \sqrt{166} \), and \( CF = \sqrt{217} \). Medians help in finding the center of a triangle, known as the centroid.
In simple words: First, we find the middle point of each side of the triangle. Then, for each median, we measure the distance from a corner to the middle point of the side across from it. We use the distance formula for these measurements.

🎯 Exam Tip: To find the length of medians, first calculate the midpoints of the sides opposite to each vertex using the midpoint formula. Then, use the distance formula between each vertex and its corresponding midpoint to get the median lengths. This is a two-step process.

Question 9. Find the point of intersection of the medians of the triangle with vertices (-1, -3, -4), (4, -2, -7), (2, 3, -8).
Answer: The point of intersection of the medians of a triangle is called its centroid. For a triangle with vertices \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \), the coordinates of the centroid G are given by the formula:
\[ G = \left( \frac { x_1+x_2+x_3 }{ 3 }, \frac { y_1+y_2+y_3 }{ 3 }, \frac { z_1+z_2+z_3 }{ 3 } \right) \]
Given the vertices: \( A(-1, -3, -4) \), \( B(4, -2, -7) \), and \( C(2, 3, -8) \).
Let's plug these values into the centroid formula:
\[ G_x = \frac { -1 + 4 + 2 }{ 3 } = \frac { 5 }{ 3 } \] \[ G_y = \frac { -3 + (-2) + 3 }{ 3 } = \frac { -2 }{ 3 } \] \[ G_z = \frac { -4 + (-7) + (-8) }{ 3 } = \frac { -19 }{ 3 } \]
So, the coordinates of the centroid are \( G \left( \frac { 5 }{ 3 }, -\frac { 2 }{ 3 }, -\frac { 19 }{ 3 } \right) \). The centroid is always located inside the triangle and represents its geometric center.
In simple words: To find where all the medians of a triangle meet (the centroid), you just add up all the x-coordinates, all the y-coordinates, and all the z-coordinates separately, and then divide each sum by three.

🎯 Exam Tip: The centroid formula is a direct application of averaging coordinates. Remember it for quick calculation of the center point of a triangle. Ensure you add all x, y, and z coordinates correctly before dividing by 3.

Question 10. Find the ratio in which the join of A(2, 1, 5) and B(3, 4, 3) is divided by the plane 2x + 2y - 2z = 1. Also, find the coordinates of the point of division.
Answer: Let the line segment joining \( A(2, 1, 5) \) and \( B(3, 4, 3) \) be divided by the plane \( 2x + 2y - 2z = 1 \) in the ratio \( k:1 \).
Using the section formula, the coordinates of the point of division, let's call it R, are: \[ R = \left( \frac { k \times 3 + 1 \times 2 }{ k+1 }, \frac { k \times 4 + 1 \times 1 }{ k+1 }, \frac { k \times 3 + 1 \times 5 }{ k+1 } \right) \] \[ R = \left( \frac { 3k+2 }{ k+1 }, \frac { 4k+1 }{ k+1 }, \frac { 3k+5 }{ k+1 } \right) \]
Since point R lies on the plane \( 2x + 2y - 2z = 1 \), its coordinates must satisfy the plane's equation. This means we can substitute the x, y, and z coordinates of R into the plane equation to find the value of k.
\( 2 \left( \frac { 3k+2 }{ k+1 } \right) + 2 \left( \frac { 4k+1 }{ k+1 } \right) - 2 \left( \frac { 3k+5 }{ k+1 } \right) = 1 \)
Multiply both sides by \( (k+1) \) to clear the denominators (assuming \( k+1 \neq 0 \)):
\( 2(3k+2) + 2(4k+1) - 2(3k+5) = 1(k+1) \)
\( 6k+4 + 8k+2 - 6k-10 = k+1 \)
Combine like terms:
\( (6k+8k-6k) + (4+2-10) = k+1 \)
\( 8k - 4 = k+1 \)
\( 8k - k = 1+4 \)
\( 7k = 5 \)
\( k = \frac { 5 }{ 7 } \)
So, the ratio in which the line segment is divided by the plane is \( k:1 = \frac { 5 }{ 7 } : 1 \), or \( 5:7 \). Since \( k \) is positive, the division is internal.
Now, we find the coordinates of the point of division R by substituting \( k = \frac { 5 }{ 7 } \) back into R's coordinates:
\( x = \frac { 3(\frac { 5 }{ 7 })+2 }{ \frac { 5 }{ 7 }+1 } = \frac { \frac { 15 }{ 7 }+ \frac { 14 }{ 7 } }{ \frac { 5 }{ 7 }+ \frac { 7 }{ 7 } } = \frac { \frac { 29 }{ 7 } }{ \frac { 12 }{ 7 } } = \frac { 29 }{ 12 } \)
\( y = \frac { 4(\frac { 5 }{ 7 })+1 }{ \frac { 5 }{ 7 }+1 } = \frac { \frac { 20 }{ 7 }+ \frac { 7 }{ 7 } }{ \frac { 12 }{ 7 } } = \frac { \frac { 27 }{ 7 } }{ \frac { 12 }{ 7 } } = \frac { 27 }{ 12 } = \frac { 9 }{ 4 } \)
\( z = \frac { 3(\frac { 5 }{ 7 })+5 }{ \frac { 5 }{ 7 }+1 } = \frac { \frac { 15 }{ 7 }+ \frac { 35 }{ 7 } }{ \frac { 12 }{ 7 } } = \frac { \frac { 50 }{ 7 } }{ \frac { 12 }{ 7 } } = \frac { 50 }{ 12 } = \frac { 25 }{ 6 } \)
The coordinates of the point of division are \( R \left( \frac { 29 }{ 12 }, \frac { 9 }{ 4 }, \frac { 25 }{ 6 } \right) \). This point is where the line connecting A and B goes through the given plane.
In simple words: We find a general point on the line using a ratio \( k:1 \). Then, we make sure this point also lies on the given flat surface (the plane) by plugging its coordinates into the plane's equation. This helps us find the ratio \( k \) and then the exact spot where the line crosses the plane.

🎯 Exam Tip: When a plane divides a line segment, the coordinates of the division point must satisfy the equation of the plane. This allows you to solve for the ratio \( k \). If \( k \) is positive, it's internal division; if negative, it's external. Always substitute the found \( k \) value back into the section formula to get the exact coordinates.

Question 11. The mid-points of the sides of a triangle are (1, 5, -1),(0, 4, -2) and (2, 3, 4). Find its vertices.
Answer: Let the vertices of the triangle be \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \).
Let the given midpoints be \( D(1, 5, -1) \), \( E(0, 4, -2) \), and \( F(2, 3, 4) \).
Assume D is the midpoint of BC, E is the midpoint of CA, and F is the midpoint of AB.
Using the midpoint formula \( \left( \frac { x_1+x_2 }{ 2 }, \frac { y_1+y_2 }{ 2 }, \frac { z_1+z_2 }{ 2 } \right) \), we set up a system of equations:
**For midpoint D(1, 5, -1) of BC:**
\( \frac { x_2+x_3 }{ 2 } = 1 \implies x_2+x_3 = 2 \) (1)
\( \frac { y_2+y_3 }{ 2 } = 5 \implies y_2+y_3 = 10 \) (2)
\( \frac { z_2+z_3 }{ 2 } = -1 \implies z_2+z_3 = -2 \) (3)
**For midpoint E(0, 4, -2) of CA:**
\( \frac { x_1+x_3 }{ 2 } = 0 \implies x_1+x_3 = 0 \) (4)
\( \frac { y_1+y_3 }{ 2 } = 4 \implies y_1+y_3 = 8 \) (5)
\( \frac { z_1+z_3 }{ 2 } = -2 \implies z_1+z_3 = -4 \) (6)
**For midpoint F(2, 3, 4) of AB:**
\( \frac { x_1+x_2 }{ 2 } = 2 \implies x_1+x_2 = 4 \) (7)
\( \frac { y_1+y_2 }{ 2 } = 3 \implies y_1+y_2 = 6 \) (8)
\( \frac { z_1+z_2 }{ 2 } = 4 \implies z_1+z_2 = 8 \) (9)
Now, we solve these systems for x, y, and z coordinates separately.
**Solving for x-coordinates:**
(1) \( x_2+x_3 = 2 \)
(4) \( x_1+x_3 = 0 \)
(7) \( x_1+x_2 = 4 \)
Adding (1), (4), and (7):
\( (x_2+x_3) + (x_1+x_3) + (x_1+x_2) = 2 + 0 + 4 \)
\( 2x_1 + 2x_2 + 2x_3 = 6 \)
\( x_1+x_2+x_3 = 3 \) (10)
Subtract (1) from (10): \( (x_1+x_2+x_3) - (x_2+x_3) = 3 - 2 \implies x_1 = 1 \)
Subtract (4) from (10): \( (x_1+x_2+x_3) - (x_1+x_3) = 3 - 0 \implies x_2 = 3 \)
Subtract (7) from (10): \( (x_1+x_2+x_3) - (x_1+x_2) = 3 - 4 \implies x_3 = -1 \)
So, the x-coordinates of the vertices are \( x_1=1, x_2=3, x_3=-1 \).
**Solving for y-coordinates:**
(2) \( y_2+y_3 = 10 \)
(5) \( y_1+y_3 = 8 \)
(8) \( y_1+y_2 = 6 \)
Adding (2), (5), and (8):
\( (y_2+y_3) + (y_1+y_3) + (y_1+y_2) = 10 + 8 + 6 \)
\( 2y_1 + 2y_2 + 2y_3 = 24 \)
\( y_1+y_2+y_3 = 12 \) (11)
Subtract (2) from (11): \( (y_1+y_2+y_3) - (y_2+y_3) = 12 - 10 \implies y_1 = 2 \)
Subtract (5) from (11): \( (y_1+y_2+y_3) - (y_1+y_3) = 12 - 8 \implies y_2 = 4 \)
Subtract (8) from (11): \( (y_1+y_2+y_3) - (y_1+y_2) = 12 - 6 \implies y_3 = 6 \)
So, the y-coordinates of the vertices are \( y_1=2, y_2=4, y_3=6 \).
**Solving for z-coordinates:**
(3) \( z_2+z_3 = -2 \)
(6) \( z_1+z_3 = -4 \)
(9) \( z_1+z_2 = 8 \)
Adding (3), (6), and (9):
\( (z_2+z_3) + (z_1+z_3) + (z_1+z_2) = -2 + (-4) + 8 \)
\( 2z_1 + 2z_2 + 2z_3 = 2 \)
\( z_1+z_2+z_3 = 1 \) (12)
Subtract (3) from (12): \( (z_1+z_2+z_3) - (z_2+z_3) = 1 - (-2) \implies z_1 = 3 \)
Subtract (6) from (12): \( (z_1+z_2+z_3) - (z_1+z_3) = 1 - (-4) \implies z_2 = 5 \)
Subtract (9) from (12): \( (z_1+z_2+z_3) - (z_1+z_2) = 1 - 8 \implies z_3 = -7 \)
So, the z-coordinates of the vertices are \( z_1=3, z_2=5, z_3=-7 \).
Combining these, the required vertices of the triangle are:
\( A(x_1, y_1, z_1) = (1, 2, 3) \)
\( B(x_2, y_2, z_2) = (3, 4, 5) \)
\( C(x_3, y_3, z_3) = (-1, 6, -7) \)
This problem shows how the properties of midpoints allow us to reconstruct the original triangle's vertices.
In simple words: We are given the middle points of the triangle's sides. We use a formula that connects these midpoints to the main corners of the triangle. By setting up and solving several simple equations, we can find the coordinates of all three original corners of the triangle.

🎯 Exam Tip: This problem involves setting up and solving a system of linear equations. Group the x, y, and z coordinates and solve them independently. The trick is to sum all three midpoint equations (e.g., \( x_2+x_3=..., x_1+x_3=..., x_1+x_2=... \)) to get \( 2(x_1+x_2+x_3) \), which simplifies the solving process.

Question 12. Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex D.
Answer: In a parallelogram, the diagonals bisect each other. This means their midpoints are the same. Let the fourth vertex be \( D(\alpha, \beta, \gamma) \).
The diagonals of parallelogram ABCD are AC and BD.
**1. Midpoint of diagonal AC:**
For \( A(3, -1, 2) \) and \( C(-1, 1, 2) \): \[ M_{AC} = \left( \frac { 3+(-1) }{ 2 }, \frac { -1+1 }{ 2 }, \frac { 2+2 }{ 2 } \right) \] \[ M_{AC} = \left( \frac { 2 }{ 2 }, \frac { 0 }{ 2 }, \frac { 4 }{ 2 } \right) \] \[ M_{AC} = (1, 0, 2) \]
**2. Midpoint of diagonal BD:**
For \( B(1, 2, -4) \) and \( D(\alpha, \beta, \gamma) \): \[ M_{BD} = \left( \frac { 1+\alpha }{ 2 }, \frac { 2+\beta }{ 2 }, \frac { -4+\gamma }{ 2 } \right) \]
Since the midpoints are the same, we equate the coordinates of \( M_{AC} \) and \( M_{BD} \):
\( \frac { 1+\alpha }{ 2 } = 1 \)
\( \implies 1+\alpha = 2 \)
\( \implies \alpha = 1 \)
\( \frac { 2+\beta }{ 2 } = 0 \)
\( \implies 2+\beta = 0 \)
\( \implies \beta = -2 \)
\( \frac { -4+\gamma }{ 2 } = 2 \)
\( \implies -4+\gamma = 4 \)
\( \implies \gamma = 8 \)
Therefore, the coordinates of the fourth vertex D are \( (1, -2, 8) \). This property of diagonals is a fundamental characteristic of parallelograms and helps locate missing vertices.
In simple words: In a shape called a parallelogram, the middle points of the two long lines that cross (diagonals) are exactly the same. We use this rule: find the middle of the known diagonal, then use that same middle point to find the missing corner of the other diagonal.

🎯 Exam Tip: The key property of a parallelogram is that its diagonals bisect each other. This means the midpoint of one diagonal is identical to the midpoint of the other diagonal. Use this fact with the midpoint formula to easily find a missing vertex.

Question 13. What is the locus of a point for which (i) x = 0 (ii) y = 0 (iii) z = 0 (iv) x = a (v) y = b (vi) z = c?
Answer: The locus of a point is the set of all points that satisfy a given condition.
(i) **\( x=0 \):** This condition means the x-coordinate of any point is zero. All such points lie on the **yz-plane**. The yz-plane is a flat surface that cuts through the x-axis at x=0.
(ii) **\( y=0 \):** This condition means the y-coordinate of any point is zero. All such points lie on the **xz-plane**. The xz-plane is a flat surface that cuts through the y-axis at y=0.
(iii) **\( z=0 \):** This condition means the z-coordinate of any point is zero. All such points lie on the **xy-plane**. The xy-plane is a flat surface that cuts through the z-axis at z=0.
(iv) **\( x=a \):** This condition means the x-coordinate is a constant value 'a'. This represents a **plane parallel to the yz-plane**, at a distance of 'a' units from it. If \( a=0 \), it becomes the yz-plane itself.
(v) **\( y=b \):** This condition means the y-coordinate is a constant value 'b'. This represents a **plane parallel to the xz-plane**, at a distance of 'b' units from it. If \( b=0 \), it becomes the xz-plane itself.
(vi) **\( z=c \):** This condition means the z-coordinate is a constant value 'c'. This represents a **plane parallel to the xy-plane**, at a distance of 'c' units from it. If \( c=0 \), it becomes the xy-plane itself.
In simple words: The "locus" is simply where a point can be if it follows certain rules. If x, y, or z is zero, the point must be on a specific flat surface (plane). If x, y, or z is a fixed number like 'a', 'b', or 'c', then the point is on a flat surface that is parallel to one of these main planes, just shifted away by that number.

🎯 Exam Tip: Remember the basic coordinate planes: \( x=0 \) is the yz-plane, \( y=0 \) is the xz-plane, and \( z=0 \) is the xy-plane. Any equation in the form \( x=a \), \( y=b \), or \( z=c \) represents a plane parallel to one of these fundamental planes.

Question 14. What is the locus of a point for which (i) x = 0, y = 0 (ii) y = 0, z = 0 (iii) z = 0, x = 0 (iv) x = a, y = b (v) y = b, z = c (vi) z = c, x = a?
Answer: The locus of a point satisfying two conditions at once is the intersection of the two geometric shapes defined by each condition separately.
(i) **\( x=0, y=0 \):** This means the point has both its x-coordinate and y-coordinate as zero. All such points lie on the **z-axis**. The z-axis is where the yz-plane and xz-plane meet.
(ii) **\( y=0, z=0 \):** This means the point has both its y-coordinate and z-coordinate as zero. All such points lie on the **x-axis**. The x-axis is where the xz-plane and xy-plane meet.
(iii) **\( z=0, x=0 \):** This means the point has both its z-coordinate and x-coordinate as zero. All such points lie on the **y-axis**. The y-axis is where the xy-plane and yz-plane meet.
(iv) **\( x=a, y=b \):** The condition \( x=a \) represents a plane parallel to the yz-plane. The condition \( y=b \) represents a plane parallel to the xz-plane. The intersection of these two planes is a **line parallel to the z-axis**.
(v) **\( y=b, z=c \):** The condition \( y=b \) represents a plane parallel to the xz-plane. The condition \( z=c \) represents a plane parallel to the xy-plane. The intersection of these two planes is a **line parallel to the x-axis**.
(vi) **\( z=c, x=a \):** The condition \( z=c \) represents a plane parallel to the xy-plane. The condition \( x=a \) represents a plane parallel to the yz-plane. The intersection of these two planes is a **line parallel to the y-axis**.
Each pair of conditions describes a line in 3D space, which is the intersection of the two planes.
In simple words: When a point has two rules, it means it must be in the place where those two rules meet. If x and y are both zero, it means the point is on the z-line. If x is a number 'a' and y is a number 'b', the point is on a line that goes straight up or down, parallel to the z-line.

🎯 Exam Tip: The intersection of any two coordinate planes (e.g., \( x=0 \) and \( y=0 \)) is always one of the coordinate axes. Similarly, the intersection of two planes parallel to coordinate planes (e.g., \( x=a \) and \( y=b \)) will always be a line parallel to one of the coordinate axes.