Q1
A point is on the \(x\)-axis. What are its \(y\)-coordinate and \(z\)-coordinates?
Answer:
If a point lies on the \(x\)-axis, its \(y\)-coordinate and \(z\)-coordinate are both zero.
Coordinates: \((x, 0, 0)\).
Thus, \(y\)-coordinate = 0 and \(z\)-coordinate = 0.
Q2
A point is in the XZ-plane. What can you say about its \(y\)-coordinate?
Answer:
If a point lies in the XZ-plane, its \(y\)-coordinate is zero.
Q3
Name the octants in which the following points lie:
\( (1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (-2, -4, -7) \)
\( (1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (-2, -4, -7) \)
Answer:
Using the sign convention \((x, y, z)\):
- \((1, 2, 3)\) : \((+,+,+)\) \(\to\) Octant I
- \((4, -2, 3)\) : \((+,-,+)\) \(\to\) Octant IV
- \((4, -2, -5)\) : \((+,-,-)\) \(\to\) Octant VIII
- \((4, 2, -5)\) : \((+,+,-)\) \(\to\) Octant V
- \((-4, 2, -5)\) : \((-,+,-)\) \(\to\) Octant VI
- \((-4, 2, 5)\) : \((-,+,+)\) \(\to\) Octant II
- \((-3, -1, 6)\) : \((-,-,+)\) \(\to\) Octant III
- \((-2, -4, -7)\) : \((-,-,-)\) \(\to\) Octant VII
Q4
Fill in the blanks:- The \(x\)-axis and \(y\)-axis taken together determine a plane known as ______.
- The coordinates of points in the XY-plane are of the form ______.
- Coordinate planes divide the space into ______ octants.
- The \(x\)-axis and \(y\)-axis taken together determine a plane known as ______.
- The coordinates of points in the XY-plane are of the form ______.
- Coordinate planes divide the space into ______ octants.
Answer:
- XY-plane
- \((x, y, 0)\)
- 8 (eight)
Q1
Find the distance between the following pairs of points:- \((2, 3, 5)\) and \((4, 3, 1)\)
- \((-3, 7, 2)\) and \((2, 4, -1)\)
- \((-1, 3, -4)\) and \((1, -3, 4)\)
- \((2, -1, 3)\) and \((-2, 1, 3)\)
- \((2, 3, 5)\) and \((4, 3, 1)\)
- \((-3, 7, 2)\) and \((2, 4, -1)\)
- \((-1, 3, -4)\) and \((1, -3, 4)\)
- \((2, -1, 3)\) and \((-2, 1, 3)\)
Answer:
Distance Formula: \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \)
- \( \sqrt{(4-2)^2 + (3-3)^2 + (1-5)^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \)
- \( \sqrt{(2-(-3))^2 + (4-7)^2 + (-1-2)^2} = \sqrt{25 + 9 + 9} = \sqrt{43} \)
- \( \sqrt{(1-(-1))^2 + (-3-3)^2 + (4-(-4))^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26} \)
- \( \sqrt{(-2-2)^2 + (1-(-1))^2 + (3-3)^2} = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5} \)
Q2
Show that the points \((-2, 3, 5), (1, 2, 3)\) and \((7, 0, -1)\) are collinear.
Answer:
Let points be \( P(-2, 3, 5), Q(1, 2, 3) \) and \( R(7, 0, -1) \).
- \( PQ = \sqrt{(1+2)^2 + (2-3)^2 + (3-5)^2} = \sqrt{9+1+4} = \sqrt{14} \)
- \( QR = \sqrt{(7-1)^2 + (0-2)^2 + (-1-3)^2} = \sqrt{36+4+16} = \sqrt{56} = 2\sqrt{14} \)
- \( PR = \sqrt{(7+2)^2 + (0-3)^2 + (-1-5)^2} = \sqrt{81+9+36} = \sqrt{126} = 3\sqrt{14} \)
Since \( PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR \), the points are collinear.
Q3
Verify the following:- \((0, 7, -10), (1, 6, -6)\) and \((4, 9, -6)\) are the vertices of an isosceles triangle.
- \((0, 7, 10), (-1, 6, 6)\) and \((-4, 9, 6)\) are the vertices of a right angled triangle.
- \((-1, 2, 1), (1, -2, 5), (4, -7, 8)\) and \((2, -3, 4)\) are the vertices of a parallelogram.
- \((0, 7, -10), (1, 6, -6)\) and \((4, 9, -6)\) are the vertices of an isosceles triangle.
- \((0, 7, 10), (-1, 6, 6)\) and \((-4, 9, 6)\) are the vertices of a right angled triangle.
- \((-1, 2, 1), (1, -2, 5), (4, -7, 8)\) and \((2, -3, 4)\) are the vertices of a parallelogram.
Answer:
- Let \( A(0, 7, -10), B(1, 6, -6), C(4, 9, -6) \).
\( AB = \sqrt{1+1+16} = \sqrt{18} \)
\( BC = \sqrt{9+9+0} = \sqrt{18} \)
\( AC = \sqrt{16+4+16} = \sqrt{36} = 6 \)
Since \( AB = BC \), it is an isosceles triangle. - Let \( A(0, 7, 10), B(-1, 6, 6), C(-4, 9, 6) \).
\( AB^2 = (1+1+16) = 18 \)
\( BC^2 = (9+9+0) = 18 \)
\( AC^2 = (16+4+16) = 36 \)
Since \( AB^2 + BC^2 = 18 + 18 = 36 = AC^2 \), it is a right-angled triangle. - Let \( A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8), D(2, -3, 4) \).
Mid-point of AC: \( \left(\frac{-1+4}{2}, \frac{2-7}{2}, \frac{1+8}{2}\right) = \left(\frac{3}{2}, -\frac{5}{2}, \frac{9}{2}\right) \)
Mid-point of BD: \( \left(\frac{1+2}{2}, \frac{-2-3}{2}, \frac{5+4}{2}\right) = \left(\frac{3}{2}, -\frac{5}{2}, \frac{9}{2}\right) \)
Since diagonals bisect each other, it is a parallelogram.
Q4
Find the equation of the set of points which are equidistant from the points \( (1, 2, 3) \) and \( (3, 2, -1) \).
Answer:
Let point \( P(x, y, z) \) be equidistant from \( A(1, 2, 3) \) and \( B(3, 2, -1) \).
\( PA = PB \Rightarrow PA^2 = PB^2 \)
\( (x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2 \)
\( (x^2 - 2x + 1) + (z^2 - 6z + 9) = (x^2 - 6x + 9) + (z^2 + 2z + 1) \)
\( -2x - 6z + 10 = -6x + 2z + 10 \)
\( 4x - 8z = 0 \Rightarrow x - 2z = 0 \)
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.
Answer:
Given \( PA + PB = 10 \). Let \( P(x, y, z) \).
\( \sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10 \)
\( \sqrt{(x-4)^2 + y^2 + z^2} = 10 - \sqrt{(x+4)^2 + y^2 + z^2} \)
Squaring both sides and simplifying (Standard ellipse derivation steps):
Final Equation: \( 9x^2 + 25y^2 + 25z^2 - 225 = 0 \)
Q1
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.
Answer:
Let the fourth vertex be \( D(x, y, z) \).
In a parallelogram, diagonals bisect each other. Mid-point of AC = Mid-point of BD.
Mid-point of AC = \( \left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right) = (1, 0, 2) \)
Mid-point of BD = \( \left(\frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2}\right) \)
Equating components:
- \( \frac{1+x}{2} = 1 \Rightarrow x = 1 \)
- \( \frac{2+y}{2} = 0 \Rightarrow y = -2 \)
- \( \frac{-4+z}{2} = 2 \Rightarrow z = 8 \)
Coordinates of D: \( (1, -2, 8) \).
Q2
Find the lengths of the medians of the triangle with vertices \( A(0, 0, 6), B(0, 4, 0) \) and \( (6, 0, 0) \).
Answer:
Vertices: \( A(0, 0, 6), B(0, 4, 0), C(6, 0, 0) \).
Median AD: D is mid-point of BC \( \left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right) = (3, 2, 0) \).
\( AD = \sqrt{(3-0)^2 + (2-0)^2 + (0-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7 \).
Median BE: E is mid-point of AC \( \left(3, 0, 3\right) \).
\( BE = \sqrt{(3-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{9+16+9} = \sqrt{34} \).
Median CF: F is mid-point of AB \( \left(0, 2, 3\right) \).
\( CF = \sqrt{(0-6)^2 + (2-0)^2 + (3-0)^2} = \sqrt{36+4+9} = \sqrt{49} = 7 \).
Q3
If the origin is the centroid of the triangle PQR with vertices \( P(2a, 2, 6), Q(-4, 3b, -10) \) and \( R(8, 14, 2c) \), then find the values of \( a, b \) and \( c \).
Answer:
Centroid formula: \( \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right) = (0, 0, 0) \).
- \( \frac{2a - 4 + 8}{3} = 0 \Rightarrow 2a + 4 = 0 \Rightarrow a = -2 \)
- \( \frac{2 + 3b + 14}{3} = 0 \Rightarrow 3b + 16 = 0 \Rightarrow b = -\frac{16}{3} \)
- \( \frac{6 - 10 + 2c}{3} = 0 \Rightarrow 2c - 4 = 0 \Rightarrow c = 2 \)
Q4
If A and B be the points \( (3, 4, 5) \) and \( (-1, 3, -7) \), respectively, find the equation of the set of points P such that \( PA^2 + PB^2 = k^2 \), where \( k \) is a constant.
Answer:
Let \( P(x, y, z) \).
\( PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2 \)
\( PB^2 = (x+1)^2 + (y-3)^2 + (z+7)^2 \)
Given \( PA^2 + PB^2 = k^2 \):
\( [(x^2-6x+9) + (y^2-8y+16) + (z^2-10z+25)] + [(x^2+2x+1) + (y^2-6y+9) + (z^2+14z+49)] = k^2 \)
\( 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2 \)
Required equation: \( 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + (109 - k^2) = 0 \)