Questions and Answers Section

Example 1: Basic Distance Formula

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

Solution:

We use the 3D distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Step 1: Substitute the coordinates of P and Q.

$$d = \sqrt{(-4-1)^2 + (1-(-3))^2 + (2-4)^2}$$

Step 2: Simplify the terms inside the parentheses.

$$d = \sqrt{(-5)^2 + (4)^2 + (-2)^2}$$

Step 3: Calculate the final distance.

$$d = \sqrt{25 + 16 + 4} = \sqrt{45} = \sqrt{9 \times 5} = \mathbf{3\sqrt{5}}$$

Example 2: Collinearity of Points

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

Solution:

Three points are collinear if the sum of the distances between two pairs of points equals the distance between the third pair. We calculate all three distances.

Step 1: Calculate the distance PQ.

$PQ = \sqrt{(1-(-2))^2 + (2-3)^2 + (3-5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9+1+4} = \mathbf{\sqrt{14}}$.

Step 2: Calculate the distance QR.

$QR = \sqrt{(7-1)^2 + (0-2)^2 + (-1-3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36+4+16} = \sqrt{56} = \mathbf{2\sqrt{14}}$.

Step 3: Calculate the distance PR.

$PR = \sqrt{(7-(-2))^2 + (0-3)^2 + (-1-5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81+9+36} = \sqrt{126} = \mathbf{3\sqrt{14}}$.

Step 4: Check the collinearity condition.

$PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14}$. This is equal to the distance PR. Since $PQ+QR=PR$, the points are collinear. ✅

Example 3: Identifying an Isosceles Triangle

Question: Show that the points A(0,7,10), B(-1,6,6) and C(-4,9,6) form an isosceles triangle.

Solution:

An isosceles triangle has at least two sides of equal length. We find the lengths of all three sides.

$AB = \sqrt{(-1-0)^2+(6-7)^2+(6-10)^2} = \sqrt{1+1+16} = \mathbf{\sqrt{18}}$.

$BC = \sqrt{(-4-(-1))^2+(9-6)^2+(6-6)^2} = \sqrt{(-3)^2+3^2+0^2} = \sqrt{9+9+0} = \mathbf{\sqrt{18}}$.

$AC = \sqrt{(-4-0)^2+(9-7)^2+(6-10)^2} = \sqrt{16+4+16} = \sqrt{36} = \mathbf{6}$.

Since the lengths of sides AB and BC are equal (both are $\sqrt{18}$), the triangle is isosceles.

Example 4: Identifying a Right-Angled Triangle

Question: Show that the points P(0,7,10), Q(-1,6,6), and R(-4,9,6) also form a right-angled triangle.

Solution:

We check if the side lengths satisfy the converse of the Pythagorean theorem, $a^2+b^2=c^2$. From the previous example, we have the side lengths squared:

• $PQ^2 = 18$

• $QR^2 = 18$

• $PR^2 = 36$

We check the sum of the squares of the two shorter sides: $PQ^2 + QR^2 = 18+18=\mathbf{36}$.

This sum is equal to the square of the longest side, $PR^2$. Since the condition is met, the triangle is right-angled.

Example 5: Locus of Equidistant Points

Question: Find the equation of the set of points equidistant from A(1,2,3) and B(3,2,-1).

Solution:

Let the point be P(x,y,z). The condition is $PA=PB$, which is equivalent to $PA^2=PB^2$ to avoid square roots.

Step 1: Set up the equation using the distance formula.

$(x-1)^2+(y-2)^2+(z-3)^2 = (x-3)^2+(y-2)^2+(z-(-1))^2$

Step 2: Cancel common terms and expand the rest.

The $(y-2)^2$ term cancels from both sides.

$(x^2-2x+1)+(z^2-6z+9) = (x^2-6x+9)+(z^2+2z+1)$

Step 3: Simplify the equation.

$-2x - 6z = -6x + 2z$

$4x - 8z = 0$

$\mathbf{x-2z=0}$. This is the equation of a plane.

Example 6: Finding a Point on an Axis

Question: Find the coordinates of the point on the y-axis which is at a distance of $5\sqrt{2}$ from the point P(3, -2, 5).

Solution:

Step 1: A point on the y-axis has coordinates A(0, y, 0).

Step 2: Set up the distance equation. $AP^2 = (5\sqrt{2})^2 = 50$.

$(3-0)^2 + (-2-y)^2 + (5-0)^2 = 50$

$9 + (y+2)^2 + 25 = 50$

Step 3: Solve for y.

$(y+2)^2 + 34 = 50 \implies (y+2)^2 = 16$.

$y+2 = \pm 4$.

• $y+2 = 4 \implies y=2$

• $y+2 = -4 \implies y=-6$

The points are (0, 2, 0) and (0, -6, 0).

Example 7: Identifying a Parallelogram

Question: Show that A(-1,2,1), B(1,-2,5), C(4,-7,8) and D(2,-3,4) are the vertices of a parallelogram.

Solution:

A quadrilateral is a parallelogram if its diagonals bisect each other, meaning they share the same midpoint.

Step 1: Find the midpoint of diagonal AC.

Midpoint AC = $(\frac{-1+4}{2}, \frac{2+(-7)}{2}, \frac{1+8}{2}) = \mathbf{(3/2, -5/2, 9/2)}$.

Step 2: Find the midpoint of diagonal BD.

Midpoint BD = $(\frac{1+2}{2}, \frac{-2+(-3)}{2}, \frac{5+4}{2}) = \mathbf{(3/2, -5/2, 9/2)}$.

Since the midpoints are identical, the figure is a parallelogram.

Example 8: Internal Division

Question: Find the coordinates of the point which divides the line segment joining (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3 internally.

Solution:

Using the internal section formula $R = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n})$ with $m=2, n=3$.

• $x = \frac{2(3)+3(1)}{2+3}=9/5$

• $y = \frac{2(4)+3(-2)}{2+3}=2/5$

• $z = \frac{2(-5)+3(3)}{2+3}=-1/5$

The point is (9/5, 2/5, -1/5).

Example 9: External Division

Question: Find the coordinates of the point which divides the line segment joining P(2,-1,4) and Q(3,1,5) in the ratio 2:1 externally.

Solution:

Using the external section formula $R = (\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}, \frac{mz_2-nz_1}{m-n})$ with $m=2, n=1$.

• $x = \frac{2(3)-1(2)}{2-1} = \frac{4}{1} = 4$

• $y = \frac{2(1)-1(-1)}{2-1} = \frac{3}{1} = 3$

• $z = \frac{2(5)-1(4)}{2-1} = \frac{6}{1} = 6$

The point is (4,3,6).

Example 10: Finding the Ratio of Division by a Plane

Question: Find the ratio in which the line segment joining (4, 8, 10) and (6, 10, –8) is divided by the YZ-plane.

Solution:

Step 1: The YZ-plane is the set of all points where the x-coordinate is 0.

Step 2: Let the ratio be k:1. The x-coordinate of the point of division is $x = \frac{k(x_2)+1(x_1)}{k+1}$.

Step 3: Set $x=0$ and solve for k:

$0 = \frac{k(6)+1(4)}{k+1} \implies 6k+4 = 0 \implies k = -2/3$.

Since k is negative, the division is external. The ratio is $|k| = 2/3$. Thus, the division is 2:3 externally.

Example 11: Centroid of a Triangle

Question: Find the centroid of a triangle whose vertices are (1,1,1), (2,0,3), and (4,1,1).

Solution:

The centroid is the average of the vertex coordinates.

$G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$

$G = (\frac{1+2+4}{3}, \frac{1+0+1}{3}, \frac{1+3+1}{3}) = \mathbf{(7/3, 2/3, 5/3)}$.

Example 12: Centroid of a Tetrahedron

Question: Find the centroid of the tetrahedron with vertices (1,1,1), (2,0,3), (4,1,1), and (3,2,1).

Solution:

The centroid of a tetrahedron is the average of the four vertices' coordinates.

$G = (\frac{1+2+4+3}{4}, \frac{1+0+1+2}{4}, \frac{1+3+1+1}{4})$

$G = (10/4, 4/4, 6/4) = \mathbf{(5/2, 1, 3/2)}$.

Example 13: Finding a Vertex from the Centroid

Question: If the origin O(0,0,0) is the centroid of a triangle with vertices A(a,1,3), B(-2,b,-5) and C(4,7,c), find a, b, and c.

Solution:

The centroid is the average of the coordinates. We set up an equation for each coordinate equal to the centroid's coordinate (0).

• x-coordinate: $\frac{a+(-2)+4}{3} = 0 \implies a+2=0 \implies \mathbf{a=-2}$.

• y-coordinate: $\frac{1+b+7}{3} = 0 \implies b+8=0 \implies \mathbf{b=-8}$.

• z-coordinate: $\frac{3+(-5)+c}{3} = 0 \implies c-2=0 \implies \mathbf{c=2}$.

Example 14: Distance from a Coordinate Plane

Question: Find the distance of the point (3, -4, 5) from the xz-plane.

Solution:

The distance of a point (x,y,z) from the xz-plane (which has the equation y=0) is the length of the perpendicular from the point to the plane. This distance is simply the absolute value of its y-coordinate.

Distance = $|-4| = \mathbf{4}$.

Example 15: Image of a Point in a Plane

Question: Find the image of the point (1, 2, 3) in the xy-plane.

Solution:

Reflection in the xy-plane means the 'mirror' is the plane where z=0. The reflection will be at the same (x,y) location but an equal distance 'below' the plane. This means the x and y coordinates remain the same, while the z-coordinate reverses its sign.

The image is (1, 2, -3).

Example 16: Finding a Vertex of a Parallelogram

Question: If the vertices of a parallelogram are A(1,2,3), B(-1,-2,-1), C(2,3,2) and D(x,y,z), find D.

Solution:

The diagonals of a parallelogram bisect each other, so the midpoint of AC is the same as the midpoint of BD.

• Midpoint of AC = $(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}) = (3/2, 5/2, 5/2)$.

• Midpoint of BD = $(\frac{-1+x}{2}, \frac{-2+y}{2}, \frac{-1+z}{2})$.

Equating the coordinates:

• $\frac{-1+x}{2}=3/2 \implies -1+x=3 \implies \mathbf{x=4}$.

• $\frac{-2+y}{2}=5/2 \implies -2+y=5 \implies \mathbf{y=7}$.

• $\frac{-1+z}{2}=5/2 \implies -1+z=5 \implies \mathbf{z=6}$.

The vertex D is (4,7,6).

Example 17: Equilateral Triangle Check

Question: Show that the points A(1,2,3), B(2,3,1), and C(3,1,2) form an equilateral triangle.

Solution:

We find the square of the side lengths to check if they are all equal.

• $AB^2 = (2-1)^2+(3-2)^2+(1-3)^2 = 1^2+1^2+(-2)^2 = 1+1+4 = \mathbf{6}$.

• $BC^2 = (3-2)^2+(1-3)^2+(2-1)^2 = 1^2+(-2)^2+1^2 = 1+4+1 = \mathbf{6}$.

• $AC^2 = (3-1)^2+(1-2)^2+(2-3)^2 = 2^2+(-1)^2+(-1)^2=4+1+1=\mathbf{6}$.

Since $AB^2=BC^2=AC^2$, all sides are equal in length, and the triangle is equilateral.

Example 18: Locus of a Point (Advanced)

Question: Find the locus of a point P such that the sum of the squares of its distances from the points A(1,0,0) and B(-1,0,0) is always 10.

Solution:

Let P be the point (x,y,z). The given condition is $PA^2+PB^2=10$.

Step 1: Write the squared distances using the distance formula.

• $PA^2 = (x-1)^2+(y-0)^2+(z-0)^2 = (x-1)^2+y^2+z^2$.

• $PB^2 = (x-(-1))^2+(y-0)^2+(z-0)^2 = (x+1)^2+y^2+z^2$.

Step 2: Substitute into the given condition.

$((x-1)^2+y^2+z^2) + ((x+1)^2+y^2+z^2) = 10$.

Step 3: Expand and simplify.

$(x^2-2x+1+y^2+z^2) + (x^2+2x+1+y^2+z^2) = 10$.

$2x^2+2y^2+2z^2+2=10$.

$2(x^2+y^2+z^2)=8$.

$\mathbf{x^2+y^2+z^2=4}$. This is the equation of a sphere centered at the origin with radius 2.

Example 19: Ratio of Division (Advanced)

Question: Find the ratio in which the join of A(2,1,5) and B(3,4,3) is divided by the plane $2x+2y-2z=1$.

Solution:

The ratio in which the plane $Ax+By+Cz+D=0$ divides the segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by the formula:

Ratio = $-\frac{Ax_1+By_1+Cz_1+D}{Ax_2+By_2+Cz_2+D}$.

Step 1: Identify the parameters.

• The plane is $2x+2y-2z-1=0$, so $A=2, B=2, C=-2, D=-1$.

• Point A: $(x_1,y_1,z_1)=(2,1,5)$.

• Point B: $(x_2,y_2,z_2)=(3,4,3)$.

Step 2: Substitute into the formula.

Ratio = $-\frac{2(2)+2(1)-2(5)-1}{2(3)+2(4)-2(3)-1} = -\frac{4+2-10-1}{6+8-6-1} = -\frac{-5}{7} = \mathbf{5/7}$.

Since the ratio is positive, the division is 5:7 internally.

Example 20: Finding a Point on a Sphere

Question: A sphere has its center at the origin and a radius of 5. Find the value of k if the point (3, k, 1) lies on the sphere.

Solution:

Step 1: Write the equation of the sphere.

The equation of a sphere centered at the origin with radius r is $x^2+y^2+z^2=r^2$.

Here, the equation is $x^2+y^2+z^2=5^2=25$.

Step 2: Substitute the point's coordinates.

Since the point (3,k,1) lies on the sphere, it must satisfy the equation:

$3^2+k^2+1^2=25$.

Step 3: Solve for k.

$9+k^2+1=25$.

$k^2+10=25$.

$k^2=15$.

$\mathbf{k=\pm\sqrt{15}}$.