Question 1: Find the distance between the following pairs of points:

(i) $\displaystyle P (1, - 1, 0) \text{ and } Q (2, 1,2)$           (ii) $\displaystyle A (3,2, - 1) \text{ and } B (-1, -1, 1)$

(i)      $\displaystyle P (1, - 1, 0) \text{ and } Q (2, 1,2)$

$\displaystyle PQ = \sqrt{ (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 }$

$\displaystyle = \sqrt{ (2-1)^2+(1+1)^2+(2-0)^2 }$

$\displaystyle = \sqrt{ 1^2 + 2^2 + 2^2 }$

$\displaystyle = \sqrt{9}$

$\displaystyle = 3 \text{ units }$

(ii)     $\displaystyle A (3,2, - 1) \text{ and } B (-1, -1, 1)$

$\displaystyle AB = \sqrt{ (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 }$

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

$\displaystyle = \sqrt{ (-4)^2 + (-3)^2 + (0)^2 }$

$\displaystyle = \sqrt{16+9+0}$

$\displaystyle = \sqrt{25}$

$\displaystyle = 5 \text{ units }$

$\\$

Question 2: Find the distance between the points $\displaystyle P$ and $\displaystyle Q$ having coordinates $\displaystyle (- 2, 3,1)$ and $\displaystyle (2, 1, 2)$

$\displaystyle PQ = \sqrt{ (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 }$

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

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

$\displaystyle = \sqrt{16+4+1}$

$\displaystyle = \sqrt{9}$

$\displaystyle = \sqrt{21} \text{ units }$

$\\$

Question 3: Using distance formula prove that the following points are collinear:

(i) $\displaystyle A (4, - 3, -1), B (5, -7,6) \text{ and } C (3,1, - 8)$

(ii) $\displaystyle P (0,7, -7), Q (1, 4, - 5) \text{ and } R (- 1,10, - 9)$

(iii) $\displaystyle A (3, - 5,1), B (-1,0, 8) \text{ and } C (7, - 10, - 6)$

(i) Given $\displaystyle A (4, - 3, -1), B (5, -7,6) \text{ and } C (3,1, - 8)$

$\displaystyle AB = \sqrt{ (5-4)^2+(-7+3)^2+(6+1)^2 }$

$\displaystyle = \sqrt{ (1)^2 + (-4)^2 + (7)^2 }$

$\displaystyle = \sqrt{1+16+49}$

$\displaystyle = \sqrt{66} \text{ units }$

$\displaystyle BC = \sqrt{ (3-5)^2+(1+7)^2+(-8-6)^2 }$

$\displaystyle = \sqrt{ (-2)^2 + (8)^2 + (-14)^2 }$

$\displaystyle = \sqrt{4+64+196}$

$\displaystyle = 2\sqrt{66} \text{ units }$

$\displaystyle AC = \sqrt{ (3-4)^2+(1+3)^2+(-8+1)^2 }$

$\displaystyle = \sqrt{ (-1)^2 + (4)^2 + (-7)^2 }$

$\displaystyle = \sqrt{1+16+49}$

$\displaystyle = \sqrt{66} \text{ units }$

Here, $\displaystyle AB + AC = \sqrt{66} + \sqrt{66} = 2\sqrt{66} = BC$

Hence, the points are collinear.

(ii) Given $\displaystyle P (0,7, -7), Q (1, 4, - 5) \text{ and } R (- 1,10, - 9)$

$\displaystyle PQ = \sqrt{ (1-0)^2+(4-7)^2+(-5+7)^2 }$

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

$\displaystyle = \sqrt{1+9+4}$

$\displaystyle = \sqrt{14} \text{ units }$

$\displaystyle QR = \sqrt{ (-1-1)^2+(10-4)^2+(-9+5)^2 }$

$\displaystyle = \sqrt{ (-2)^2 + (6)^2 + (-4)^2 }$

$\displaystyle = \sqrt{4+36+16}$

$\displaystyle = 2\sqrt{14} \text{ units }$

$\displaystyle PR = \sqrt{ (-1-0)^2+(10-7)^2+(-9+7)^2 }$

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

$\displaystyle = \sqrt{1+9+4}$

$\displaystyle = \sqrt{14} \text{ units }$

Here, $\displaystyle PQ + PR = \sqrt{14} + \sqrt{14} = 2\sqrt{14} = QR$

Hence, the points are collinear.

(iii) Given $\displaystyle A (3, - 5,1), B (-1,0, 8) \text{ and } C (7, - 10, - 6)$

$\displaystyle AB = \sqrt{ (-1-3)^2+(0+5)^2+(8-1)^2 }$

$\displaystyle = \sqrt{ (-4)^2 + (5)^2 + (7)^2 }$

$\displaystyle = \sqrt{16+25+49}$

$\displaystyle = 3\sqrt{10} \text{ units }$

$\displaystyle BC = \sqrt{ (7+1)^2+(-10-0)^2+(-6-8)^2 }$

$\displaystyle = \sqrt{ (8)^2 + (-10)^2 + (-14)^2 }$

$\displaystyle = \sqrt{64+100+196}$

$\displaystyle = 6\sqrt{10} \text{ units }$

$\displaystyle AC = \sqrt{ (7-3)^2+(-10+5)^2+(-6-1)^2 }$

$\displaystyle = \sqrt{ (4)^2 + (-5)^2 + (-7)^2 }$

$\displaystyle = \sqrt{16+25+49}$

$\displaystyle = 3\sqrt{10} \text{ units }$

Here, $\displaystyle AB + AC = 3\sqrt{10} + 3\sqrt{10} = 6\sqrt{10} = BC$

Hence, the points are collinear.

$\\$

Question 4: Determine the points in (i) $\displaystyle xy-plane$ (ii) $\displaystyle yz-plane$ and (iii) $\displaystyle zx-plane$ which are equidistant from the points $\displaystyle A (1, - 1, 0), B (2, 1, 2) \text{ and } C (3, 2, - 1).$

(i)     We know that the z-coordinate of every point on the xy-plane is zero.

So let $\displaystyle P(x, y, 0)$ be the point on the xy-plane such that $\displaystyle PA = PB = PB$

Now, $\displaystyle PA = PB$

$\displaystyle \Rightarrow PA^2 = PB^2$

$\displaystyle \Rightarrow (x-1)^2+(y+1)^2 + (0-0)^2 = (x-2)^2+(y-1)^2 + (0-2)^2$

$\displaystyle \Rightarrow x^2 - 2x + 1 + y^2 + 2y + 1 = x^2 - 4x + 4 + y^2 - 2y + 1 + 4$

$\displaystyle \Rightarrow -2x + 4x + 2y + 2y + 2 - 9 = 0$

$\displaystyle \Rightarrow 2x+4y-7=0$

$\displaystyle \Rightarrow 2x+4y = 7 \text{ ... ... ... ... ... i)}$

Now, $\displaystyle PB = PC$

$\displaystyle \Rightarrow PB^2 = PC^2$

$\displaystyle \Rightarrow (x-2)^2+(y-1)^2 + (0-2)^2 = (x-3)^2+(y-2)^2 + (0+1)^2$

$\displaystyle \Rightarrow x^2 - 4x + 4 + y^2 - 2y + 1+4 = x^2 - 6x + 9 + y^2 - 4y + 4 + 1$

$\displaystyle \Rightarrow -4x + 6x - 2y + 4y + 9 - 14 = 0$

$\displaystyle \Rightarrow 2x+2y-5=0$

$\displaystyle \Rightarrow 2x+2y = 5 \text{ ... ... ... ... ... ii)}$

Solving equation i) and ii) simultaneously. we get $\displaystyle x = \frac{3}{2} \text{ and } y = 1$

$\displaystyle \text{Hence the required point is } \Big( \frac{3}{2}, 1, 0 \Big)$

(ii)    We know that the x-coordinate of every point on the yz-plane is zero.

So let $\displaystyle P(0, y, z)$ be the point on the xy-plane such that $\displaystyle PA = PB = PB$

Now, $\displaystyle PA = PB$

$\displaystyle \Rightarrow PA^2 = PB^2$

$\displaystyle \Rightarrow (0-1)^2+(y+1)^2 + (z-0)^2 = (0-2)^2+(y-1)^2 + (z-2)^2$

$\displaystyle \Rightarrow 1+y^2+2y+1+z^2=4+y^2-2y+1+z^2-4z+4$

$\displaystyle \Rightarrow 2y+2 = -2y-4z+9$

$\displaystyle \Rightarrow 2y+2y -4z= 9-2$

$\displaystyle \Rightarrow 4y-4z=7 \text{ ... ... ... ... ... i)}$

Now, $\displaystyle PB = PC$

$\displaystyle \Rightarrow PB^2 = PC^2$

$\displaystyle \Rightarrow (0-2)^2+(y-1)^2 + (z-2)^2 = (0-3)^2+(y-2)^2 + (z+1)^2$

$\displaystyle \Rightarrow 4+y^2-2y+1+z^2+4-4z=9+y^2-4y+4+z^2+2z+1$

$\displaystyle \Rightarrow -2y-4z+9 = -4y+2z+14$

$\displaystyle \Rightarrow -2y+4y -4z-2z= 14-9$

$\displaystyle \Rightarrow 2y-6z=5 \text{ ... ... ... ... ... ii)}$

Solving equation i) and ii) simultaneously. we get $\displaystyle y = \frac{11}{8} \text{ and } z = \frac{-3}{8}$

$\displaystyle \text{Hence the required point is } \Big(0, \frac{11}{8}, \frac{-3}{8} \Big)$

(iii)    We know that the y-coordinate of every point on the zx-plane is zero.

So let $\displaystyle P(x, 0, z)$ be the point on the xy-plane such that $\displaystyle PA = PB = PB$

Now, $\displaystyle PA = PB$

$\displaystyle \Rightarrow PA^2 = PB^2$

$\displaystyle \Rightarrow (x-1)^2+(0+1)^2 + (z-0)^2 = (x-2)^2+(0-1)^2 + (z-2)^2$

$\displaystyle \Rightarrow x^2 + 1 - 2x +1+z^2=x^2 - 4x + 4+1+z^2-4z+4$

$\displaystyle \Rightarrow -2x+2 = -4x-4z+9$

$\displaystyle \Rightarrow 2x-4z = 7 \text{ ... ... ... ... ... i)}$

Now, $\displaystyle PB = PC$

$\displaystyle \Rightarrow PB^2 = PC^2$

$\displaystyle \Rightarrow (x-2)^2+(0-1)^2 + (z-2)^2 = (x-3)^2+(0-2)^2 + (z+1)^2$

$\displaystyle \Rightarrow x^2 + 4 - 4x +1+z^2+4-4z=x^2 - 6x + 9+4+z^2+2z+1$

$\displaystyle \Rightarrow -4x+6x -4z-2z = 14-9$

$\displaystyle \Rightarrow 2x-6z = 5 \text{ ... ... ... ... ... ii)}$

Solving equation i) and ii) simultaneously. we get $\displaystyle x = \frac{11}{2} \text{ and } z = 1$

$\displaystyle \text{Hence the required point is } \Big(\frac{11}{2}, 0, 1 \Big)$

$\\$

Question 5: Determine the point on z-axis which is equidistant from the points $\displaystyle (1,5,7) \text{ and } (5, 1, - 4).$

Let M be the point on the z-axis.

Then the coordinates of $M$ will be $(0, 0, z)$

Let M be equidistant from the points $\displaystyle (1,5,7) \text{ and } (5, 1, - 4).$

$\displaystyle AM = \sqrt{ (0-1)^2 + (0-5)^2 + ( z-7)^2 }$

$\displaystyle = \sqrt{ (-1)^2 + (-5)^2 + ( z-7)^2 }$

$\displaystyle = \sqrt{ 1 + 25 + z^2 - 14z + 49 }$

$\displaystyle = \sqrt{ z^2 - 14z + 75 }$

$\displaystyle BM = \sqrt{ (0-5)^2 + (0-1)^2 + ( z+4)^2 }$

$\displaystyle = \sqrt{ (-5)^2 + (-1)^2 + ( z+4)^2 }$

$\displaystyle = \sqrt{ 25 + 1 + z^2 +8z + 42 }$

$\displaystyle = \sqrt{ z^2 +8z + 42 }$

Since $\displaystyle AM = BM$

$\displaystyle \therefore \sqrt{ z^2 - 14z + 75 } = \sqrt{ z^2 +8z + 42 }$

Squaring both sides $\displaystyle z^2 - 14z + 75 = z^2 +8z + 42$

$\displaystyle \Rightarrow - 14z - 8z = 42- 75$

$\displaystyle \Rightarrow - 22z = - 33$

$\displaystyle \Rightarrow z = \frac{33}{22} = \frac{3}{2}$

$\displaystyle \text{Thus the coordinates of M are } \Big( 0, 0, \frac{3}{2} \Big)$

$\\$

Question 6: Find the point on y-axis which is equidistant from the points $\displaystyle (3, 1, 2) \text{ and } (5, 5, 2).$

Let the point of y-axis be $\displaystyle M(0, y, 0)$ which is equidistant from the points $\displaystyle P(3, 1, 2) \text{ and } Q(5, 5, 2).$

Therefore $\displaystyle PM = QM$

$\displaystyle \therefore \sqrt{ (0-3)^2+(y-1)^2+(0-2)^2 } = \sqrt{ (0-5)^2+(y-5)^2+(0-2)^2 }$

$\displaystyle \Rightarrow \sqrt{ (-3)^2+y^2 + 2y + 1+(-2)^2 } = \sqrt{ (-5)^2+y^2 + 10y + 25+(-2)^2 }$

$\displaystyle \Rightarrow \sqrt{ 9+y^2 + 2y + 1+4 } = \sqrt{ 25+y^2 + 10y + 25+4 }$

$\displaystyle \Rightarrow \sqrt{ y^2 + 2y + 14 } = \sqrt{ y^2 + 10y + 54 }$

Squaring both sides we get

$\displaystyle \Rightarrow y^2 + 2y + 14 = y^2 + 10y + 54$

$\displaystyle \Rightarrow 2y-10y = 54-14$

$\displaystyle \Rightarrow -8y = 40$

$\displaystyle \Rightarrow y = -5$

Therefore the required point on the y-axis is $\displaystyle (0, -5, 0)$

$\\$

Question 7: Find the points on $\displaystyle z-axis$ which are at a distance $\displaystyle \sqrt{21}$ from the point $\displaystyle (1 , 2, 3) .$

Let the point of z-axis be $\displaystyle A(0, 0, z)$

$\displaystyle AP = \sqrt{21}$

$\displaystyle \Rightarrow \sqrt{ (0-1)^2 + (0-2)^2 + (z-3)^2 } = \sqrt{21}$

$\displaystyle \Rightarrow (-1)^2 + (-2)^2 + ( z-3)^2 = 21$

$\displaystyle \Rightarrow 1 + 4 + (z-3)^2 = 21$

$\displaystyle \Rightarrow (z-3)^2 = 21 - 5$

$\displaystyle \Rightarrow (z-3)^2 = 16$

$\displaystyle \Rightarrow (z-3) = \pm 4$

$\displaystyle \Rightarrow z = 7 \text{ or } z = -1$

Hence the coordinates of the required point are $\displaystyle ( 0, 0, 7)$ and $\displaystyle (0, 0, -1)$

$\\$

Question 8: Prove that the triangle formed by joining the three points whose coordinates are $\displaystyle (1,2,3), (2,3,1) \text{ and } (3, 1,2)$ is an equilateral triangle.

Let $\displaystyle A(1,2,3), B(2,3,1) \text{ and } C(3, 1,2)$ are the coordinates of the $\displaystyle \triangle ABC$

$\displaystyle AB = \sqrt{ (2-1)^2 + ( 3-2)^2 + ( 1 -3)^2}$

$\displaystyle = \sqrt{ (1)^2 + (1)^2 + (-2)^2}$

$\displaystyle = \sqrt{ 1 + 1 + 4}$

$\displaystyle = \sqrt{6}$

$\displaystyle BC = \sqrt{ (3-2)^2 + ( 1-3)^2 + ( 2-1)^2}$

$\displaystyle = \sqrt{ (1)^2 + (-2)^2 + (1)^2}$

$\displaystyle = \sqrt{ 1 + 4 + 1}$

$\displaystyle = \sqrt{6}$

$\displaystyle CA = \sqrt{ (3-1)^2 + (1-2)^2 + (2 -3)^2}$

$\displaystyle = \sqrt{ (2)^2 + (-1)^2 + (-1)^2}$

$\displaystyle = \sqrt{ 4 + 1 + 1}$

$\displaystyle = \sqrt{6}$

Now, $\displaystyle AB = BC = CA$

Therefore $\displaystyle \triangle ABC$ is an equilateral triangle.

$\\$

Question 9: Show that the points $\displaystyle (0,7, 10), (- 1, 6,6) \text{ and } (- 4,9,6)$ are the vertices of an isosceles right-angled triangle.

Let $\displaystyle A(0,7, 10), B(- 1, 6,6) \text{ and } C(- 4,9,6)$ be the coordinates of $\triangle ABC$

$\displaystyle AB = \sqrt{ (0+1)^2+(7-6)^2 + (10-2)^2 } = \sqrt{18} = 3\sqrt{2}$

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

$\displaystyle CA = \sqrt{ (0+4)^2+(7-9)^2 + (10-6)^2 } = \sqrt{36} = 6$

Clearly, $\displaystyle AB^2 + BC^2 = CA^2 \Rightarrow \angle ABC = 90^{\circ} \text{ and } AB = BC$

Therefore the $\displaystyle \triangle ABC$ is a right-angles isosceles triangle.

$\\$

Question 10: Show that the points $\displaystyle A (3,3,3),B(0,6,3), C(1.,7,7) \text{ and } D(4,4,7)$ are the vertices of a square.

Let $\displaystyle A (3,3,3),B(0,6,3), C(1.,7,7) \text{ and } D(4,4,7)$ are the vertices of a quadrilateral ABCD.

$\displaystyle AB = \sqrt{ (0-3)^2+(6-3)^2 + (3-3)^2 } = \sqrt{18} = 3\sqrt{2}$

$\displaystyle BC = \sqrt{ (1-0)^2+(7-6)^2 + (7-3)^2 } = \sqrt{18} = 3\sqrt{2}$

$\displaystyle CD = \sqrt{ (4-1)^2+(4-7)^2 + (7-7)^2 } = \sqrt{18} = 3\sqrt{2}$

$\displaystyle DA = \sqrt{ (4-3)^2+(4-3)^2 + (7-3)^2 } = \sqrt{18} = 3\sqrt{2}$

Therefore $\displaystyle AB = BC = CD = DA.$ The four sides are equal.

$\displaystyle AC = \sqrt{ (1-3)^2+(7-3)^2 + (7-3)^2 } = \sqrt{36} = 6$

$\displaystyle BD = \sqrt{ (4-0)^2+(4-6)^2 + (7-3)^2 } = \sqrt{36} = 6$

Therefore $\displaystyle AC = BD.$The diagonals are equal.

Hence $\displaystyle ABCD$ is a square.

$\\$

Question 11: Prove that the point $\displaystyle A(1,3,0), B(-5,5,2),C(-9,-1,2) \text{ and } D(-9,-9,0)$ taken in order are the vertices of a parallelogram. Also, show that, $\displaystyle ABCD$ is not a rectangle.

Let $\displaystyle A(1,3,0), B(-5,5,2),C(-9,-1,2) \text{ and } D(-9,-9,0)$ are the vertices of a quadrilateral ABCD.

$\displaystyle AB = \sqrt{ (-5-1)^2+(5-3)^2 + (2-0)^2 } = \sqrt{44} = 2\sqrt{11}$

$\displaystyle BC = \sqrt{ (-9+5)^2+(-1-5)^2 + (2-2)^2 } = \sqrt{52} = 2\sqrt{13}$

$\displaystyle CD = \sqrt{ (-3+9)^2+(-3+1)^2 + (0-2)^2 } = \sqrt{44} = 2\sqrt{11}$

$\displaystyle DA = \sqrt{ (1+3)^2+(3+3)^2 + (0-0)^2 } = \sqrt{52} = 2\sqrt{13}$

Therefore $\displaystyle AB = CD \text{ and } BC = DA.$ Since the opposite sides are equal, therefore, ABCD is a parallelogram.

$\displaystyle AC = \sqrt{ (-9-1)^2+(-1-3)^2 + (2-0)^2 } = \sqrt{100+16+4}= \sqrt{120}$

$\displaystyle BD = \sqrt{ (-3+5)^2+(-3-5)^2 + (0-2)^2 } = \sqrt{4+64+4}= \sqrt{72}$

Therefore $\displaystyle AC \neq BD.$The diagonals are not equal.

Hence $\displaystyle ABCD$ is not a rectangle.

$\\$

Question 12: Show that the points $\displaystyle A(1,3,4), B(-1,6,10), C(-7,4,7) \text{ and } D(-5,1,1)$ are the vertices of a rhombus.

Let $\displaystyle A(1,3,4), B(-1,6,10), C(-7,4,7) \text{ and } D(-5,1,1)$ are the vertices of a quadrilateral ABCD.

$\displaystyle AB = \sqrt{ (-1-1)^2+(6-3)^2 + (10-4)^2 } = \sqrt{4+9+36} = \sqrt{49} = 7$

$\displaystyle BC = \sqrt{ (-7+1)^2+(4-6)^2 + (7-10)^2 } = \sqrt{36+4+9} = \sqrt{49} = 7$

$\displaystyle CD = \sqrt{ (-5+7)^2+(1-4)^2 + (1-7)^2 } = \sqrt{4+9+36} = \sqrt{49} = 7$

$\displaystyle DA = \sqrt{ (1+5)^2+(3-1)^2 + (4-1)^2 } = \sqrt{36+4+9} = \sqrt{49} = 7$

Therefore $\displaystyle AB = BC=CD+DA$ Since all sides are equal, therefore, ABCD is a rhombus.

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Question 13: Prove that the tetrahedron with vertices at the points $\displaystyle O (0,0,0), A (0,1,1),B(1, 0, 1) \text{ and } C (1, 1, 0)$ is a regular one.

The faces of a regular tetrahedron are equilateral triangles.

$\displaystyle OA = \sqrt{ (0-0)^2+(0-1)^2 + (0-1)^2 } = \sqrt{0+1+1} = \sqrt{2}$

$\displaystyle OB = \sqrt{ (1-0)^2+(0-0)^2 + (1-0)^2 } = \sqrt{1+0+1} = \sqrt{2}$

$\displaystyle OC = \sqrt{ (1-0)^2+(0-1)^2 + (1-1)^2 } = \sqrt{1+1+0} = \sqrt{2}$

Hence this face is an equilateral triangle. Similarly, the other faces are equilateral triangles.

Hence the tetrahedron is a regular one.

$\\$

Question 14: Show that the points $\displaystyle (3,2,2), (-1,4,2), (0,5,6), (2, 1,2)$ lie on a sphere whose center is $\displaystyle (1,3,4).$ Find also its radius.

Let $\displaystyle (3,2,2), (-1,4,2), (0,5,6), (2, 1,2)$ lie on a sphere whose center is $\displaystyle (1,3,4).$

Since AP, BP, CP and DP are radii, therefore AP = BP = CP = DP.

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

$\displaystyle = \sqrt{ (-2)^2 + (1)^2 + (2)^2}$

$\displaystyle = \sqrt{ 4 + 1 + 4}$

$\displaystyle = \sqrt{9} = 3$

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

$\displaystyle = \sqrt{ (2)^2 + (-1)^2 + (2)^2}$

$\displaystyle = \sqrt{ 4 + 1 + 4}$

$\displaystyle = \sqrt{9} = 3$

$\displaystyle CP = \sqrt{ (1-0)^2 + (3-5)^2 + (4-6)^2}$

$\displaystyle = \sqrt{ (1)^2 + (-2)^2 + (-2)^2}$

$\displaystyle = \sqrt{ 1 + 4 + 4}$

$\displaystyle = \sqrt{9} = 3$

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

$\displaystyle = \sqrt{ (-1)^2 + (2)^2 + (2)^2}$

$\displaystyle = \sqrt{ 1 + 4 + 4}$

$\displaystyle = \sqrt{9} = 3$

Now, $\displaystyle AP = BP = CP=DP$

Hence $\displaystyle (3,2,2), (-1,4,2), (0,5,6), (2, 1,2)$ lie on a sphere whose center is $\displaystyle (1,3,4).$ and radius is $\displaystyle 3 .$

$\\$

Question 15: Find the coordinates of the point which is equidistant from the four points $\displaystyle O (0,0,0), A (2,0,0), B (0,3,0) \text{ and } C (0, 0, 8).$

Let P(x, y, z) be the point which is equidistant from the four points $\displaystyle O (0,0,0), A (2,0,0), B (0,3,0) \text{ and } C (0, 0, 8).$

The $\displaystyle OP = AP$

$\displaystyle \Rightarrow OP^2 = AP^2$

$\displaystyle \therefore x^2 + y^2 + z^2 = (x-2)^2 + y^2 + z^2$

$\displaystyle \Rightarrow x^2 = ( x- 2)^2$

$\displaystyle \Rightarrow x^2 = x^2 - 4x + 4$

$\displaystyle \Rightarrow 4x = 4$

$\displaystyle \therefore x = 1$

Similarly, we have

$\displaystyle OP=BP$

$\displaystyle \Rightarrow OP^2 = BP^2$

$\displaystyle \therefore x^2 + y^2 + z^2 = x^2 + ( y-3)^2 + z^2$

$\displaystyle \Rightarrow y^2 = y^2 - 6y + 9$

$\displaystyle \Rightarrow 6y = 9$

$\displaystyle \Rightarrow y = \frac{3}{2}$

Similarly, we also have

$\displaystyle OP = CP$

$\displaystyle \Rightarrow OP^2 = CP^2$

$\displaystyle \Rightarrow x^2 + y^2 + z^2 = x^2 + y^2 + ( z-8)^2$

$\displaystyle \Rightarrow z^2 = z^2 - 16z + 64$

$\displaystyle \Rightarrow 16 z= 64$

$\displaystyle \Rightarrow z = 4$

$\displaystyle \text{Thus the required point is } P ( 1, \frac{3}{2}, 4)$

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Question 16: If $\displaystyle A(-2,2,3) \text{ and } B(13,-3,13)$ are two points. Find the locus of a point $\displaystyle P$ which moves in such away that $\displaystyle 3PA=2PB.$

Let the coordinate of point $\displaystyle P$ be $\displaystyle ( x, y, z)$

Given:

$\displaystyle 3PA = 2PB$

$\displaystyle \Rightarrow 3( \sqrt{(x+2)^2 + ( y-2)^2 + ( z-3)^2 } ) = 2 ( \sqrt{ (x-13)^2 + ( y+3)^2 + ( z-13)^2} )$

$\displaystyle \Rightarrow 3( \sqrt{x^2 + 2x + 4 + y^2 + 4 - 4 y + z^2+9 - 6z } ) = 2 ( \sqrt{ x^2 - 26x + 169 +y^2 + 6y + 9 + z^2- 26z+ 169} )$

Squaring both sides we get:

$\displaystyle 9( x^2 + y^2 + z^2+ 4x- 4y - 6z+ 17 ) = 4 ( x^2 + y^2 + z^2 - 26x +6y -26z +347 )$

$\displaystyle \Rightarrow 9x^2 + 9y^2 + 9z^2+ 36x- 36y - 54z+ 153 = 4x^2 + 4y^2 + 4z^2 - 104x +24y -104z +1388$

$\displaystyle \Rightarrow 5x^2 + 5y^2 + 5z^2 + 140 x - 60y +50z - 1235 = 0$

This is the locus of the point $\displaystyle P.$

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Question 17: Find the locus of $\displaystyle P$ if $\displaystyle PA^2 + PB^2 =2k^2,$ where $\displaystyle A \text{ and } B$ are the points $\displaystyle (3, 4,5) \text{ and } (-1,3,-7).$

Let the coordinate of point $\displaystyle P$ be $\displaystyle ( x, y, z)$

Given:

$\displaystyle PA^2 +PB^2 = 2k^2$

$\displaystyle ( \sqrt{(x-3)^2 + ( y-4)^2 + ( z-5)^2 } ) + ( \sqrt{ (x+1)^2 + ( y-3)^2 + ( z+7)^2} ) = 2k^2$

$\displaystyle ( \sqrt{x^2 -6x + 9 + y^2 + 16 - 8 y + z^2+25 - 10z } ) + ( \sqrt{ x^2 + 2x + 1 +y^2 - 6y + 9 + z^2+14z+ 49} ) = 2k^2$

Squaring both sides we get:

$\displaystyle ( x^2 + y^2 + z^2-6x- 8y - 10z+ 50 ) + ( x^2 + y^2 + z^2 +2x -6y +14z +59 ) = 2k^2$

$\displaystyle \Rightarrow 2x^2 + 2y^2 + 2z^2 -4 x - 14y +4z +109-2k^2 = 0$

This is the locus of the point $\displaystyle P.$

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Question 18: Show that the points $\displaystyle (a , b , c), (b , c , a) \text{ and } (c, a, b)$ are the vertices of an equilateral triangle.

Let $\displaystyle (a , b , c), (b , c , a) \text{ and } (c, a, b)$ be the vertices of $\displaystyle \triangle ABC.$ Then,

$\displaystyle AB = \sqrt{ (b-a)^2+(c-b)^2+(a-c)^2 }$

$\displaystyle = \sqrt{ b^2-2ab+a^2+c^2-2bc+b^2+a^2-2ca+c^2 }$

$\displaystyle = \sqrt{ 2a^2+2b^2+2c^2-2ab-2bc-2ca } = \sqrt{ 2 (a^2+b^2+c^2-ab-bc-ca) }$

$\displaystyle BC = \sqrt{ (c-b)^2+(a-c)^2+(b-a)^2 }$

$\displaystyle = \sqrt{ c^2 - 2bc + c^2 + a^2 - 2ac+c^2+b^2 - 2ba+a^2 }$

$\displaystyle = \sqrt{ 2a^2+2b^2+2c^2-2ab-2bc-2ca } = \sqrt{ 2 (a^2+b^2+c^2-ab-bc-ca) }$

$\displaystyle BC = \sqrt{ (a-c)^2+(b-a)^2+(c-b)^2 }$

$\displaystyle = \sqrt{ a^2 - 2ac + c^2 + b^2 - 2ba+a^2+c^2 - 2bc+b^2 }$

$\displaystyle = \sqrt{ 2a^2+2b^2+2c^2-2ab-2bc-2ca } = \sqrt{ 2 (a^2+b^2+c^2-ab-bc-ca) }$

$\displaystyle \therefore AB = BC = CA$

Therefore $\displaystyle \triangle ABC$ is an equilateral triangle.

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Question 19: Are the points $\displaystyle A(3,6,9), B(10,20,30) \text{ and } C(25,-41, 5),$ the vertices of a right-angled triangle?

Let $\displaystyle A(3,6,9), B(10,20,30) \text{ and } C(25,-41, 5),$ the vertices of a $\displaystyle \triangle ABC$

$\displaystyle AB = \sqrt{ (10-30)^2 + ( 20-6)^2 + ( 30-9)^2 }$

$\displaystyle = \sqrt{ (7)^2 + ( 14)^2 + ( 21)^2 }$

$\displaystyle = \sqrt{ 49 + 196+ 441 }$

$\displaystyle = \sqrt{686} = 7\sqrt{14}$

$\displaystyle BC = \sqrt{ (25-10)^2 + ( -41-20)^2 + ( 5-30)^2 }$

$\displaystyle = \sqrt{ (15)^2 + ( -61)^2 + ( -25)^2 }$

$\displaystyle = \sqrt{ 225 + 3721+ 625 }$

$\displaystyle = \sqrt{4571}$

$\displaystyle CA = \sqrt{ (3-25)^2 + ( 6+41)^2 + ( 9-5)^2 }$

$\displaystyle = \sqrt{ (-22)^2 + ( 47)^2 + ( -4)^2 }$

$\displaystyle = \sqrt{ 484 + 2209+ 16 }$

$\displaystyle = \sqrt{2709} = 3\sqrt{301}$

$\displaystyle AB^2 + BC^2 = (7\sqrt{14} )^2+(\sqrt{4571})^2 = 686+4571 = 5257$

$\displaystyle CA^2 = 2709$

$\displaystyle \therefore AB^2 + BC^2 \neq CA^2$

Therefore the points are not vertices of a right angles triangle.

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Question 20: Verify the following:

(i) $\displaystyle (0, 7, -10), (1,6, -6) \text{ and } (4,9, -6)$ are vertices of an isosceles triangle.

(ii) $\displaystyle (0,7,10), (-1,6,6) \text{ and } (-4,9,6)$ are vertices of a right-angled triangle.

(iii) $\displaystyle (-1,2,1), (1, -2,5), (4, -7,8) \text{ and } (2, -3,4)$ are vertices of a parallelogram.

(iv) $\displaystyle (5, -1,1), (7 , -4,7), (1, -6,10) \text{ and } (-1, - 3,4)$ are the vertices of a rhombus.

(i) Let $\displaystyle (0, 7, -10), (1,6, -6) \text{ and } (4,9, -6)$ be the vertices of a $\displaystyle \triangle ABC$

$\displaystyle AB = \sqrt{ (1-0)^2 + ( 6-7)^2 + ( -6+10)^2 }$

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

$\displaystyle = \sqrt{ 1 + 1+ 16 }$

$\displaystyle = \sqrt{18} = 3\sqrt{2}$

$\displaystyle BC = \sqrt{ (4-1)^2 + ( 9-6)^2 + ( -6+6)^2 }$

$\displaystyle = \sqrt{ (3)^2 + ( 3)^2 + (0)^2 }$

$\displaystyle = \sqrt{ 9+9 }$

$\displaystyle = \sqrt{18} = 3\sqrt{2}$

$\displaystyle CA = \sqrt{ (0-4)^2 + ( 7-9)^2 + ( -10+6)^2 }$

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

$\displaystyle = \sqrt{ 16 + 4+ 16 }$

$\displaystyle = \sqrt{36} = 6$

Therefore we see $\displaystyle AB = BC$

Thus the given points are the vertices of an isosceles triangle.

(ii) Let $\displaystyle (0,7,10), (-1,6,6) \text{ and } (-4,9,6)$ be the vertices of a $\displaystyle \triangle ABC$

$\displaystyle AB = \sqrt{ (-1-0)^2 + ( 6-7)^2 + ( 6-10)^2 }$

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

$\displaystyle = \sqrt{ 1 + 1+ 16 }$

$\displaystyle = \sqrt{18} = 3\sqrt{2}$

$\displaystyle BC = \sqrt{ (-4+1)^2 + ( 9-6)^2 + ( -6+6)^2 }$

$\displaystyle = \sqrt{ (-3)^2 + ( 3)^2 + (0)^2 }$

$\displaystyle = \sqrt{ 9+9 }$

$\displaystyle = \sqrt{18} = 3\sqrt{2}$

$\displaystyle CA = \sqrt{ (-4-0)^2 + ( 9-7)^2 + ( 6-10)^2 }$

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

$\displaystyle = \sqrt{ 16 + 4+ 16 }$

$\displaystyle = \sqrt{36} = 6$

Therefore we see $\displaystyle AC^2 = AB^2 + BC^2$

Thus the given points are the vertices of a right-angled  triangle.

(iii) Let $\displaystyle (-1,2,1), (1, -2,5), (4, -7,8) \text{ and } (2, -3,4)$ be the vertices of a quadrilateral $\displaystyle ABCD$

$\displaystyle AB = \sqrt{ (1+1)^2 + ( -2-2)^2 + ( 5-1)^2 }$

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

$\displaystyle = \sqrt{ 4 + 16+ 16 }$

$\displaystyle = \sqrt{36} = 6$

$\displaystyle BC = \sqrt{ (4-1)^2 + ( -7+2)^2 + ( 8-5)^2 }$

$\displaystyle = \sqrt{ (3)^2 + ( -5)^2 + (3)^2 }$

$\displaystyle = \sqrt{ 9+25 + 9 }$

$\displaystyle = \sqrt{43}$

$\displaystyle CD = \sqrt{ (2-4)^2 + ( -3+7)^2 + ( 4-8)^2 }$

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

$\displaystyle = \sqrt{ 4 + 16+ 16 }$

$\displaystyle = \sqrt{36} = 6$

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

$\displaystyle = \sqrt{ (-3)^2 + ( 5)^2 + (-3)^2 }$

$\displaystyle = \sqrt{ 9+25 + 9 }$

$\displaystyle = \sqrt{43}$

Therefore we see $\displaystyle AB = CD \text{ and } BC = DA$

Since. each pair of opposite sides are equal. the quadrilateral $\displaystyle ABCD$

(iv) Let $\displaystyle (5, -1,1), (7 , -4,7), (1, -6,10) \text{ and } (-1, - 3,4)$ be the vertices of a quadrilateral $\displaystyle ABCD$

$\displaystyle AB = \sqrt{ (5-7)^2 + (-1+4)^2 + ( 1-7)^2 }$

$\displaystyle = \sqrt{ (-2)^2 + ( 3)^2 + ( -6)^2 }$

$\displaystyle = \sqrt{ 4 + 9+ 36 }$

$\displaystyle = \sqrt{49} = 7$

$\displaystyle BC = \sqrt{ (7-1)^2 + ( -4+6)^2 + ( 7-10)^2 }$

$\displaystyle = \sqrt{ (6)^2 + ( 2)^2 + (-3)^2 }$

$\displaystyle = \sqrt{ 36+4 + 9 }$

$\displaystyle = \sqrt{49} = 7$

$\displaystyle CD = \sqrt{ (7-1)^2 + ( -4+6)^2 + ( 7-10)^2 }$

$\displaystyle = \sqrt{ (6)^2 + ( 2)^2 + ( -3)^2 }$

$\displaystyle = \sqrt{ 36 + 4+ 9 }$

$\displaystyle = \sqrt{49} = 7$

$\displaystyle DA = \sqrt{ (-1-5)^2 + ( -3+1)^2 + ( 4-1)^2 }$

$\displaystyle = \sqrt{ (-6)^2 + ( -2)^2 + (3)^2 }$

$\displaystyle = \sqrt{ 36+4 + 9 }$

$\displaystyle = \sqrt{49} = 7$

Therefore we see $\displaystyle AB = BC=CD=DA$

Since. each sides are equal. the quadrilateral $\displaystyle ABCD$ is a rhombus

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Question 21: Find the locus of the points which are equidistant from the points $\displaystyle (1,2,3) \text{ and } (3, 2,-1).$

Let $\displaystyle P(x, y, z)$ be any point that is equidistant from the points $\displaystyle (1,2,3) \text{ and } (3, 2,-1).$

Therefore $\displaystyle PA = PB$

$\displaystyle \Rightarrow \sqrt{ (x-1)^2 + ( y-2)^2 + ( z-3)^2 } = \sqrt{ (x-3)^2 + ( y-2)^2 + (z+1)^2 }$

$\displaystyle \Rightarrow x^2-2x+1+y^2-4y+4+z^2+9-6z = x^2-6x+9+y^2 - 4y + 4 + z^2 + 2z + 1$

$\displaystyle \Rightarrow -2x - 4y - 6z + 14 = - 6x - 4y+2z+14$

$\displaystyle \Rightarrow -2x+6x-4y+4y-6z-2z= 14 - 14$

$\displaystyle \Rightarrow 4x-8z=0$

$\displaystyle \Rightarrow x - 2z=0$

Hence the locus is $\displaystyle x - 2z = 0$

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Question 22: Find the locus of the point, the sum of whose distances from the points $\displaystyle A(4,-0,0) \text{ and } B (- 4,0, 0)$ is equal to $\displaystyle 10.$

Let P(x, y, z) be any point, the sum of whose distances from the points $\displaystyle A(4,-0,0) \text{ and } B (- 4,0, 0)$ is equal to $\displaystyle 10.$

Therefore PA + PB = 10

$\displaystyle \Rightarrow \sqrt{ (x-4)^2 + ( y-0)^2 + ( z-0)^2 } + \sqrt{ (x+4)^2 + ( y-0)^2 + (z-0)^2 } = 10$

$\displaystyle \Rightarrow \sqrt{ (x+4)^2 + ( y-0)^2 + (z-0)^2 } = 10 - \sqrt{ (x-4)^2 + ( y-0)^2 + ( z-0)^2 }$

$\displaystyle \Rightarrow \sqrt{ x^2+8x+16 + y^2 + z^2 } = 10 - \sqrt{ x^2-8x+16 + y^2+z^2 }$

Squaring both sides we get

$\displaystyle \Rightarrow x^2+8x+16 + y^2 + z^2 = 100 + x^2-8x+16 + y^2+z^2 - 20 \sqrt{ x^2-8x+16 + y^2+z^2 }$

$\displaystyle \Rightarrow 16x-100 = - 20 \sqrt{ x^2-8x+16 + y^2+z^2 }$

$\displaystyle \Rightarrow 4x-25 = - 5 \sqrt{ x^2-8x+16 + y^2+z^2 }$

Squaring both sides we get

$\displaystyle \Rightarrow 16x^2 + 625 - 200 x = 25 ( x^2-8x+16 + y^2+z^2 )$

$\displaystyle \Rightarrow 16x^2 + 625 - 200 x = 25 x^2-200x+400 + 25y^2+25z^2$

$\displaystyle \Rightarrow 9x^2+25y^2+25z^2-225=0$

Hence the locus is $\displaystyle 9x^2+25y^2+25z^2-225=0$

$\\$

Question 23: Show that the points $\displaystyle A (1,2, 3), B(-1, -2, -1),C (2, 3,2) \text{ and } , D (4,7 ,6)$ and the vertices of a parallelogram $\displaystyle ABCD,$ but not a rectangle.

Let $\displaystyle A (1,2, 3), B(-1, -2, -1),C (2, 3,2) \text{ and } , D (4,7 ,6)$ are the vertices of a quadrilateral ABCD.

$\displaystyle AB = \sqrt{ (-1-1)^2+(-2-2)^2 + (-1-3)^2 } = \sqrt{4+16+16 } = \sqrt{36} = 6$

$\displaystyle BC = \sqrt{ (2+1)^2+(3+2)^2 + (2+1)^2 } = \sqrt{9+25+9 } = \sqrt{43}$

$\displaystyle CD = \sqrt{ (4-2)^2+(7-3)^2 + (6-2)^2 } = \sqrt{4+16+16 } = \sqrt{36} = 6$

$\displaystyle DA = \sqrt{ (1-4)^2+(2-7)^2 + (3-6)^2 } = \sqrt{9+25+9 } = \sqrt{43}$

Therefore $\displaystyle AB = CD \text{ and } BC = DA.$ The opposite sides are equal. Therefore ABCD is a parallelogram.

$\displaystyle AC = \sqrt{ (2-1)^2+(3-2)^2 + (2-3)^2 } =\sqrt{1+1+1 } = \sqrt{3}$

$\displaystyle BD = \sqrt{ (4+1)^2+(7+2)^2 + (6+1)^2 } = \sqrt{25+81+49 } = \sqrt{155}$

Therefore $\displaystyle AC \neq BD.$The diagonals are not equal.

Hence $\displaystyle ABCD$ is not a rectangle.

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Question 24: Find the equation of the set of the points P such that its distances from the points $\displaystyle A(3,4, -5) \text{ and } B (-2,1,4)$ are equal.

Let $\displaystyle P(x, y, z)$ be any point that is equidistant from the points $\displaystyle A(3,4, -5) \text{ and } B (-2,1,4)$

Therefore $\displaystyle PA = PB$

$\displaystyle \Rightarrow \sqrt{ (x-3)^2 + ( y-4)^2 + ( z+5)^2 } = \sqrt{ (x+2)^2 + ( y-1)^2 + (z-4)^2 }$

$\displaystyle \Rightarrow \sqrt{ x^2-6x+9+y^2-8y+16+z^2+25-10z } = \sqrt{ x^2-4x+4+y^2 - 2y + 1 + z^2 - 8z + 16 }$

Squaring both sides we get

$\displaystyle \Rightarrow x^2-6x+9+y^2-8y+16+z^2+25-10z = x^2-4x+4+y^2 - 2y + 1 + z^2 - 8z + 16$

$\displaystyle \Rightarrow -10x-6y+18z+29=0$

$\displaystyle \Rightarrow \therefore 10x+6y-18z-29=0$

Hence, the required equation is $\displaystyle 10x+6y-18z-29=0$