Q: Find the equation of the set of points which are equidistant from the points \$(1, 2, 3)\$ and \$(3, 2, -1)\$

Let \$P (x, y, z)\$ be the point that is equidistant from points \$A(1, 2, 3)\$ and \$B(3, 2, -1)\$ .i.e. \$PA = PB\$\$\displaystyle \Rightarrow PA^{2}=PB^{2}\$\$\displaystyle \Rightarrow \left ( x-1 \right )^{2}+\left ( y-2 \right )^{2}+\left ( z-3 \right )^{2}=\left ( x-3 \right )^{2}+\left ( y-2 \right )^{2}+\left ( z+1 \right )^{2}\$\$\displaystyle \Rightarrow x^{2}-2x+1+y^{2}-4y+4+z^{2}-6z+9=x^2-6x+9+y^{2}-4y+4+z^{2}+2z+1\$\$\displaystyle \Rightarrow -2x-6z+14=-6x+2z+14\$\$\displaystyle \Rightarrow -2x-6z+6x-2z=0\$\$\displaystyle \Rightarrow 4x-8z=0\$\$\displaystyle \Rightarrow x-2z=0\$Thus the required equation is \$x-2z=0 \$.

Q: 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\$.

Let the coordinates of \$P\$ be \$(x, y, z)\$.The coordinates of points \$A\$ and \$B\$ are \$(4, 0, 0)\$ and \$(-4, 0, 0)\$ respectively.It is given that \$PA + PB = 10\$\$\displaystyle \Rightarrow \sqrt{\left ( x-4 \right )^{2}+y^{2}+z^{2}}+\sqrt{\left ( x+4 \right )^{2}+y^{2}+z^{2}}=10\$\$\displaystyle \Rightarrow \sqrt{\left ( x-4 \right )^{2}+y^{2}+z^{2}}=10-\sqrt{\left ( x+4 \right )^{2}+y^{2}+z^{2}}\$On squaring both sides, we obtain\$\displaystyle \Rightarrow \left ( x-4 \right )^{2}+y^{2}+z^{2}=100+ \left( x+4 \right )^{2}+y^{2}+z^{2}-20\sqrt{\left ( x+4 \right )^{2}+y^{2}+z^{2}}\$\$\displaystyle \Rightarrow x^{2}-8x+16+y^{2}+z^{2}=100-20\sqrt{x^{2}+8x+16+y^{2}+z^{2}}+x^{2}+8x+16+y^{2}+z^{2} \$\$\displaystyle \Rightarrow 20\sqrt{x^{2}+8x+16+y^{2}+z^{2}}=100+16x\$\$\displaystyle \Rightarrow 5\sqrt{x^{2}+8x+16+y^{2}+z^{2}}=\left ( 25+4x \right )\$On squaring both sides again, we obtain\$\displaystyle 25\left ( x^{2} +8x+16+y^{2}+z^{2}\right )=625+16x^{2}+200x\$\$\displaystyle \Rightarrow 25x^{2}+200x+400+25y^{2}+25z^{2}=625+16x^{2}+200x\$\$\displaystyle \Rightarrow 9x^{2}+25y^{2}+25z^{2}-225=0\$Thus the required equation is \$\displaystyle 9x^{2}+25y^{2}+25z^{2}-225=0\$.

Q: Find the distance between the following pairs of points:(i) \$(2, 3, 5)\$ and \$(4, 3, 1)\$(ii)\$ (-3, 7, 2)\$and \$((2, 4, -1)\$(iii) \$(-1, 3, -4)\$ and \$(1, -3, 4)\$(iv) \$(2, -1, 3)\$ and \$(-2, 1, 3)\$

The distance between points \$P(x_{1}, y _{1}, z_{1})\$ and \$Q(x_{2}, y _{2}, z_{2})\$ is given by\$PQ =\$ \$\displaystyle \sqrt{\left ( x_{2} -x_{1}\right )^{2}+\left ( y_{2} -y_{1}\right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}\$(i) Distance between points \$(2, 3, 5)\$ and \$(4, 3, 1)\$\$=\$ \$\displaystyle \sqrt{\left ( 4-2 \right )^{2}+\left ( 3-3 \right )^{2}+\left ( 1-5 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{\left ( 2 \right )^{2}+\left ( 0 \right )^{2}+\left ( -4 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{4+16}\$\$=\$ \$\displaystyle \sqrt{20}\$\$=\$ \$\displaystyle 2\sqrt{5}\$(ii) Distance between points \$(-3, 7, 2)\$ and \$(2, 4, -1)\$\$=\$ \$\displaystyle \sqrt{\left ( 2+3 \right )^{2}+\left ( 4-7 \right )^{2}+\left ( -1-2 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{\left ( 5 \right )^{2}+\left ( -3 \right )^{2}+\left ( -3 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{25+9+9}\$\$=\$ \$\displaystyle \sqrt{43}\$(iii) Distance between points \$(-1, 3, -4)\$ and \$(1, -3, 4)\$\$ =\$ \$\displaystyle \sqrt{\left ( 1+1 \right )^{2}+\left ( -3-3 \right )^{2}+\left ( 4+4 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{\left ( 2 \right )^{2}+\left ( -6 \right )^{2}+\left ( 8 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{4+36+64}=\sqrt{104}\$\$=2\sqrt{26}\$(iv) Distance between points \$(2, -1, 3)\$ and \$(-2, 1, 3)\$\$=\$ \$\displaystyle \sqrt{\left ( -2-2 \right )^{2}+\left ( 1+1 \right )^{2}+\left ( 3-3 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{\left ( -4 \right )^{2}+\left ( 2 \right )^{2}+\left ( 0 \right )^{2}}\$\$=\$ \$\displaystyle \sqrt{16+4}\$\$=\$ \$\displaystyle \sqrt{20}\$\$=\$ \$\displaystyle 2\sqrt{5}\$

Q: The x-coordinate of a point on the line joining the points \$P(2,2,1)\$ and \$Q(5,1,-2)\$ is \$4\$. Find its z-coordinate.

\$P(2,2,1),Q(5,1,-2)\$\$\therefore\$ Equation of line through \$P\$ & \$Q\$,\$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \\ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r\$\$\therefore P\$ be point of line \$\Rightarrow P\equiv (-3r+2,r+2,3r+1)\$\$ \therefore -3r+2=4\$ (\$\because \$ since x-coordinate is \$4\$)\$\Rightarrow r=\cfrac { -2 }{ 3 } \$\$\therefore\$ z-coordinate \$=3r+1=-1\$

Q: Find the lengths of the medians of the triangle with vertices \$A (0, 0, 6), B (0, 4, 0)\$ and \$(6, 0, 0).\$

Let \$AD, BE\$ and \$CF\$ be the medians of the given \$\triangle ABC\$.Since \$AD\$ is the median, \$D\$ is the mid-point of \$BC\$.\$\displaystyle \therefore \$ coordinates of point \$D =\$ \$\displaystyle \left ( \frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2} \right )=\left ( 3, 2, 0 \right )\$\$AD=\$ \$\displaystyle \sqrt{\left ( 0-3 \right )^{2}+\left ( 0-2 \right )^{2}+\left ( 6-0 \right )^{2}}=\sqrt{9+4+36}=\sqrt{49}=7\$Since \$BE\$ is the median, \$E\$ is the mid-point of \$AC\$.\$\displaystyle \therefore \$ coordinates of point \$E =\$ \$\displaystyle \left ( \frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2} \right )=\left ( 3,0,3 \right )\$\$BE =\$ \$\displaystyle \sqrt{\left ( 3-0 \right )^{2}+\left ( 0-4 \right )^{2}+\left ( 3-0 \right )^{2}}=\sqrt{9+16+9}=\sqrt{34}\$Since \$CF\$ is the median, \$F\$ is the mid point of \$AB\$.\$\displaystyle \therefore \$ coordinates of point \$F =\$ \$\displaystyle \left ( \frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2} \right )=\left ( 0,2,3 \right )\$Length of \$CF =\$ \$\displaystyle \sqrt{\left ( 6-0 \right )^{2}+\left ( 0-2 \right )^{2}+\left ( 0-3 \right )^{2}}=\sqrt{36+4+9}=\sqrt{49}=7\$Thus the lengths of the medians of \$\triangle ABC\$ are \$7, \sqrt{34} \$ and \$7\$ units.

Q: If the extremities of a diagonal of a square are \$(1, -2, 3)\$ and \$(2, -3, 5)\$, then the length of its side is:

The extremities of a diagonal of a square are given by \$(1,-2,3)\$ and \$(2,-3,5)\$.The length of this diagonal will be given by \$ \sqrt {({1}^{2} + {1}^{2} + {2}^{2} )} = \sqrt{6 }\$Length of a side \$ \times \sqrt{2} = \$ Length of diagonalSo, the length of the side of the square will be given by \$ = \sqrt {3 } \$

Q: Find the coordinates of a point equidistant from four points \$O(0,0,0),A(\ell,0,0),B(0,m,0)\$ and \$C(0,0,n)\$.

\$P(x,y,z)\$\$\begin{array}{l} PO=\sqrt { { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } } \\ PA=\sqrt { { { \left( { l-x } \right) }^{ 2 } }+{ { \left( { -y } \right) }^{ 2 } }+{ { \left( { -z } \right) }^{ 2 } } } =\sqrt { { { \left( { l-x } \right) }^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } } \\ PB=\sqrt { { x^{ 2 } }+{ { \left( { m-y } \right) }^{ 2 } }+{ z^{ 2 } } } \\ PC=\sqrt { { x^{ 2 } }+{ y^{ 2 } }+{ { \left( { n-z } \right) }^{ 2 } } } \\ PO=PA\Rightarrow \sqrt { { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } } =\sqrt { { { \left( { l-x } \right) }^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } } \\ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } }={ \left( { l-x } \right) ^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } \\ \Rightarrow x=\pm \left( { l-x } \right) \\ \Rightarrow x=l-x \\ \Rightarrow 2x=l \\ \Rightarrow x=\frac { l }{ 2 } \\ similiarly,\, \, \, y=\frac { m }{ 2 } ,z=\frac { n }{ 2 } \\ P\left( { \frac { l }{ 2 } ,\frac { m }{ 2 } ,\frac { n }{ 2 } } \right) \end{array}\$

Q: The point in the \$xy -\$ plane which is equidistant from \$(2, 0, 3), (0, 3, 2)\$ and \$(0,0, 1)\$ is

Let the points are \$A(2,0,3),B(0,3,2)\$ and \$D(0,0,1)\$We know that Z-coordinate of every point an xy-plane is zero so let \$p(x,y,0) \$ be a point on xy-plane such that \$PA=PB=PC\$Now, \$PA=PB\$\$\Rightarrow PA^2=PB^2\$\$\Rightarrow (x-2)^2+(y-0)^2+(0-3)^2=(x-0)^2+(y-3)^2+(0-2)\$\$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0\$......(i)and , \$PB=PC\$\$\Rightarrow PB^2=PC^2\$\$\Rightarrow (x-0)^2+(y-3)^2(0-2)^2=(x-0)^2+(y-0)^2+(0-1)^2\$\$\Rightarrow -6y+12=0\$\$\Rightarrow y=2\$......(ii)Putting \$y=2\$ in equation (i), we get \$x=3\$Hence, the required point is \$(3,2,0)\$

Q: In a \$\triangle {ABC}\$, side \$AB\$ has the equation \$2x+3y=29\$ and the side \$AC\$ has the equation \$x+2y=16\$. If the mid point of \$BC\$ is \$(5,6)\$, then the equation of \$BC\$ is

Let co-ordinates of \$B\$ be \$(x_1, y_1)\$ & \$C\$ be \$(x_2, y_2)\$\$\therefore\$ \$(5,6)\$ is the mid point,so, \$\dfrac{x_1 + x_2}{2} = 5, \dfrac{y_1 + y_2}{2} = 6\$\$\Rightarrow x_1 + x_2 = 10, y_1 + y_2 = 12\$\$B(x_1, y_1)\$ lies on the line \$2x + 3y = 29\$\$\therefore 2x_1 + 3y_1 = 29\$ ----(1)\$C(x_2, y_2)\$ lies on the line \$x + 2y = 16,\$\$\therefore x_2 + 2y_2 = 16\$ ----(2)\$\therefore\$ putting \$x_1, y_1\$ in the form of \$x_2, y_2\$ in (1)\$2(10 - x_2) + 3(12 - y_2) = 29\$ {\$x_1 = 10 - x_2, y_1 = 12 - y_2\$}\$\Rightarrow 20 - 2x_2 + 36 - 3y_2 = 29\$\$\Rightarrow 2x_2 + 3y_2 = 27\$ ----(3)on subtracting \$(3)\$ and \$(2)\$ \$\times\$ \$2\$\$-y_2 = -5\$\$y_2 = 5\$Putting \$y_2 \, in (2)\$\$x_2 + 2(5) = 16\$\$x_2 = 6\$\$x_1 = 10 - x_2\$ \$= 4\$\$y_1 = 12 - 5 = 7\$Equation : \$\dfrac{x - x_1}{x_2 - x_1} = \dfrac{y - y_1}{y_2 - y_1}\$\$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}\$\$\Rightarrow -x + 4 = y - 7\$\$\Rightarrow x + y = 11\$

Question 1

Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$

Accepted Answer

Let $P (x, y, z)$ be the point that is equidistant from points $A(1, 2, 3)$ and $B(3, 2, -1)$ .i.e.  $PA = PB$$\displaystyle \Rightarrow PA^{2}=PB^{2}$$\displaystyle \Rightarrow \left ( x-1 \right )^{2}+\left ( y-2 \right )^{2}+\left ( z-3 \right )^{2}=\left ( x-3 \right )^{2}+\left ( y-2 \right )^{2}+\left ( z+1 \right )^{2}$$\displaystyle \Rightarrow x^{2}-2x+1+y^{2}-4y+4+z^{2}-6z+9=x^2-6x+9+y^{2}-4y+4+z^{2}+2z+1$$\displaystyle \Rightarrow -2x-6z+14=-6x+2z+14$$\displaystyle \Rightarrow -2x-6z+6x-2z=0$$\displaystyle \Rightarrow 4x-8z=0$$\displaystyle \Rightarrow x-2z=0$Thus the required equation is $x-2z=0 $.

Question 2

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 $10$.

Accepted Answer

Let the coordinates of $P$ be $(x, y, z)$.The coordinates of points $A$ and $B$ are $(4, 0, 0)$ and $(-4, 0, 0)$ respectively.It is given that $PA + PB = 10$$\displaystyle \Rightarrow \sqrt{\left ( x-4 \right )^{2}+y^{2}+z^{2}}+\sqrt{\left ( x+4 \right )^{2}+y^{2}+z^{2}}=10$$\displaystyle \Rightarrow \sqrt{\left ( x-4 \right )^{2}+y^{2}+z^{2}}=10-\sqrt{\left ( x+4 \right )^{2}+y^{2}+z^{2}}$On squaring both sides, we obtain$\displaystyle \Rightarrow \left ( x-4 \right )^{2}+y^{2}+z^{2}=100+ \left( x+4 \right )^{2}+y^{2}+z^{2}-20\sqrt{\left ( x+4 \right )^{2}+y^{2}+z^{2}}$$\displaystyle \Rightarrow x^{2}-8x+16+y^{2}+z^{2}=100-20\sqrt{x^{2}+8x+16+y^{2}+z^{2}}+x^{2}+8x+16+y^{2}+z^{2} $$\displaystyle \Rightarrow 20\sqrt{x^{2}+8x+16+y^{2}+z^{2}}=100+16x$$\displaystyle \Rightarrow 5\sqrt{x^{2}+8x+16+y^{2}+z^{2}}=\left ( 25+4x \right )$On squaring both sides again, we obtain$\displaystyle 25\left ( x^{2} +8x+16+y^{2}+z^{2}\right )=625+16x^{2}+200x$$\displaystyle \Rightarrow 25x^{2}+200x+400+25y^{2}+25z^{2}=625+16x^{2}+200x$$\displaystyle \Rightarrow 9x^{2}+25y^{2}+25z^{2}-225=0$Thus the required equation is $\displaystyle  9x^{2}+25y^{2}+25z^{2}-225=0$.

Question 3

Find the distance between the following pairs of points:(i) $(2, 3, 5)$ and $(4, 3, 1)$(ii)$ (-3, 7, 2)$and $((2, 4, -1)$(iii) $(-1, 3, -4)$ and $(1, -3, 4)$(iv) $(2, -1, 3)$ and $(-2, 1, 3)$

Accepted Answer

The distance between points $P(x_{1}, y _{1}, z_{1})$ and $Q(x_{2}, y _{2}, z_{2})$  is given by$PQ =$ $\displaystyle \sqrt{\left ( x_{2} -x_{1}\right )^{2}+\left ( y_{2} -y_{1}\right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$(i) Distance between points $(2, 3, 5)$ and $(4, 3, 1)$$=$ $\displaystyle \sqrt{\left ( 4-2 \right )^{2}+\left ( 3-3 \right )^{2}+\left ( 1-5 \right )^{2}}$$=$ $\displaystyle \sqrt{\left ( 2 \right )^{2}+\left ( 0 \right )^{2}+\left ( -4 \right )^{2}}$$=$ $\displaystyle \sqrt{4+16}$$=$ $\displaystyle \sqrt{20}$$=$ $\displaystyle 2\sqrt{5}$(ii) Distance between points $(-3, 7, 2)$ and $(2, 4, -1)$$=$ $\displaystyle \sqrt{\left ( 2+3 \right )^{2}+\left ( 4-7 \right )^{2}+\left ( -1-2 \right )^{2}}$$=$ $\displaystyle \sqrt{\left ( 5 \right )^{2}+\left ( -3 \right )^{2}+\left ( -3 \right )^{2}}$$=$ $\displaystyle \sqrt{25+9+9}$$=$ $\displaystyle \sqrt{43}$(iii) Distance between points $(-1, 3, -4)$ and $(1, -3, 4)$$ =$ $\displaystyle \sqrt{\left ( 1+1 \right )^{2}+\left ( -3-3 \right )^{2}+\left ( 4+4 \right )^{2}}$$=$ $\displaystyle \sqrt{\left ( 2 \right )^{2}+\left ( -6 \right )^{2}+\left ( 8 \right )^{2}}$$=$ $\displaystyle \sqrt{4+36+64}=\sqrt{104}$$=2\sqrt{26}$(iv) Distance between points $(2, -1, 3)$ and $(-2, 1, 3)$$=$ $\displaystyle \sqrt{\left ( -2-2 \right )^{2}+\left ( 1+1 \right )^{2}+\left ( 3-3 \right )^{2}}$$=$ $\displaystyle \sqrt{\left ( -4 \right )^{2}+\left ( 2 \right )^{2}+\left ( 0 \right )^{2}}$$=$ $\displaystyle \sqrt{16+4}$$=$ $\displaystyle \sqrt{20}$$=$ $\displaystyle 2\sqrt{5}$

Question 4

The x-coordinate of a point on the line joining the points $P(2,2,1)$ and $Q(5,1,-2)$ is $4$. Find its z-coordinate.

Accepted Answer

$P(2,2,1),Q(5,1,-2)$$\therefore$ Equation of line through $P$ & $Q$,$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r$$\therefore P$ be point of line $\Rightarrow P\equiv (-3r+2,r+2,3r+1)$$ \therefore -3r+2=4$ ($\because $ since x-coordinate is $4$)$\Rightarrow r=\cfrac { -2 }{ 3 } $$\therefore$ z-coordinate $=3r+1=-1$

Question 5

Find the lengths of the medians of the triangle with vertices $A (0, 0, 6), B (0, 4, 0)$ and $(6, 0, 0).$

Accepted Answer

Let $AD, BE$ and $CF$ be the medians of the given $\triangle ABC$.Since $AD$ is the median, $D$ is the mid-point of $BC$.$\displaystyle \therefore $ coordinates of point $D =$ $\displaystyle \left ( \frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2} \right )=\left ( 3, 2, 0 \right )$$AD=$  $\displaystyle \sqrt{\left ( 0-3 \right )^{2}+\left ( 0-2 \right )^{2}+\left ( 6-0 \right )^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$Since $BE$ is the median, $E$ is the mid-point of $AC$.$\displaystyle \therefore $ coordinates of point $E =$ $\displaystyle \left ( \frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2} \right )=\left ( 3,0,3 \right )$$BE =$ $\displaystyle \sqrt{\left ( 3-0 \right )^{2}+\left ( 0-4 \right )^{2}+\left ( 3-0 \right )^{2}}=\sqrt{9+16+9}=\sqrt{34}$Since $CF$ is the median, $F$ is the mid point of $AB$.$\displaystyle \therefore $ coordinates of point $F =$ $\displaystyle \left ( \frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2} \right )=\left ( 0,2,3 \right )$Length of $CF =$ $\displaystyle \sqrt{\left ( 6-0 \right )^{2}+\left ( 0-2 \right )^{2}+\left ( 0-3 \right )^{2}}=\sqrt{36+4+9}=\sqrt{49}=7$Thus the lengths of the medians of $\triangle ABC$ are $7, \sqrt{34} $ and $7$ units.

Question 6

Verify the following(i) $(0, 7, -10 ), (1, 6, -6)$ and $(4, 9, -6)$ are the vertices of an isosceles triangle(ii) $(0, 7, 10), (-1, 6, 6)$ and $(-4, 9, 6)$ are the vertices of a right angled triangle(iii) $(-1, 2, 1), (1, -2, 5), (4, -7, 8)$ and $(2, -3, 4)$ are the vertices of a parallelogram

Accepted Answer

(i) Let point$(0, 7, -10), (1, 6, -6)$ and $(4, 9, -6)$ be denoted by $A, B$ and $C$ respectively$\displaystyle AB=\sqrt{\left ( 1-0 \right )^{2}+\left ( 6-7 \right )^{2}+\left ( -6+10 \right )^{2}}$$=\displaystyle \sqrt{\left ( 1 \right )^{2}+\left ( -1 \right )^{2}+\left ( 4 \right )^{2}}$$=\displaystyle \sqrt{1+1+16}$$=\displaystyle \sqrt{18}$$\Rightarrow AB=\displaystyle 3\sqrt{2}$$\displaystyle BC=\sqrt{\left ( 4-1 \right )^{2}+\left ( 9-6 \right )^{2}+\left ( -6+6 \right )^{2}}$=$\displaystyle \sqrt{\left ( 3 \right )^{2}+\left ( 3 \right )^{2}}$=$\displaystyle \sqrt{9+9}=\sqrt{18}$$\Rightarrow BC=3\sqrt{2}$$\displaystyle CA=\sqrt{\left ( 0-4 \right )^{2}+\left ( 7-9 \right )^{2}+\left ( -10+6 \right )^{2}}$=$\displaystyle \sqrt{\left ( -4 \right )^{2}+\left ( -2 \right )^{2}+\left ( -4 \right )^{2}}$=$\displaystyle \sqrt{16+4+16}=\sqrt{36}=6$Here $AB = BC$ $\displaystyle \neq $ $CA$Thus the given points are the vertices of an isosceles triangle(ii) Let $(0, 7, 10), (-1, 6, 6)$ and $(-4, 9, 6)$ be denoted by $A, B$ and $C$ respectively$\displaystyle AB =  \sqrt{\left ( -1-0 \right )^{2}+\left ( 6-7 \right )^{2}+\left ( 6-10 \right )^{2}}$$\displaystyle =\sqrt{\left ( -1 \right )^{2}+\left ( -1 \right )^{2}+\left ( -4 \right )^{2}}$$\displaystyle =\sqrt{1+1+16}=\sqrt{18}$$\displaystyle =3\sqrt{2}$   $\displaystyle BC=\sqrt{\left ( -4+1 \right )^{2}+\left ( 9-6 \right )^{2}+\left ( 6-6 \right )^{2}}$$\displaystyle =\sqrt{\left ( -3 \right )^{2}+\left ( 3 \right )^{2}+\left ( 0 \right )^{2}}$$\displaystyle =\sqrt{9+9}=\sqrt{18}$$\displaystyle =3\sqrt{2}$$\displaystyle CA=\sqrt{\left ( 0+4 \right )^{2}+\left ( 7-9 \right )^{2}+\left ( 10-6 \right )^{2}}$$\displaystyle =\sqrt{\left ( 4 \right )^{2}+\left ( -2 \right )^{2}+\left ( 4 \right )^{2}}$$\displaystyle =\sqrt{16+4+16}$$\displaystyle =\sqrt{36}$$=6$Now $\displaystyle AB^{2}+BC^{2}=\left ( 3\sqrt{2} \right )^{2}+\left ( 3\sqrt{2} \right )^{2}=18+18=36=AC^{2}$Therefore by pythagoras theorem $ABC$ is a right triangleHence the given points are the vertices of a right-angled triangle(iii) Let $(-1, 2, 1), (1, -2, 5), (4, -7, 8)$ and $(2, -3, 4)$ be denoted by $A, B, C$ and $D$ respectively$\displaystyle AB=\sqrt{\left ( 1+1 \right )^{2}+\left ( -2-2 \right )^{2}+\left ( 5-1 \right )^{2}}$$\displaystyle =\sqrt{4+16+16}$$\displaystyle AB= \sqrt{36}$$AB=6$$BC=\sqrt { \left( 4-1 \right) ^{ 2 }+\left( -7+2 \right) ^{ 2 }+\left( 8-5 \right) ^{ 2 } } $$\displaystyle =\sqrt{9+25+9}=\sqrt{43}$$\displaystyle CD= \sqrt{\left ( 2-4 \right )^{2}+\left ( -3+7 \right )^{2}+\left ( 4-8 \right )^{2}}$$\displaystyle =\sqrt{4+16+16}$$\displaystyle =\sqrt{36}$$CD=6$$DA = \displaystyle \sqrt{\left ( -1-2 \right )^{2}+\left ( 2+3 \right )^{2}+\left ( 1-4 \right )^{2}}$$\displaystyle DA=\sqrt{9+25+9}=\sqrt{43}$Here $AB = CD = 6$, $ BC = AD = \displaystyle \sqrt{43}$Hence the opposite sides of quadrilateral $ABCD$ whose vertices are taken in order are equalTherefore $ABCD$ is a parallelogramHence the given points are the vertices of a parallelogram

Question 7

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$, then the length of its side is:

Accepted Answer

The extremities of a diagonal of a square are given by $(1,-2,3)$ and $(2,-3,5)$.The length of this diagonal will be given by $ \sqrt {({1}^{2} + {1}^{2} + {2}^{2} )}  =  \sqrt{6 }$Length of a side $ \times \sqrt{2} = $ Length of diagonalSo, the length of the side of the square will be given by $ = \sqrt {3 } $

Question 8

Find the coordinates of a point equidistant from four points $O(0,0,0),A(\ell,0,0),B(0,m,0)$ and $C(0,0,n)$.

Accepted Answer

$P(x,y,z)$$\begin{array}{l} PO=\sqrt { { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } }  \ PA=\sqrt { { { \left( { l-x } \right)  }^{ 2 } }+{ { \left( { -y } \right)  }^{ 2 } }+{ { \left( { -z } \right)  }^{ 2 } } } =\sqrt { { { \left( { l-x } \right)  }^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } }  \ PB=\sqrt { { x^{ 2 } }+{ { \left( { m-y } \right)  }^{ 2 } }+{ z^{ 2 } } }  \ PC=\sqrt { { x^{ 2 } }+{ y^{ 2 } }+{ { \left( { n-z } \right)  }^{ 2 } } }  \ PO=PA\Rightarrow \sqrt { { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } } =\sqrt { { { \left( { l-x } \right)  }^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } }  \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } }={ \left( { l-x } \right) ^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } } \ \Rightarrow x=\pm \left( { l-x } \right)  \ \Rightarrow x=l-x \ \Rightarrow 2x=l \ \Rightarrow x=\frac { l }{ 2 }  \ similiarly,\, \, \, y=\frac { m }{ 2 } ,z=\frac { n }{ 2 }  \ P\left( { \frac { l }{ 2 } ,\frac { m }{ 2 } ,\frac { n }{ 2 }  } \right)  \end{array}$

Question 9

The point in the $xy -$ plane which is equidistant from $(2, 0, 3), (0, 3, 2)$ and $(0,0, 1)$ is

Accepted Answer

Let the points are $A(2,0,3),B(0,3,2)$ and $D(0,0,1)$We know that Z-coordinate of every point an xy-plane is zero so let $p(x,y,0) $ be a point on xy-plane such that $PA=PB=PC$Now, $PA=PB$$\Rightarrow PA^2=PB^2$$\Rightarrow (x-2)^2+(y-0)^2+(0-3)^2=(x-0)^2+(y-3)^2+(0-2)$$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0$......(i)and , $PB=PC$$\Rightarrow PB^2=PC^2$$\Rightarrow (x-0)^2+(y-3)^2(0-2)^2=(x-0)^2+(y-0)^2+(0-1)^2$$\Rightarrow -6y+12=0$$\Rightarrow y=2$......(ii)Putting $y=2$ in equation (i), we get $x=3$Hence, the required point is $(3,2,0)$

Question 10

In a $\triangle {ABC}$, side $AB$ has the equation $2x+3y=29$ and the side $AC$ has the equation $x+2y=16$. If the mid point of $BC$ is $(5,6)$, then the equation of $BC$ is

Accepted Answer

Let co-ordinates of $B$ be $(x_1, y_1)$ & $C$ be $(x_2, y_2)$$\therefore$ $(5,6)$ is the mid point,so, $\dfrac{x_1 + x_2}{2} = 5, \dfrac{y_1 + y_2}{2} = 6$$\Rightarrow x_1 + x_2 = 10, y_1 + y_2 = 12$$B(x_1, y_1)$ lies on the line $2x + 3y = 29$$\therefore 2x_1 + 3y_1 = 29$  ----(1)$C(x_2, y_2)$ lies on the line $x + 2y = 16,$$\therefore x_2 + 2y_2 = 16$  ----(2)$\therefore$ putting $x_1, y_1$ in the form of $x_2, y_2$ in (1)$2(10 - x_2) + 3(12 - y_2) = 29$  {$x_1 = 10 - x_2, y_1 = 12 - y_2$}$\Rightarrow 20 - 2x_2 + 36 - 3y_2 = 29$$\Rightarrow 2x_2 + 3y_2 = 27$  ----(3)on subtracting $(3)$ and $(2)$ $\times$ $2$$-y_2 = -5$$y_2 = 5$Putting $y_2 \,  in (2)$$x_2 + 2(5) = 16$$x_2 = 6$$x_1 = 10 - x_2$      $= 4$$y_1 = 12 - 5 = 7$Equation : $\dfrac{x - x_1}{x_2 - x_1} = \dfrac{y - y_1}{y_2 - y_1}$$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$$\Rightarrow -x + 4 = y - 7$$\Rightarrow x + y = 11$

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Distance Formula in 3D Geometry

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Distance Formula in 3D Geometry

Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, - 1)$

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 $10$ .

Find the distance between the following pairs of points:
(i) $(2, 3, 5)$ and $(4, 3, 1)$
(ii) $(- 3, 7, 2)$ and $((2, 4, - 1)$
(iii) $(- 1, 3, - 4)$ and $(1, - 3, 4)$
(iv) $(2, - 1, 3)$ and $(- 2, 1, 3)$

The x-coordinate of a point on the line joining the points $P (2, 2, 1)$ and $Q (5, 1, - 2)$ is $4$ . Find its z-coordinate.

Find the lengths of the medians of the triangle with vertices $A (0, 0, 6), B (0, 4, 0)$ and $(6, 0, 0) .$

If the extremities of a diagonal of a square are $(1, - 2, 3)$ and $(2, - 3, 5)$ , then the length of its side is:

Find the coordinates of a point equidistant from four points $O (0, 0, 0), A (ℓ, 0, 0), B (0, m, 0)$ and $C (0, 0, n)$ .

The point in the $x y -$ plane which is equidistant from $(2, 0, 3), (0, 3, 2)$ and $(0, 0, 1)$ is

In a $△ A B C$ , side $A B$ has the equation $2 x + 3 y = 29$ and the side $A C$ has the equation $x + 2 y = 16$ . If the mid point of $B C$ is $(5, 6)$ , then the equation of $B C$ is

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Distance Formula in 3D Geometry

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Quick Summary With Stories

Distance Formula in 3D Geometry

Find the equation of the set of points which are equidistant from the points (1,2,3) and (3,2,−1)

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.

Find the distance between the following pairs of points: (i) (2,3,5) and (4,3,1) (ii)(−3,7,2)and ((2,4,−1) (iii) (−1,3,−4) and (1,−3,4) (iv) (2,−1,3) and (−2,1,3)

The x-coordinate of a point on the line joining the points P(2,2,1) and Q(5,1,−2) is 4. Find its z-coordinate.

Find the lengths of the medians of the triangle with vertices A(0,0,6),B(0,4,0) and (6,0,0).

Verify the following (i) (0,7,−10),(1,6,−6) and (4,9,−6) are the vertices of an isosceles triangle (ii) (0,7,10),(−1,6,6) and (−4,9,6) are the vertices of a right angled triangle (iii) (−1,2,1),(1,−2,5),(4,−7,8) and (2,−3,4) are the vertices of a parallelogram

If the extremities of a diagonal of a square are (1,−2,3) and (2,−3,5), then the length of its side is:

Find the coordinates of a point equidistant from four points O(0,0,0),A(ℓ,0,0),B(0,m,0) and C(0,0,n).

The point in the xy− plane which is equidistant from (2,0,3),(0,3,2) and (0,0,1) is

In a △ABC, side AB has the equation 2x+3y=29 and the side AC has the equation x+2y=16. If the mid point of BC is (5,6), then the equation of BC is

Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, - 1)$

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 $10$ .

Find the distance between the following pairs of points:
(i) $(2, 3, 5)$ and $(4, 3, 1)$
(ii) $(- 3, 7, 2)$ and $((2, 4, - 1)$
(iii) $(- 1, 3, - 4)$ and $(1, - 3, 4)$
(iv) $(2, - 1, 3)$ and $(- 2, 1, 3)$

The x-coordinate of a point on the line joining the points $P (2, 2, 1)$ and $Q (5, 1, - 2)$ is $4$ . Find its z-coordinate.

Find the lengths of the medians of the triangle with vertices $A (0, 0, 6), B (0, 4, 0)$ and $(6, 0, 0) .$

If the extremities of a diagonal of a square are $(1, - 2, 3)$ and $(2, - 3, 5)$ , then the length of its side is:

Find the coordinates of a point equidistant from four points $O (0, 0, 0), A (ℓ, 0, 0), B (0, m, 0)$ and $C (0, 0, n)$ .

The point in the $x y -$ plane which is equidistant from $(2, 0, 3), (0, 3, 2)$ and $(0, 0, 1)$ is

In a $△ A B C$ , side $A B$ has the equation $2 x + 3 y = 29$ and the side $A C$ has the equation $x + 2 y = 16$ . If the mid point of $B C$ is $(5, 6)$ , then the equation of $B C$ is