Q: Find the ratio in which the YZ-plane divides the line segment formed by joining the points \$(-2,4,7)\$ and \$(3,-5,8)\$.

Let the YZ plane divide the line segment joining points \$(-2,4,7)\$ and \$(3,-5,8)\$ in the ratio \$k:1\$.Hence, by section formula, the coordinates of point of intersection are given by :\$(\dfrac{k(3)-2}{k+1}, \dfrac{k(-5)+4}{k+1}, \dfrac{k(8)+7}{k+1})\$On the YZ plane, the x-coordinate of any point is zero.\$\dfrac{3k-2}{k+1}=0\$\$3k-2=0\$\$k=\dfrac{2}{3}\$Thus, the YZ plane divides the line segment formed by joining the given points in the ratio \$2:3\$.

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: 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) \$

The \$x\$-coordinate \$y\$-coordinate and\$z\$-coordinate of point \$(1, 2, 3)\$ are all positive.Therefore this point lies in octant IThe \$x\$-coordinate \$y\$-coordinate and \$z\$-coordinate of point \$(4, -2, 3)\$ are positivenegative and positive respectively. Therefore this point lies in octant IVThe \$x\$-coordinate \$y\$-coordinate and \$z\$-coordinate of point \$(4, -2, -5)\$ are positivenegative and negative respectively. Therefore this point lies in octant VIIIThe \$x\$-coordinate, \$y\$-coordinate and \$z\$-coordinate of point \$(4, 2, -5)\$ are positivepositive and negative respectively. Therefore this point lies in octant VThe \$x\$-coordinate, \$y\$-coordinate and \$z\$-coordinate of point \$(-4, 2, -5)\$ are negativepositive, and negative respectively. Therefore this point lies in octant VIThe \$x\$-coordinate, \$y\$-coordinate and \$z\$-coordinate of point \$(-4, 2, 5)\$ are negativepositive and positive respectively. Therefore this point lies in octant IIThe \$x\$-coordinate \$y\$-coordinate and \$z\$-coordinate of point \$(-3, -1, 6)\$ are negativenegative and positive respectively. Therefore this point lies in octant IIIThe \$x\$-coordinate, \$y\$-coordinate and \$z\$-coordinate of point \$(2, -4, -7)\$ are positivenegative and negative respectively. Therefore this point lies in octant VIII

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: The ratio in which the line joining points \$(2,4,5)\$ and \$(3,5,-4)\$ divide YZ -plane is

Let the ratio be \$\lambda:1\$The dividing plane is \$YZ\$ plane.So, the co-ordinate of \$x\$ after dividing the line \$= 0\$Now According to internal division rule of line segment \$x=\dfrac{3\times \lambda - 1\times 2}{\lambda + 1}\$\$\Rightarrow \dfrac{3\lambda - 2}{\lambda + 1}=0\$\$\Rightarrow 3\lambda - 2=0 \ \Rightarrow \lambda=\dfrac{2}{3}\$.So. Ratio \$=2:3\$

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 ratio in which the XY - plane divides AB if is (1,2,3) and B is (-3,4,-5). Also find the positive vector of the point of division.

We have, Let \$AB\$ be the line segment joining the points \$A\left( 1,2,3 \right)\$ and \$B\left( -3,4,-5 \right)\$. Let \$XZ-plane\$ divide line \$AB\$.at \$P\left( x,y,z \right)\$ in the ratio \$k:1\$. Co- ordinate of point \$P\left( x,y,z \right)\$ that divide line segment joining point \$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\$ and \$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\$ in the ratio \$m:n\$ is \$=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\,\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)\$ Here, \$m:n=k:1\$ \$ A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,2,3 \right) \$ \$ B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( -3,4,-5 \right) \$ Then the co-ordinate of \$ P\left( x,y,z \right)=\left( \dfrac{k\times \left( -3 \right)+1\left( 1 \right)}{k+1},\dfrac{k\times \left( 4 \right)+1\times \left( 2 \right)}{k+1},\dfrac{k\times \left( -5 \right)+1\times \left( 3 \right)}{k+1} \right) \$ \$ =\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1} \right) \$ Since, point \$P\left( x,y,z \right)\$ lie on the \$XZ-plane\$ Then, Its Y-coordinate will be zero. \$P\left( 0,y,z \right)=\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1} \right)\$ Then, Comparing and we get, \$ \dfrac{-3k+1}{k+1}=0 \$ \$ -3k+1=0 \$ \$ -3k=-1 \$ \$ k=\dfrac{1}{3} \$ Then, \$k:1=1:3\$ Hence, this is the answer.

Q: The point \$P\$ is on the \$y\$-axis. If \$P\$ is equidistant from \$(1,2, 3)\$ and \$(2,3, 4)\$, then \$P_{y}=\$

Let the point be \$(0,y,0)\$Since, it is equidistant from the \$2\$ points.\$\Rightarrow (1+ (y-2)^{2} + 9) = (4 + (y-3)^{2} + 16) \$\$\Rightarrow (10+y^2-4y+4)= (20+y^2-6y+9)\$\$\Rightarrow y^2-4y+14=y^2-6y+29\$\$\Rightarrow 2y=15\$\$\Rightarrow y =\$ \$ \displaystyle \dfrac{15}{2} \$

Question 1

Find the ratio in which the YZ-plane divides the line segment formed by joining the points $(-2,4,7)$ and $(3,-5,8)$.

Accepted Answer

Let the YZ plane divide the line segment joining points $(-2,4,7)$ and $(3,-5,8)$ in the ratio $k:1$.Hence, by section formula, the coordinates of point of intersection are given by :$(\dfrac{k(3)-2}{k+1}, \dfrac{k(-5)+4}{k+1}, \dfrac{k(8)+7}{k+1})$On the YZ plane, the x-coordinate of any point is zero.$\dfrac{3k-2}{k+1}=0$$3k-2=0$$k=\dfrac{2}{3}$Thus, the YZ plane divides the line segment formed by joining the given points in the ratio $2:3$.

Question 2

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 3

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 4

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 5

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)  $

Accepted Answer

The $x$-coordinate $y$-coordinate and$z$-coordinate of point $(1, 2, 3)$ are all positive.Therefore this point lies in octant IThe $x$-coordinate $y$-coordinate and $z$-coordinate of point $(4, -2, 3)$ are positivenegative and positive respectively. Therefore this point lies in octant IVThe $x$-coordinate $y$-coordinate and $z$-coordinate of point $(4, -2, -5)$ are positivenegative and negative respectively. Therefore this point lies in octant VIIIThe $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(4, 2, -5)$ are positivepositive and negative respectively. Therefore this point lies in octant VThe $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(-4, 2, -5)$ are negativepositive, and negative respectively. Therefore this point lies in octant VIThe $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(-4, 2, 5)$ are negativepositive and positive respectively. Therefore this point lies in octant IIThe $x$-coordinate $y$-coordinate and $z$-coordinate of point $(-3, -1, 6)$ are negativenegative and positive respectively. Therefore this point lies in octant IIIThe $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(2, -4, -7)$ are positivenegative and negative respectively. Therefore this point lies in octant VIII

Question 6

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 7

The ratio in which the line joining points $(2,4,5)$ and $(3,5,-4)$ divide YZ -plane is

Accepted Answer

Let the ratio be $\lambda:1$The dividing plane is $YZ$ plane.So, the co-ordinate of $x$ after dividing the line $= 0$Now According to internal division rule of line segment $x=\dfrac{3\times \lambda - 1\times 2}{\lambda + 1}$$\Rightarrow \dfrac{3\lambda - 2}{\lambda + 1}=0$$\Rightarrow 3\lambda - 2=0 \ \Rightarrow \lambda=\dfrac{2}{3}$.So. Ratio $=2:3$

Question 8

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 9

Find the ratio in which the XY - plane divides AB if is (1,2,3) and B is (-3,4,-5). Also find the positive vector of the point of division.

Accepted Answer

We have,&#10;&#10;Let $AB$ be the line segment joining the points&#10;&#10;$A\left( 1,2,3 \right)$ and $B\left( -3,4,-5 \right)$.&#10;&#10;Let $XZ-plane$ divide line $AB$.at $P\left( x,y,z&#10;\right)$ in the ratio $k:1$.&#10;&#10;Co- ordinate of point $P\left( x,y,z \right)$ that divide&#10;line segment joining point $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and&#10;$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ in the ratio $m:n$ is&#10;&#10;$=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\,\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n}&#10;\right)$&#10;&#10;Here,&#10;&#10;$m:n=k:1$&#10;&#10;$ A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left(&#10;1,2,3 \right) $&#10;&#10;$ B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left(&#10;-3,4,-5 \right) $&#10;&#10;Then the co-ordinate of&#10;&#10;$ P\left( x,y,z \right)=\left( \dfrac{k\times \left( -3&#10;\right)+1\left( 1 \right)}{k+1},\dfrac{k\times \left( 4 \right)+1\times \left(&#10;2 \right)}{k+1},\dfrac{k\times \left( -5 \right)+1\times \left( 3 \right)}{k+1}&#10;\right) $&#10;&#10;$ =\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1}&#10;\right) $&#10;&#10;Since, point $P\left( x,y,z \right)$ lie on the $XZ-plane$&#10;&#10;Then, Its Y-coordinate will be zero.&#10;&#10;$P\left( 0,y,z \right)=\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1}&#10;\right)$&#10;&#10;Then, Comparing and we get,&#10;&#10;$ \dfrac{-3k+1}{k+1}=0 $&#10;&#10;$ -3k+1=0 $&#10;&#10;$ -3k=-1 $&#10;&#10;$ k=\dfrac{1}{3} $&#10;&#10;Then, $k:1=1:3$&#10;&#10;Hence, this is the&#10;answer.

Question 10

The point $P$ is on the $y$-axis. If $P$ is equidistant from $(1,2, 3)$ and $(2,3, 4)$, then $P_{y}=$

Accepted Answer

Let the point be $(0,y,0)$Since, it is equidistant from the $2$ points.$\Rightarrow  (1+ (y-2)^{2} + 9) = (4 + (y-3)^{2} + 16) $$\Rightarrow (10+y^2-4y+4)= (20+y^2-6y+9)$$\Rightarrow y^2-4y+14=y^2-6y+29$$\Rightarrow 2y=15$$\Rightarrow y =$ $ \displaystyle \dfrac{15}{2} $