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

Question 1

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 2

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 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 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 5

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 6

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	imes \lambda - 1	imes 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 7

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) 	herefore  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 	herefore P  be point of line  \Rightarrow P\equiv (-3r+2,r+2,3r+1)  	herefore -3r+2=4  ( \because   since x-coordinate is  4 ) \Rightarrow r=\cfrac { -2 }{ 3 }  	herefore  z-coordinate  =3r+1=-1

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   	imes \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,

Let  AB  be the line segment joining the points

A\left( 1,2,3 ight)  and  B\left( -3,4,-5 ight) .

Let  XZ-plane  divide line  AB .at  P\left( x,y,z
ight)  in the ratio  k:1 .

Co- ordinate of point  P\left( x,y,z ight)  that divide
line segment joining point  A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} ight)  and
 B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} ight)  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}
ight)

Here,

m:n=k:1

A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} ight)=\left(
1,2,3 ight)

B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} ight)=\left(
-3,4,-5 ight)

Then the co-ordinate of

P\left( x,y,z ight)=\left( \dfrac{k	imes \left( -3
ight)+1\left( 1 ight)}{k+1},\dfrac{k	imes \left( 4 ight)+1	imes \left(
2 ight)}{k+1},\dfrac{k	imes \left( -5 ight)+1	imes \left( 3 ight)}{k+1}
ight)

=\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1}
ight)

Since, point  P\left( x,y,z ight)  lie on the  XZ-plane

Then, Its Y-coordinate will be zero.

P\left( 0,y,z ight)=\left( \dfrac{-3k+1}{k+1},\dfrac{4k+2}{k+1},\dfrac{-5k+3}{k+1}
ight)

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.

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}