\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{The ratio in which the line joining }(2,4,5)\text{ and }(3,5,-9)\text{ is}
\displaystyle \text{divided by the }yz\text{-plane is:}
\displaystyle \text{(a) }2:3\qquad\text{(b) }3:2\qquad\text{(c) }2:3\qquad\text{(d) }4:-3
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }yz\text{-plane is }x=0.
\displaystyle \text{Let the }yz\text{-plane divide the line joining }A(2,4,5)\text{ and }B(3,5,-9)
\displaystyle \text{externally in the ratio }m:n.
\displaystyle \text{Using the external division formula for the }x\text{-coordinate,}
\displaystyle 0=\frac{3m-2n}{m-n}.
\displaystyle \therefore 3m-2n=0.
\displaystyle \therefore \frac{m}{n}=\frac{2}{3}.
\displaystyle \therefore \text{The required ratio is }2:3.
\displaystyle \text{Hence, option (a) or (c) is correct; the given options contain a duplication.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The ratio in which the line joining the points }(a,b,c)\text{ and}
\displaystyle (-a,-c,-b)\text{ is divided by the }xy\text{-plane is:}
\displaystyle \text{(a) }a:b\qquad\text{(b) }b:c\qquad\text{(c) }c:a\qquad\text{(d) }c:b
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }xy\text{-plane is }z=0.
\displaystyle \text{Let the }xy\text{-plane divide the line joining }A(a,b,c)\text{ and }B(-a,-c,-b)
\displaystyle \text{internally in the ratio }m:n.
\displaystyle \text{Using the section formula for the }z\text{-coordinate,}
\displaystyle 0=\frac{m(-b)+nc}{m+n}.
\displaystyle \therefore -mb+nc=0.
\displaystyle \therefore mb=nc.
\displaystyle \therefore \frac{m}{n}=\frac{c}{b}.
\displaystyle \therefore \text{The required ratio is }c:b.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }P(0,1,2),\ Q(4,-2,1)\text{ and }O(0,0,0)\text{ are three points, then}
\displaystyle \angle POQ\text{ is:}
\displaystyle \text{(a) }\frac{\pi}{6}\qquad\text{(b) }\frac{\pi}{4}\qquad\text{(c) }\frac{\pi}{3}\qquad\text{(d) }\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{OP}=(0,1,2),\qquad\overrightarrow{OQ}=(4,-2,1).
\displaystyle \overrightarrow{OP}\cdot\overrightarrow{OQ}=0(4)+1(-2)+2(1).
\displaystyle =-2+2=0.
\displaystyle \therefore \overrightarrow{OP}\perp\overrightarrow{OQ}.
\displaystyle \therefore \angle POQ=\frac{\pi}{2}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the extremities of a diagonal of a square are }(1,-2,3)\text{ and }(2,-3,5),
\displaystyle \text{then the length of its side is:}
\displaystyle \text{(a) }\sqrt6\qquad\text{(b) }\sqrt3\qquad\text{(c) }\sqrt5\qquad\text{(d) }\sqrt7
\displaystyle \text{Answer:}
\displaystyle \text{Length of the diagonal}
\displaystyle =\sqrt{(2-1)^2+(-3+2)^2+(5-3)^2}.
\displaystyle =\sqrt{1^2+(-1)^2+2^2}=\sqrt6.
\displaystyle \text{If }s\text{ is the side of a square, then diagonal}=s\sqrt2.
\displaystyle \therefore s=\frac{\sqrt6}{\sqrt2}=\sqrt3.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The points }(5,-4,2),(4,-3,1),(7,6,4)\text{ and }(8,-7,5)\text{ are the}
\displaystyle \text{vertices of:}
\displaystyle \text{(a) A rectangle}\qquad\text{(b) A square}\qquad\text{(c) A parallelogram}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(5,-4,2),\ B(4,-3,1),\ C(7,6,4)\text{ and }D(8,-7,5).
\displaystyle AB^2=(4-5)^2+(-3+4)^2+(1-2)^2=3.
\displaystyle AC^2=(7-5)^2+(6+4)^2+(4-2)^2=108.
\displaystyle AD^2=(8-5)^2+(-7+4)^2+(5-2)^2=27.
\displaystyle BC^2=(7-4)^2+(6+3)^2+(4-1)^2=99.
\displaystyle BD^2=(8-4)^2+(-7+3)^2+(5-1)^2=48.
\displaystyle CD^2=(8-7)^2+(-7-6)^2+(5-4)^2=171.
\displaystyle \text{These distances do not satisfy the conditions for a rectangle, square or parallelogram.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In three-dimensional space, the equation }x^2-5x+6=0\text{ represents:}
\displaystyle \text{(a) Points}\qquad\text{(b) Planes}\qquad\text{(c) Curves}\qquad\text{(d) Pair of straight lines}
\displaystyle \text{Answer:}
\displaystyle x^2-5x+6=0.
\displaystyle (x-2)(x-3)=0.
\displaystyle \therefore x=2\quad\text{or}\quad x=3.
\displaystyle \text{In three-dimensional space, }x=2\text{ and }x=3\text{ represent planes parallel to the }yz\text{-plane.}
\displaystyle \therefore \text{The equation represents a pair of parallel planes.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }(3,4,-1)\text{ and }(-1,2,3)\text{ be the end points of a diameter}
\displaystyle \text{of a sphere. Then the radius of the sphere is equal to:}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }6\qquad\text{(d) }7
\displaystyle \text{Answer:}
\displaystyle \text{Diameter}=\sqrt{(-1-3)^2+(2-4)^2+(3+1)^2}.
\displaystyle =\sqrt{(-4)^2+(-2)^2+4^2}.
\displaystyle =\sqrt{16+4+16}=\sqrt{36}=6.
\displaystyle \therefore \text{Radius}=\frac{6}{2}=3.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The }XOZ\text{-plane divides the line joining }(2,3,1)\text{ and }(6,7,1)
\displaystyle \text{in the ratio:}
\displaystyle \text{(a) }3:7\qquad\text{(b) }2:7\qquad\text{(c) }-3:7\qquad\text{(d) }-2:7
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }XOZ\text{-plane is }y=0.
\displaystyle \text{Let the plane divide the line joining the given points in the ratio }m:n.
\displaystyle \text{Using the section formula for the }y\text{-coordinate,}
\displaystyle 0=\frac{7m+3n}{m+n}.
\displaystyle \therefore 7m+3n=0.
\displaystyle \therefore 7m=-3n.
\displaystyle \therefore \frac{m}{n}=-\frac{3}{7}.
\displaystyle \therefore \text{The required ratio is }-3:7.
\displaystyle \text{The negative sign indicates that the division is external.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{What is the locus of a point for which }y=0,\ z=0\text{?}
\displaystyle \text{(a) }x\text{-axis}\qquad\text{(b) }y\text{-axis}\qquad\text{(c) }z\text{-axis}\qquad\text{(d) }yz\text{-plane}
\displaystyle \text{Answer:}
\displaystyle \text{For every point on the locus, }y=0\text{ and }z=0.
\displaystyle \therefore \text{A general point on the locus is }(x,0,0).
\displaystyle \text{Since }x\text{ can take any real value, the locus is the }x\text{-axis.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The coordinates of the foot of the perpendicular drawn from the point}
\displaystyle P(3,4,5)\text{ on the }yz\text{-plane are:}
\displaystyle \text{(a) }(3,4,0)\qquad\text{(b) }(0,4,5)\qquad\text{(c) }(3,0,5)\qquad\text{(d) }(3,0,0)
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }yz\text{-plane is }x=0.
\displaystyle \text{The perpendicular to the }yz\text{-plane is parallel to the }x\text{-axis.}
\displaystyle \therefore \text{The }y\text{- and }z\text{-coordinates remain unchanged.}
\displaystyle \text{Thus, the foot of the perpendicular is }(0,4,5).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The coordinates of the foot of the perpendicular from the point}
\displaystyle P(6,7,8)\text{ on the }x\text{-axis are:}
\displaystyle \text{(a) }(6,0,0)\qquad\text{(b) }(0,7,0)\qquad\text{(c) }(0,0,8)\qquad\text{(d) }(0,7,8)
\displaystyle \text{Answer:}
\displaystyle \text{A general point on the }x\text{-axis is }(x,0,0).
\displaystyle \text{For the foot of the perpendicular from }P(6,7,8),\text{ the }x\text{-coordinate remains }6.
\displaystyle \therefore \text{The foot of the perpendicular is }(6,0,0).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The perpendicular distance of the point }P(6,7,8)\text{ from the }xy\text{-plane is:}
\displaystyle \text{(a) }8\qquad\text{(b) }7\qquad\text{(c) }6\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }xy\text{-plane is }z=0.
\displaystyle \text{The perpendicular distance of }(x,y,z)\text{ from the }xy\text{-plane is }|z|.
\displaystyle \therefore \text{Distance}=|8|=8.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The length of the perpendicular drawn from }P(3,4,5)\text{ to the }y\text{-axis is:}
\displaystyle \text{(a) }10\qquad\text{(b) }\sqrt{34}\qquad\text{(c) }\sqrt{113}\qquad\text{(d) }5\sqrt2
\displaystyle \text{Answer:}
\displaystyle \text{The foot of the perpendicular from }P(3,4,5)\text{ on the }y\text{-axis is }(0,4,0).
\displaystyle \therefore \text{Required distance}=\sqrt{(3-0)^2+(4-4)^2+(5-0)^2}.
\displaystyle =\sqrt{9+25}=\sqrt{34}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The perpendicular distance of the point }P(3,3,4)\text{ from the }x\text{-axis is:}
\displaystyle \text{(a) }3\sqrt2\qquad\text{(b) }5\qquad\text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \text{The foot of the perpendicular from }P(3,3,4)\text{ on the }x\text{-axis is }(3,0,0).
\displaystyle \therefore \text{Required distance}=\sqrt{(3-3)^2+(3-0)^2+(4-0)^2}.
\displaystyle =\sqrt{9+16}=\sqrt{25}=5.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The length of the perpendicular drawn from the point }P(a,b,c)\text{ to the}
\displaystyle z\text{-axis is:}
\displaystyle \text{(a) }\sqrt{a^2+b^2}\qquad\text{(b) }\sqrt{b^2+c^2}\qquad\text{(c) }\sqrt{a^2+c^2}\qquad\text{(d) }\sqrt{a^2+b^2+c^2}
\displaystyle \text{Answer:}
\displaystyle \text{The foot of the perpendicular from }P(a,b,c)\text{ on the }z\text{-axis is }(0,0,c).
\displaystyle \therefore \text{Required distance}=\sqrt{(a-0)^2+(b-0)^2+(c-c)^2}.
\displaystyle =\sqrt{a^2+b^2}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A plane is parallel to the }yz\text{-plane, so it is perpendicular to the:}
\displaystyle \text{(a) }x\text{-axis}\qquad\text{(b) }y\text{-axis}\qquad\text{(c) }z\text{-axis}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{The }yz\text{-plane is perpendicular to the }x\text{-axis.}
\displaystyle \text{A plane parallel to the }yz\text{-plane has the same normal direction.}
\displaystyle \therefore \text{It is also perpendicular to the }x\text{-axis.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The point }(-2,-3,-4)\text{ lies in the:}
\displaystyle \text{(a) First octant}\qquad\text{(b) Seventh octant}\qquad\text{(c) Second octant}\qquad\text{(d) Eighth octant}
\displaystyle \text{Answer:}
\displaystyle \text{For the point }(-2,-3,-4),
\displaystyle x<0,\qquad y<0,\qquad z<0.
\displaystyle \text{The sign combination }(-,-,-)\text{ represents the seventh octant.}
\displaystyle \therefore \text{The point lies in the seventh octant.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If the distance between the points }(a,0,1)\text{ and }(0,1,2)\text{ is}
\displaystyle \sqrt{27},\text{ then the value of }a\text{ is:}
\displaystyle \text{(a) }5\qquad\text{(b) }\pm5\qquad\text{(c) }-5\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \sqrt{(a-0)^2+(0-1)^2+(1-2)^2}=\sqrt{27}.
\displaystyle \sqrt{a^2+1+1}=\sqrt{27}.
\displaystyle a^2+2=27.
\displaystyle \therefore a^2=25.
\displaystyle \therefore a=\pm5.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The }x\text{-axis is the intersection of the planes:}
\displaystyle \text{(a) }xy\text{ and }xz\qquad\text{(b) }yz\text{ and }zx\qquad\text{(c) }xy\text{ and }yz\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{For a point on the }x\text{-axis, }y=0\text{ and }z=0.
\displaystyle y=0\text{ represents the }xz\text{-plane.}
\displaystyle z=0\text{ represents the }xy\text{-plane.}
\displaystyle \therefore \text{The }x\text{-axis is the intersection of the }xy\text{- and }xz\text{-planes.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The equation of the }y\text{-axis is considered as:}
\displaystyle \text{(a) }x=0,\ y=0\qquad\text{(b) }y=0,\ z=0\qquad\text{(c) }z=0,\ x=0\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{A general point on the }y\text{-axis is }(0,y,0).
\displaystyle \therefore x=0\text{ and }z=0.
\displaystyle \therefore \text{The equations of the }y\text{-axis are }x=0,\ z=0.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \text{CASE BASED / SOURCE BASED QUESTIONS}

\displaystyle \textbf{Question 21: }\text{A three-dimensional Cartesian coordinate system consists of the origin }O
\displaystyle \text{and three mutually perpendicular coordinate axes, called the }x\text{-axis, }y\text{-axis and }z\text{-axis.}
\displaystyle \text{These axes are also known as the abscissa, ordinate and applicate axes, respectively.}
\displaystyle \text{The three coordinate planes divide the three-dimensional space into eight regions called octants.}

\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) In which octant does the point }(3,-2,-5)\text{ lie?}
\displaystyle \text{(a) II}\qquad\text{(b) VI}\qquad\text{(c) VII}\qquad\text{(d) VIII}

\displaystyle \text{(ii) If a point lies on the }z\text{-axis, then its coordinates are:}
\displaystyle \text{(a) }(x,y,z)\qquad\text{(b) }(x,y,0)\qquad\text{(c) }(0,y,0)\qquad\text{(d) }(0,0,z)

\displaystyle \text{(iii) The locus of a point for which }x=0\text{ is:}
\displaystyle \text{(a) }xy\text{-plane}\qquad\text{(b) }yz\text{-plane}\qquad\text{(c) }zx\text{-plane}\qquad\text{(d) None of these}

\displaystyle \text{(iv) A plane parallel to the }yz\text{-plane is perpendicular to the:}
\displaystyle \text{(a) }x\text{-axis}\qquad\text{(b) }y\text{-axis}\qquad\text{(c) }z\text{-axis}\qquad\text{(d) None of these}

\displaystyle \text{Answer:}
\displaystyle \text{(i) For }(3,-2,-5),\quad x>0,\ y<0,\ z<0.
\displaystyle \text{The sign combination }(+,-,-)\text{ corresponds to the VIII octant.}
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(ii) A point on the }z\text{-axis has }x=0\text{ and }y=0.
\displaystyle \therefore \text{Its coordinates are }(0,0,z).
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(iii) The equation }x=0\text{ represents the }yz\text{-plane.}
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(iv) The }yz\text{-plane is perpendicular to the }x\text{-axis.}
\displaystyle \text{Therefore, a plane parallel to the }yz\text{-plane is also perpendicular to the }x\text{-axis.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The coordinates of a point }P\text{ in space are }(a,b,c).
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) The distance of point }P\text{ from the }xy\text{-plane is:}
\displaystyle \text{(a) }|a|\qquad\text{(b) }|b|\qquad\text{(c) }|c|\qquad\text{(d) }\sqrt{a^2+b^2}

\displaystyle \text{(ii) The coordinates of the foot of the perpendicular drawn from }P\text{ to the }yz\text{-plane are:}
\displaystyle \text{(a) }(0,a,b)\qquad\text{(b) }(0,a,c)\qquad\text{(c) }(0,b,a)\qquad\text{(d) }(0,b,c)

\displaystyle \text{(iii) The coordinates of the foot of the perpendicular drawn from }P\text{ to the }z\text{-axis are:}
\displaystyle \text{(a) }\left(0,0,\sqrt{b^2+c^2}\right)\qquad\text{(b) }(0,0,c)
\displaystyle \text{(c) }\left(0,0,\sqrt{a^2+b^2}\right)\qquad\text{(d) }\left(0,0,\sqrt{c^2+a^2}\right)

\displaystyle \text{(iv) The distance between }P\text{ and its image in the }zx\text{-plane is:}
\displaystyle \text{(a) }2|b|\qquad\text{(b) }2|a|\qquad\text{(c) }2|c|\qquad\text{(d) }2|a+c|
\displaystyle \text{Answer:}

\displaystyle \text{(i) The equation of the }xy\text{-plane is }z=0.
\displaystyle \text{The perpendicular distance of }P(a,b,c)\text{ from the }xy\text{-plane is }|c|.
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(ii) The equation of the }yz\text{-plane is }x=0.
\displaystyle \text{Hence, only the }x\text{-coordinate changes to zero, while }y\text{ and }z\text{ remain unchanged.}
\displaystyle \therefore \text{The foot of the perpendicular is }(0,b,c).
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(iii) A general point on the }z\text{-axis is }(0,0,z).
\displaystyle \text{The foot of the perpendicular from }P(a,b,c)\text{ to the }z\text{-axis is }(0,0,c).
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(iv) The }zx\text{-plane is given by }y=0.
\displaystyle \text{The image of }P(a,b,c)\text{ in the }zx\text{-plane is }P'(a,-b,c).
\displaystyle PP'=\sqrt{(a-a)^2+(b+b)^2+(c-c)^2}.
\displaystyle =\sqrt{4b^2}=2|b|.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 23: }\text{Write the distance of the point }P(2,3,5)\text{ from the }xy\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance from the }xy\text{-plane}=|z|=|5|=5\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Write the distance of the point }P(3,4,5)\text{ from the }z\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Distance from the }z\text{-axis}=\sqrt{x^2+y^2}.
\displaystyle =\sqrt{3^2+4^2}=\sqrt{25}=5\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the distance between the points }P(a,2,1)\text{ and }Q(1,-1,1)
\displaystyle \text{is }5\text{ units, find the value of }a.
\displaystyle \text{Answer:}
\displaystyle \sqrt{(a-1)^2+(2+1)^2+(1-1)^2}=5.
\displaystyle (a-1)^2+9=25.
\displaystyle (a-1)^2=16.
\displaystyle a-1=\pm4.
\displaystyle \therefore a=5\text{ or }-3.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The coordinates of the midpoints of sides }AB,\ BC\text{ and }CA\text{ of}
\displaystyle \triangle ABC\text{ are }D(1,2,-3),\ E(3,0,1)\text{ and }F(-1,1,-4)\text{ respectively.}
\displaystyle \text{Write the coordinates of its centroid.}
\displaystyle \text{Answer:}
\displaystyle \text{The centroid of }\triangle ABC\text{ is also the centroid of }\triangle DEF.
\displaystyle G=\left(\frac{1+3-1}{3},\frac{2+0+1}{3},\frac{-3+1-4}{3}\right).
\displaystyle =\left(\frac{3}{3},\frac{3}{3},\frac{-6}{3}\right).
\displaystyle \therefore G=(1,1,-2).
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Write the coordinates of the third vertex of a triangle having centroid at}
\displaystyle \text{the origin and two vertices }(3,-5,7)\text{ and }(3,0,1).
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }(x,y,z).
\displaystyle \text{Since the centroid is }(0,0,0),
\displaystyle \frac{3+3+x}{3}=0,\qquad\frac{-5+0+y}{3}=0,\qquad\frac{7+1+z}{3}=0.
\displaystyle 6+x=0,\qquad -5+y=0,\qquad 8+z=0.
\displaystyle \therefore x=-6,\qquad y=5,\qquad z=-8.
\displaystyle \therefore \text{The third vertex is }(-6,5,-8).
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{What is the locus of a point }(x,y,z)\text{ for }y=0,\ z=0\text{?}
\displaystyle \text{Answer:}
\displaystyle y=0,\qquad z=0.
\displaystyle \text{The }x\text{-coordinate can take any real value.}
\displaystyle \therefore \text{The locus of the point is the }x\text{-axis.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Find the point on the }y\text{-axis which is at a distance of }\sqrt{10}
\displaystyle \text{units from the point }(1,2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }y\text{-axis be }(0,y,0).
\displaystyle \sqrt{(0-1)^2+(y-2)^2+(0-3)^2}=\sqrt{10}.
\displaystyle 1+(y-2)^2+9=10.
\displaystyle (y-2)^2=0.
\displaystyle \therefore y=2.
\displaystyle \therefore \text{The required point is }(0,2,0).
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If the origin is the centroid of a triangle }ABC\text{ having vertices}
\displaystyle A(a,1,3),\ B(-2,b,-5)\text{ and }C(4,7,c),\text{ find the values of }a,b,c.
\displaystyle \text{Answer:}
\displaystyle \text{Since the centroid is }(0,0,0),
\displaystyle \frac{a-2+4}{3}=0,\qquad\frac{1+b+7}{3}=0,\qquad\frac{3-5+c}{3}=0.
\displaystyle a+2=0,\qquad b+8=0,\qquad c-2=0.
\displaystyle \therefore a=-2,\qquad b=-8,\qquad c=2.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Find the distance of the point }(a,b,c)\text{ from the }zx\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }zx\text{-plane is }y=0.
\displaystyle \therefore \text{The perpendicular distance of }(a,b,c)\text{ from the }zx\text{-plane is }|b|.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If }L\text{ and }M\text{ are the feet of the perpendiculars from }P(3,4,5)
\displaystyle \text{on the }xy\text{- and }yz\text{-planes respectively, find }LM.
\displaystyle \text{Answer:}
\displaystyle \text{The foot of the perpendicular on the }xy\text{-plane is }L=(3,4,0).
\displaystyle \text{The foot of the perpendicular on the }yz\text{-plane is }M=(0,4,5).
\displaystyle LM=\sqrt{(3-0)^2+(4-4)^2+(0-5)^2}.
\displaystyle =\sqrt{9+25}=\sqrt{34}.
\displaystyle \therefore LM=\sqrt{34}\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Find the value of }\beta\text{ so that the distance between the points}
\displaystyle (7,1,-3)\text{ and }(4,5,\beta)\text{ is }13.
\displaystyle \text{Answer:}
\displaystyle \sqrt{(7-4)^2+(1-5)^2+(-3-\beta)^2}=13.
\displaystyle 9+16+(\beta+3)^2=169.
\displaystyle (\beta+3)^2=144.
\displaystyle \beta+3=\pm12.
\displaystyle \therefore \beta=9\text{ or }-15.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Are the points }A(3,6,9),\ B(10,20,30)\text{ and }C(25,-41,5)
\displaystyle \text{vertices of a right-angled triangle? Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle AB^2=(10-3)^2+(20-6)^2+(30-9)^2.
\displaystyle =7^2+14^2+21^2=686.
\displaystyle AC^2=(25-3)^2+(-41-6)^2+(5-9)^2.
\displaystyle =22^2+(-47)^2+(-4)^2=2709.
\displaystyle BC^2=(25-10)^2+(-41-20)^2+(5-30)^2.
\displaystyle =15^2+(-61)^2+(-25)^2=4571.
\displaystyle AB^2+AC^2=686+2709=3395\ne4571=BC^2.
\displaystyle AB^2+BC^2\ne AC^2\quad\text{and}\quad AC^2+BC^2\ne AB^2.
\displaystyle \therefore \text{The given points do not form a right-angled triangle.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Find the midpoint of }(1,4,6)\text{ and }(5,8,10).
\displaystyle \text{Answer:}
\displaystyle \text{Midpoint}=\left(\frac{1+5}{2},\frac{4+8}{2},\frac{6+10}{2}\right).
\displaystyle =\left(3,6,8\right).
\displaystyle \therefore \text{The required midpoint is }(3,6,8).
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Find the distance between the point }A(3,4,5)\text{ and the origin }(0,0,0).
\displaystyle \text{Answer:}
\displaystyle \text{Distance}=\sqrt{(3-0)^2+(4-0)^2+(5-0)^2}.
\displaystyle =\sqrt{9+16+25}=\sqrt{50}=5\sqrt2.
\displaystyle \therefore \text{The required distance is }5\sqrt2\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{A point is on the }x\text{-axis. What are its }y\text{-coordinate and }z\text{-coordinate?}
\displaystyle \text{Answer:}
\displaystyle \text{A general point on the }x\text{-axis is }(x,0,0).
\displaystyle \therefore y=0\qquad\text{and}\qquad z=0.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{A point is in the }xz\text{-plane. What can you say about its }y\text{-coordinate?}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }xz\text{-plane is }y=0.
\displaystyle \therefore \text{The }y\text{-coordinate of the point is }0.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Name the octant in which the point }(7,-1,5)\text{ lies.}
\displaystyle \text{Answer:}
\displaystyle \text{For the point }(7,-1,5),\quad x>0,\quad y<0,\quad z>0.
\displaystyle \text{The sign combination }(+,-,+)\text{ corresponds to the IV octant.}
\displaystyle \therefore \text{The point lies in the IV octant.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Find the coordinates of a point on the }y\text{-axis which is at a distance}
\displaystyle \text{of }5\sqrt2\text{ from the point }P(3,-2,5).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }y\text{-axis be }(0,y,0).
\displaystyle \sqrt{(0-3)^2+(y+2)^2+(0-5)^2}=5\sqrt2.
\displaystyle 9+(y+2)^2+25=50.
\displaystyle (y+2)^2=16.
\displaystyle y+2=\pm4.
\displaystyle \therefore y=2\text{ or }-6.
\displaystyle \therefore \text{The required points are }(0,2,0)\text{ and }(0,-6,0).
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Write the plane containing the }x\text{-axis and }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The plane containing the }x\text{-axis and }y\text{-axis is the }xy\text{-plane.}
\displaystyle \therefore \text{Its equation is }z=0.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Find the distance of the point }P(3,4,5)\text{ from the }yz\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }yz\text{-plane is }x=0.
\displaystyle \text{Distance from the }yz\text{-plane}=|x|=|3|=3\text{ units}.
\displaystyle \\

\displaystyle \text{SHORT ANSWER TYPE QUESTIONS}

\displaystyle \textbf{Question 43: }\text{Show that the points }(a,b,c),(b,c,a)\text{ and }(c,a,b)\text{ are}
\displaystyle \text{vertices of an equilateral triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(a,b,c),\quad B(b,c,a),\quad C(c,a,b).
\displaystyle AB^2=(b-a)^2+(c-b)^2+(a-c)^2.
\displaystyle BC^2=(c-b)^2+(a-c)^2+(b-a)^2.
\displaystyle CA^2=(a-c)^2+(b-a)^2+(c-b)^2.
\displaystyle \therefore AB^2=BC^2=CA^2.
\displaystyle \therefore AB=BC=CA.
\displaystyle \therefore \triangle ABC\text{ is an equilateral triangle.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Using the distance formula, prove that the points }A(4,-3,-1),
\displaystyle B(5,-7,6)\text{ and }C(3,1,-8)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(5-4)^2+(-7+3)^2+(6+1)^2}.
\displaystyle =\sqrt{1+16+49}=\sqrt{66}.
\displaystyle BC=\sqrt{(3-5)^2+(1+7)^2+(-8-6)^2}.
\displaystyle =\sqrt{4+64+196}=\sqrt{264}=2\sqrt{66}.
\displaystyle AC=\sqrt{(3-4)^2+(1+3)^2+(-8+1)^2}.
\displaystyle =\sqrt{1+16+49}=\sqrt{66}.
\displaystyle \text{Thus, }AB+AC=\sqrt{66}+\sqrt{66}=2\sqrt{66}=BC.
\displaystyle \therefore A\text{ lies between }B\text{ and }C.
\displaystyle \therefore A,\ B\text{ and }C\text{ are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Find the point on the }y\text{-axis which is equidistant from the points}
\displaystyle (3,1,2)\text{ and }(5,5,2).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }y\text{-axis be }P(0,y,0).
\displaystyle \text{Since }P\text{ is equidistant from }A(3,1,2)\text{ and }B(5,5,2),
\displaystyle PA=PB.
\displaystyle (0-3)^2+(y-1)^2+(0-2)^2=(0-5)^2+(y-5)^2+(0-2)^2.
\displaystyle 9+(y-1)^2=25+(y-5)^2.
\displaystyle 9+y^2-2y+1=25+y^2-10y+25.
\displaystyle 8y=40.
\displaystyle \therefore y=5.
\displaystyle \therefore \text{The required point is }(0,5,0).
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{Find the points on the }z\text{-axis which are at a distance }\sqrt{21}\text{ from}
\displaystyle \text{the point }(1,2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }z\text{-axis be }P(0,0,z).
\displaystyle \sqrt{(0-1)^2+(0-2)^2+(z-3)^2}=\sqrt{21}.
\displaystyle 1+4+(z-3)^2=21.
\displaystyle (z-3)^2=16.
\displaystyle z-3=\pm4.
\displaystyle \therefore z=7\text{ or }-1.
\displaystyle \therefore \text{The required points are }(0,0,7)\text{ and }(0,0,-1).
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Show that the points }(0,7,10),(-1,6,6)\text{ and }(-4,9,6)\text{ are}
\displaystyle \text{the vertices of an isosceles right-angled triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(0,7,10),\quad B(-1,6,6),\quad C(-4,9,6).
\displaystyle AB^2=(-1-0)^2+(6-7)^2+(6-10)^2=1+1+16=18.
\displaystyle BC^2=(-4+1)^2+(9-6)^2+(6-6)^2=9+9=18.
\displaystyle AC^2=(-4-0)^2+(9-7)^2+(6-10)^2=16+4+16=36.
\displaystyle \therefore AB^2=BC^2=18\quad\Rightarrow\quad AB=BC.
\displaystyle \text{Hence, }\triangle ABC\text{ is isosceles.}
\displaystyle AB^2+BC^2=18+18=36=AC^2.
\displaystyle \therefore \angle ABC=90^\circ.
\displaystyle \therefore \triangle ABC\text{ is an isosceles right-angled triangle.}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{Three vertices of a parallelogram }ABCD\text{ are }A(3,-1,2),
\displaystyle B(1,2,-4)\text{ and }C(-1,1,2).\text{ Find the coordinates of the fourth vertex.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the fourth vertex be }D(x,y,z).
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore A+C=B+D.
\displaystyle D=A+C-B.
\displaystyle D=(3,-1,2)+(-1,1,2)-(1,2,-4).
\displaystyle D=(3-1-1,-1+1-2,2+2+4).
\displaystyle \therefore D=(1,-2,8).
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Find the equation of the set of points }P\text{ such that its distance}
\displaystyle \text{from the points }A(3,4,-5)\text{ and }B(-2,1,4)\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point of the required locus.}
\displaystyle \text{Since }P\text{ is equidistant from }A\text{ and }B,\quad PA=PB.
\displaystyle (x-3)^2+(y-4)^2+(z+5)^2=(x+2)^2+(y-1)^2+(z-4)^2.
\displaystyle x^2-6x+9+y^2-8y+16+z^2+10z+25
\displaystyle =x^2+4x+4+y^2-2y+1+z^2-8z+16.
\displaystyle -6x-8y+10z+50=4x-2y-8z+21.
\displaystyle \therefore 10x+6y-18z-29=0.
\displaystyle \therefore \text{The required equation is }10x+6y-18z-29=0.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Using section formula, prove that the three points }(-4,6,10),
\displaystyle (2,4,6)\text{ and }(14,0,-2)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-4,6,10),\quad B(2,4,6),\quad C(14,0,-2).
\displaystyle \text{Suppose }B\text{ divides }AC\text{ internally in the ratio }m:n.
\displaystyle 2=\frac{14m-4n}{m+n}.
\displaystyle 2m+2n=14m-4n.
\displaystyle 12m=6n\quad\Rightarrow\quad m:n=1:2.
\displaystyle \text{The point dividing }AC\text{ internally in the ratio }1:2\text{ is}
\displaystyle \left(\frac{1(14)+2(-4)}{3},\frac{1(0)+2(6)}{3},\frac{1(-2)+2(10)}{3}\right).
\displaystyle =\left(\frac{6}{3},\frac{12}{3},\frac{18}{3}\right)=(2,4,6).
\displaystyle \text{This point is }B.
\displaystyle \therefore A,\ B\text{ and }C\text{ are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{Find the coordinates of the point which divides the line segment joining}
\displaystyle \text{the points }(1,-2,3)\text{ and }(3,4,-5)\text{ in the ratio }2:3\text{ internally.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,-2,3)\text{ and }B(3,4,-5).
\displaystyle \text{Using the section formula, the required point is}
\displaystyle \left(\frac{2(3)+3(1)}{2+3},\frac{2(4)+3(-2)}{2+3},\frac{2(-5)+3(3)}{2+3}\right).
\displaystyle =\left(\frac{9}{5},\frac{2}{5},-\frac{1}{5}\right).
\displaystyle \therefore \text{The required point is }\left(\frac95,\frac25,-\frac15\right).
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Given that }P(3,2,-4),Q(5,4,-6)\text{ and }R(9,8,-10)\text{ are}
\displaystyle \text{collinear, find the ratio in which }Q\text{ divides }PR.
\displaystyle \text{Answer:}
\displaystyle \text{Let }Q\text{ divide }PR\text{ internally in the ratio }m:n.
\displaystyle 5=\frac{9m+3n}{m+n}.
\displaystyle 5m+5n=9m+3n.
\displaystyle 2n=4m.
\displaystyle \therefore m:n=1:2.
\displaystyle \text{Checking with the }y\text{-coordinate:}
\displaystyle \frac{8(1)+2(2)}{1+2}=\frac{12}{3}=4.
\displaystyle \text{Checking with the }z\text{-coordinate:}
\displaystyle \frac{-10(1)+(-4)(2)}{1+2}=\frac{-18}{3}=-6.
\displaystyle \therefore Q\text{ divides }PR\text{ internally in the ratio }1:2.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Find the ratio in which the }yz\text{-plane divides the line segment}
\displaystyle \text{joining the points }(-2,4,7)\text{ and }(3,-5,8).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,4,7),\ B(3,-5,8)\text{ and let the }yz\text{-plane divide }AB\text{ in the ratio }m:n.
\displaystyle \text{Since the point of division lies on the }yz\text{-plane, its }x\text{-coordinate is }0.
\displaystyle \therefore \frac{3m-2n}{m+n}=0.
\displaystyle 3m-2n=0.
\displaystyle 3m=2n.
\displaystyle \therefore m:n=2:3.
\displaystyle \therefore \text{The }yz\text{-plane divides the line segment internally in the ratio }2:3.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{Find the coordinates of the points which trisect the line segment joining}
\displaystyle \text{the points }P(4,2,-6)\text{ and }Q(10,-16,6).
\displaystyle \text{Answer:}
\displaystyle \text{Let the points of trisection be }R\text{ and }S.
\displaystyle R\text{ divides }PQ\text{ internally in the ratio }1:2.
\displaystyle R=\left(\frac{1(10)+2(4)}{3},\frac{1(-16)+2(2)}{3},\frac{1(6)+2(-6)}{3}\right).
\displaystyle =\left(\frac{18}{3},\frac{-12}{3},\frac{-6}{3}\right)=(6,-4,-2).
\displaystyle S\text{ divides }PQ\text{ internally in the ratio }2:1.
\displaystyle S=\left(\frac{2(10)+1(4)}{3},\frac{2(-16)+1(2)}{3},\frac{2(6)+1(-6)}{3}\right).
\displaystyle =\left(\frac{24}{3},\frac{-30}{3},\frac{6}{3}\right)=(8,-10,2).
\displaystyle \therefore \text{The points of trisection are }(6,-4,-2)\text{ and }(8,-10,2).
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Verify that }(-1,2,1),(1,-2,5),(4,-7,8)\text{ and }(2,-3,4)
\displaystyle \text{are the vertices of a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-1,2,1),\ B(1,-2,5),\ C(4,-7,8)\text{ and }D(2,-3,4).
\displaystyle \text{Midpoint of }AC=\left(\frac{-1+4}{2},\frac{2-7}{2},\frac{1+8}{2}\right).
\displaystyle =\left(\frac{3}{2},-\frac{5}{2},\frac{9}{2}\right).
\displaystyle \text{Midpoint of }BD=\left(\frac{1+2}{2},\frac{-2-3}{2},\frac{5+4}{2}\right).
\displaystyle =\left(\frac{3}{2},-\frac{5}{2},\frac{9}{2}\right).
\displaystyle \therefore \text{The diagonals }AC\text{ and }BD\text{ bisect each other.}
\displaystyle \therefore ABCD\text{ is a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Given the points }A(1,2,-3)\text{ and }B(3,-2,1),\text{ find the locus of }P
\displaystyle \text{if }AP^2-BP^2=18.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point on the locus.}
\displaystyle AP^2=(x-1)^2+(y-2)^2+(z+3)^2.
\displaystyle BP^2=(x-3)^2+(y+2)^2+(z-1)^2.
\displaystyle \text{Given, }AP^2-BP^2=18.
\displaystyle (x-1)^2+(y-2)^2+(z+3)^2
\displaystyle -(x-3)^2-(y+2)^2-(z-1)^2=18.
\displaystyle 4x-8y+8z=18.
\displaystyle \therefore 2x-4y+4z=9.
\displaystyle \therefore \text{The locus of }P\text{ is }2x-4y+4z=9.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Find the locus of a point which is equidistant from the points }(3,2,1)
\displaystyle \text{and }(1,2,3).\text{ What surface does it represent?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point on the locus.}
\displaystyle \text{Since }P\text{ is equidistant from }A(3,2,1)\text{ and }B(1,2,3),
\displaystyle PA=PB.
\displaystyle (x-3)^2+(y-2)^2+(z-1)^2=(x-1)^2+(y-2)^2+(z-3)^2.
\displaystyle x^2-6x+9+z^2-2z+1=x^2-2x+1+z^2-6z+9.
\displaystyle -4x+4z=0.
\displaystyle \therefore x-z=0.
\displaystyle \therefore \text{The locus is the plane }x-z=0.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Find the point on the }z\text{-axis which is equidistant from the points}
\displaystyle (1,5,7)\text{ and }(5,1,-4).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }z\text{-axis be }P(0,0,z).
\displaystyle \text{Since }P\text{ is equidistant from }A(1,5,7)\text{ and }B(5,1,-4),
\displaystyle PA=PB.
\displaystyle 1^2+5^2+(z-7)^2=5^2+1^2+(z+4)^2.
\displaystyle (z-7)^2=(z+4)^2.
\displaystyle z^2-14z+49=z^2+8z+16.
\displaystyle -22z=-33.
\displaystyle \therefore z=\frac{3}{2}.
\displaystyle \therefore \text{The required point is }\left(0,0,\frac{3}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Show that the points }(0,7,-10),(1,6,-6)\text{ and }(4,9,-6)
\displaystyle \text{are the vertices of an isosceles triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(0,7,-10),\quad B(1,6,-6),\quad C(4,9,-6).
\displaystyle AB^2=(1-0)^2+(6-7)^2+(-6+10)^2.
\displaystyle =1+1+16=18.
\displaystyle AC^2=(4-0)^2+(9-7)^2+(-6+10)^2.
\displaystyle =16+4+16=36.
\displaystyle BC^2=(4-1)^2+(9-6)^2+(-6+6)^2.
\displaystyle =9+9=18.
\displaystyle \therefore AB^2=BC^2=18.
\displaystyle \therefore AB=BC.
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Find the image of:}
\displaystyle \text{(i) }(-2,3,4)\text{ in the }yz\text{-plane.}
\displaystyle \text{(ii) }(-5,4,-3)\text{ in the }xz\text{-plane.}
\displaystyle \text{(iii) }(5,2,-7)\text{ in the }xy\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) In reflection in the }yz\text{-plane, the }x\text{-coordinate changes sign.}
\displaystyle (-2,3,4)\longrightarrow(2,3,4).
\displaystyle \therefore \text{The image is }(2,3,4).
\displaystyle \text{(ii) In reflection in the }xz\text{-plane, the }y\text{-coordinate changes sign.}
\displaystyle (-5,4,-3)\longrightarrow(-5,-4,-3).
\displaystyle \therefore \text{The image is }(-5,-4,-3).
\displaystyle \text{(iii) In reflection in the }xy\text{-plane, the }z\text{-coordinate changes sign.}
\displaystyle (5,2,-7)\longrightarrow(5,2,7).
\displaystyle \therefore \text{The image is }(5,2,7).
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{Name the octants in which the following points lie:}
\displaystyle \text{(i) }(5,2,3)\quad\text{(ii) }(-5,4,3)\quad\text{(iii) }(4,-3,5)\quad\text{(iv) }(7,4,-3)
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(5,2,3):\quad x>0,\ y>0,\ z>0.
\displaystyle \therefore \text{The point lies in the first octant.}
\displaystyle \text{(ii) }(-5,4,3):\quad x<0,\ y>0,\ z>0.
\displaystyle \therefore \text{The point lies in the second octant.}
\displaystyle \text{(iii) }(4,-3,5):\quad x>0,\ y<0,\ z>0.
\displaystyle \therefore \text{The point lies in the fourth octant.}
\displaystyle \text{(iv) }(7,4,-3):\quad x>0,\ y>0,\ z<0.
\displaystyle \therefore \text{The point lies in the fifth octant.}
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Show that the points }A(3,3,3),B(0,6,3),C(1,7,7)\text{ and}
\displaystyle D(4,4,7)\text{ are vertices of a square.}
\displaystyle \text{Answer:}
\displaystyle AB^2=(0-3)^2+(6-3)^2+(3-3)^2=9+9=18.
\displaystyle BC^2=(1-0)^2+(7-6)^2+(7-3)^2=1+1+16=18.
\displaystyle CD^2=(4-1)^2+(4-7)^2+(7-7)^2=9+9=18.
\displaystyle DA^2=(3-4)^2+(3-4)^2+(3-7)^2=1+1+16=18.
\displaystyle \therefore AB=BC=CD=DA=\sqrt{18}.
\displaystyle AC^2=(1-3)^2+(7-3)^2+(7-3)^2=4+16+16=36.
\displaystyle BD^2=(4-0)^2+(4-6)^2+(7-3)^2=16+4+16=36.
\displaystyle \therefore AC=BD=6.
\displaystyle \text{Thus, all four sides are equal and the diagonals are equal.}
\displaystyle \therefore ABCD\text{ is a square.}
\displaystyle \\

\displaystyle \text{LONG ANSWER TYPE QUESTIONS}

\displaystyle \textbf{Question 63: }\text{Find the ratio in which the plane }2x+2y-2z=1\text{ divides the line}
\displaystyle \text{segment joining the points }A(2,1,5)\text{ and }B(3,4,3).\text{ Also find the coordinates}
\displaystyle \text{of the point of division.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the plane divide }AB\text{ internally at }P\text{ in the ratio }m:n.
\displaystyle \therefore P=\left(\frac{3m+2n}{m+n},\frac{4m+n}{m+n},\frac{3m+5n}{m+n}\right).
\displaystyle \text{Since }P\text{ lies on the plane }2x+2y-2z=1,
\displaystyle 2\left(\frac{3m+2n}{m+n}\right)+2\left(\frac{4m+n}{m+n}\right)
\displaystyle -2\left(\frac{3m+5n}{m+n}\right)=1.
\displaystyle \frac{6m+4n+8m+2n-6m-10n}{m+n}=1.
\displaystyle \frac{8m-4n}{m+n}=1.
\displaystyle 8m-4n=m+n.
\displaystyle 7m=5n.
\displaystyle \therefore m:n=5:7.
\displaystyle \therefore \text{The plane divides }AB\text{ internally in the ratio }5:7.
\displaystyle \text{Using the section formula,}
\displaystyle P=\left(\frac{5(3)+7(2)}{12},\frac{5(4)+7(1)}{12},\frac{5(3)+7(5)}{12}\right).
\displaystyle =\left(\frac{29}{12},\frac{27}{12},\frac{50}{12}\right).
\displaystyle =\left(\frac{29}{12},\frac94,\frac{25}{6}\right).
\displaystyle \therefore \text{The required ratio is }5:7\text{ and the point of division is}
\displaystyle \left(\frac{29}{12},\frac94,\frac{25}{6}\right).
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{The midpoints of the sides of a triangle are }(1,5,-1),(0,4,-2)
\displaystyle \text{and }(2,3,4).\text{ Find its vertices.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }D(1,5,-1),\ E(0,4,-2)\text{ and }F(2,3,4)\text{ be the midpoints of}
\displaystyle BC,\ CA\text{ and }AB\text{ respectively.}
\displaystyle \text{Let }A(x_1,y_1,z_1),\ B(x_2,y_2,z_2),\ C(x_3,y_3,z_3).
\displaystyle \text{Since }D\text{ is the midpoint of }BC,
\displaystyle B+C=2D.
\displaystyle \text{Similarly, }C+A=2E\quad\text{and}\quad A+B=2F.
\displaystyle (C+A)+(A+B)-(B+C)=2E+2F-2D.
\displaystyle 2A=2(E+F-D).
\displaystyle \therefore A=E+F-D.
\displaystyle A=(0,4,-2)+(2,3,4)-(1,5,-1).
\displaystyle \therefore A=(1,2,3).
\displaystyle \text{Similarly, }B=F+D-E.
\displaystyle B=(2,3,4)+(1,5,-1)-(0,4,-2).
\displaystyle \therefore B=(3,4,5).
\displaystyle \text{Also, }C=D+E-F.
\displaystyle C=(1,5,-1)+(0,4,-2)-(2,3,4).
\displaystyle \therefore C=(-1,6,-7).
\displaystyle \therefore \text{The vertices of the triangle are }(1,2,3),(3,4,5)\text{ and }(-1,6,-7).
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Find the lengths of the medians of the triangle with vertices }A(0,0,6),
\displaystyle B(0,4,0)\text{ and }C(6,0,0).
\displaystyle \text{Answer:}
\displaystyle \text{Let }D,E\text{ and }F\text{ be the midpoints of }BC,CA\text{ and }AB\text{ respectively.}
\displaystyle D=\left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right)=(3,2,0).
\displaystyle E=\left(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right)=(3,0,3).
\displaystyle F=\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)=(0,2,3).
\displaystyle \text{The median from }A\text{ is }AD.
\displaystyle AD=\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}.
\displaystyle =\sqrt{9+4+36}=\sqrt{49}=7.
\displaystyle \text{The median from }B\text{ is }BE.
\displaystyle BE=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}.
\displaystyle =\sqrt{9+16+9}=\sqrt{34}.
\displaystyle \text{The median from }C\text{ is }CF.
\displaystyle CF=\sqrt{(0-6)^2+(2-0)^2+(3-0)^2}.
\displaystyle =\sqrt{36+4+9}=\sqrt{49}=7.
\displaystyle \therefore \text{The lengths of the medians are }7,\sqrt{34}\text{ and }7\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{Write the coordinates of the point }P\text{ which is five-sixths of the way}
\displaystyle \text{from }A(-2,0,6)\text{ to }B(10,-6,-12).
\displaystyle \text{Answer:}
\displaystyle \text{Since }P\text{ is five-sixths of the way from }A\text{ to }B,
\displaystyle AP=\frac{5}{6}AB.
\displaystyle \therefore AP:PB=5:1.
\displaystyle \text{Using the section formula,}
\displaystyle P=\left(\frac{5(10)+1(-2)}{5+1},\frac{5(-6)+1(0)}{5+1},  \frac{5(-12)+1(6)}{5+1}\right).
\displaystyle =\left(\frac{48}{6},\frac{-30}{6},\frac{-54}{6}\right).
\displaystyle =(8,-5,-9).
\displaystyle \therefore \text{The coordinates of }P\text{ are }(8,-5,-9).
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{Find the equation of the set of points }P\text{, the sum of whose distances}
\displaystyle \text{from }A(4,0,0)\text{ and }B(-4,0,0)\text{ is equal to }10.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point on the required locus.}
\displaystyle \text{Given, }PA+PB=10.
\displaystyle \sqrt{(x-4)^2+y^2+z^2}+\sqrt{(x+4)^2+y^2+z^2}=10.
\displaystyle \text{Let }S=x^2+y^2+z^2+16.
\displaystyle \therefore \sqrt{S-8x}+\sqrt{S+8x}=10.
\displaystyle \sqrt{S-8x}=10-\sqrt{S+8x}.
\displaystyle S-8x=100+S+8x-20\sqrt{S+8x}.
\displaystyle 20\sqrt{S+8x}=100+16x.
\displaystyle 5\sqrt{S+8x}=25+4x.
\displaystyle 25(S+8x)=(25+4x)^2.
\displaystyle 25S+200x=625+200x+16x^2.
\displaystyle 25S=625+16x^2.
\displaystyle 25(x^2+y^2+z^2+16)=625+16x^2.
\displaystyle 25x^2+25y^2+25z^2+400=625+16x^2.
\displaystyle 9x^2+25y^2+25z^2=225.
\displaystyle \therefore \frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{9}=1.
\displaystyle \therefore \text{The required locus is an ellipsoid of revolution.}
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{Show that the three points }A(2,3,4),B(-1,2,-3)\text{ and}
\displaystyle C(-4,1,-10)\text{ are collinear and find the ratio in which }C\text{ divides }AB.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(-1-2)^2+(2-3)^2+(-3-4)^2}.
\displaystyle =\sqrt{9+1+49}=\sqrt{59}.
\displaystyle BC=\sqrt{(-4+1)^2+(1-2)^2+(-10+3)^2}.
\displaystyle =\sqrt{9+1+49}=\sqrt{59}.
\displaystyle AC=\sqrt{(-4-2)^2+(1-3)^2+(-10-4)^2}.
\displaystyle =\sqrt{36+4+196}=\sqrt{236}=2\sqrt{59}.
\displaystyle \therefore AB+BC=\sqrt{59}+\sqrt{59}=2\sqrt{59}=AC.
\displaystyle \therefore A,B,C\text{ are collinear, with }B\text{ lying between }A\text{ and }C.
\displaystyle \text{Hence }C\text{ divides the line }AB\text{ externally.}
\displaystyle AC:CB=2\sqrt{59}:\sqrt{59}=2:1.
\displaystyle \therefore C\text{ divides }AB\text{ externally in the ratio }2:1.
\displaystyle \\

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