\displaystyle \textbf{Question 1. }\text{The ratio in which the line joining }(2,4,5)\text{ and }(3,5,-9) \\ \text{ is 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 }yz\text{-plane is }x=0
\displaystyle \text{Let the ratio be }m:n
\displaystyle \frac{3m+2n}{m+n}=0
\displaystyle 3m+2n=0
\displaystyle m:n=-2:3
\displaystyle \therefore \text{Correct option is (c)}
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\displaystyle \textbf{Question 2. }\text{The ratio in which the line joining the points }(a,b,c)\text{ and } \\ (-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 }xy\text{-plane is }z=0
\displaystyle \text{Let the ratio be }m:n
\displaystyle \frac{m(-b)+n(c)}{m+n}=0
\displaystyle -bm+cn=0
\displaystyle bm=cn
\displaystyle m:n=c:b
\displaystyle \therefore \text{Correct option is (d)}
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\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 }\angle POQ=
\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),\quad \overrightarrow{OQ}=(4,-2,1)
\displaystyle \overrightarrow{OP}\cdot\overrightarrow{OQ}=0(4)+1(-2)+2(1)=0
\displaystyle \therefore \overrightarrow{OP}\perp \overrightarrow{OQ}
\displaystyle \therefore \angle POQ=\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (d)}
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\displaystyle \textbf{Question 4. }\text{If the extremities of the diagonal of a square are }(1,-2,3)\text{ and }(2,-3,5), \\ \text{ then the length of the 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 diagonal}=\sqrt{(2-1)^2+(-3+2)^2+(5-3)^2}
\displaystyle =\sqrt{1+1+4}=\sqrt6
\displaystyle \text{For a square, diagonal }= \text{side}\times\sqrt2
\displaystyle \therefore \text{Side}=\frac{\sqrt6}{\sqrt2}=\sqrt3
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 5. }\text{The points }(5,-4,2),(4,-3,1),(7,6,4)\text{ and }(8,-7,5)\text{ are the 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 \textbf{Question. }\text{Determine the geometric shape formed by the vertices.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given points be }A=(5,-4,2),\ B=(4,-3,1),\ C=(7,6,4)\text{ and }D=(8,-7,5).
\displaystyle \text{A fundamental property of a parallelogram is that its diagonals bisect each other.}
\displaystyle \text{Therefore, the midpoints of the diagonals must be identical.}
\displaystyle \text{We check all possible pairings of diagonals.}
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\displaystyle \textbf{1. Check if }ABCD\text{ is a parallelogram} 
\displaystyle \text{Diagonals are }AC\text{ and }BD.
\displaystyle \text{Midpoint of }AC=\left(\frac{5+7}{2},\frac{-4+6}{2},\frac{2+4}{2}\right)=(6,1,3)
\displaystyle \text{Midpoint of }BD=\left(\frac{4+8}{2},\frac{-3-7}{2},\frac{1+5}{2}\right)=(6,-5,3)
\displaystyle \text{Since }(6,1,3)\neq(6,-5,3),\ ABCD\text{ is not a parallelogram.}
\\
\displaystyle \textbf{2. Check if }ABDC\text{ is a parallelogram} 
\displaystyle \text{Diagonals are }AD\text{ and }BC.
\displaystyle \text{Midpoint of }AD=\left(\frac{5+8}{2},\frac{-4-7}{2},\frac{2+5}{2}\right)=(6.5,-5.5,3.5)
\displaystyle \text{Midpoint of }BC=\left(\frac{4+7}{2},\frac{-3+6}{2},\frac{1+4}{2}\right)=(5.5,1.5,2.5)
\displaystyle \text{Since the midpoints do not match, }ABDC\text{ is not a parallelogram.}
\\
\displaystyle \textbf{3. Check if }ADBC\text{ is a parallelogram} 
\displaystyle \text{Diagonals are }AB\text{ and }CD.
\displaystyle \text{Midpoint of }AB=\left(\frac{5+4}{2},\frac{-4-3}{2},\frac{2+1}{2}\right)=(4.5,-3.5,1.5)
\displaystyle \text{Midpoint of }CD=\left(\frac{7+8}{2},\frac{6-7}{2},\frac{4+5}{2}\right)=(7.5,-0.5,4.5)
\displaystyle \text{Since the midpoints do not match, }ADBC\text{ is not a parallelogram.}
\\
\displaystyle \textbf{Conclusion:} 
\displaystyle \text{Since the vertices cannot form a parallelogram in any order, they cannot form a} \\ \text{rectangle or a square either.}
\displaystyle \therefore \text{ The correct option is }(d).
\\

\displaystyle \textbf{Question 6. }\text{In a 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 x=2\quad \text{or}\quad x=3
\displaystyle \text{Each equation represents a plane parallel to the }yz\text{-plane.}
\displaystyle \therefore \text{The equation represents two planes.}
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 7. }\text{Let }(3,4,-1)\text{ and }(-1,2,3)\text{ be the end points of a diameter of a} \\ \text{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{Length of diameter}=\sqrt{(-1-3)^2+(2-4)^2+(3+1)^2}
\displaystyle =\sqrt{16+4+16}=\sqrt{36}=6
\displaystyle \therefore \text{Radius}=\frac{6}{2}=3
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 8. }\text{XOZ-plane divides the join of }(2,3,1)\text{ and }(6,7,1)\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{XOZ-plane is }y=0
\displaystyle \text{Let the ratio be }m:n
\displaystyle \text{Using section formula for }y\text{-coordinate,}
\displaystyle \frac{7m+3n}{m+n}=0
\displaystyle 7m+3n=0
\displaystyle \therefore m:n=-3:7
\displaystyle \therefore \text{Correct option is (c)}
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\displaystyle \textbf{Question 9. }\text{What is the locus of a point for which }y=0,\ z=0?
\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 y=0,\quad z=0
\displaystyle \text{Only }x\text{ varies freely.}
\displaystyle \therefore \text{The locus is the }x\text{-axis}
\displaystyle \therefore \text{Correct option is (a)}
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\displaystyle \textbf{Question 10. }\text{The coordinates of the foot of the perpendicular drawn from} \\ \text{the point }P(3,4,5)\text{ on}
\displaystyle \text{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 }yz\text{-plane is }x=0
\displaystyle \text{Hence foot of perpendicular is obtained by putting }x=0
\displaystyle \therefore \text{Foot of perpendicular}=(0,4,5)
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 11. }\text{The coordinates of the foot of the perpendicular from a point }P(6,7,8) \\ \text{ on }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{Any point on }x\text{-axis has coordinates }(x,0,0)
\displaystyle \text{Hence foot of perpendicular from }(6,7,8)\text{ is }(6,0,0)
\displaystyle \therefore \text{Correct option is (a)}
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\displaystyle \textbf{Question 12. }\text{The perpendicular distance of the point }P(6,7,8)\text{ from }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{Distance from }xy\text{-plane is }|z|
\displaystyle \therefore \text{Distance}=|8|=8
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 13. }\text{The length of the perpendicular drawn from the point }P(3,4,5)\text{ on }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{Distance from }y\text{-axis}=\sqrt{x^2+z^2}
\displaystyle =\sqrt{3^2+5^2}
\displaystyle =\sqrt{9+25}
\displaystyle =\sqrt{34}
\displaystyle \therefore \text{Correct option is (b)}
\\

\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{Distance from }x\text{-axis}=\sqrt{y^2+z^2}
\displaystyle =\sqrt{3^2+4^2}
\displaystyle =\sqrt{9+16}
\displaystyle =5
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 15. }\text{The length of the perpendicular drawn from the point }P(a,b,c) \\ \text{ from }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{Distance from }z\text{-axis}=\sqrt{x^2+y^2}
\displaystyle \therefore \text{Distance from }(a,b,c)\text{ to }z\text{-axis is }\sqrt{a^2+b^2}
\displaystyle \therefore \text{Correct option is (a)}
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