Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

For the orthogonal section $${C_1}P$$  and $${C_2}P$$  are pendicular where $${C_1}$$ and $${C_2}$$ are centres of sphere of radii $$4$$ and $$3$$ respectively.
Now $${C_1}P = 4$$   and $${C_2}P = 3,$$   so $$\tan \,\theta = \frac{3}{4}$$
$$\therefore $$  Radius of circle of intersection
$$OP = {C_1}P\,\sin \,\theta = 4 \times \frac{3}{5} = \frac{{12}}{5}$$

Equation of PN :-
$$\eqalign{ & \frac{{x + 1}}{1} = \frac{{y - 3}}{{ - 2}} = \frac{{z - 4}}{0} = \lambda \cr & {\text{N}}\left( {\lambda - 1,\, - 2\lambda + 3 - 4} \right) \cr & {\text{It lies on }}x - 2y = 0 \cr & \Rightarrow \lambda - 1 + 4\lambda - 6 = 0 \cr & \Rightarrow \lambda = \frac{7}{5} \cr & {\text{N}}\left( {\frac{2}{5},\,\frac{1}{5},\,4} \right) \cr} $$
N is mid point of PP’
$$\eqalign{ & \therefore \alpha - 1 = \frac{4}{5},\,\,\beta + 3 = \frac{2}{5},\,\,r + 4 = 8 \cr & \Rightarrow \alpha = \frac{9}{5},\,\,\beta = \frac{{ - 13}}{5},\,\,r = 4 \cr} $$
$$\therefore $$ Image is $$\left( {\frac{9}{5},\,\frac{{ - 13}}{5},\,4} \right)$$

Direction cosines of the normal to the plane $$3x + y + z = 5$$     are $$3,\,1,\,1$$
Direction cosines of the normal to the plane $$x - 2y - z = 6$$    are $$1,\,- 2,\, - 1$$
Sum of the product of direction cosines $$ = 3 \times 1 + 1 \times \left( { - 2} \right) + 1 \times \left( { - 1} \right) = 0$$
Hence, normls to the two planes are perpendicular to each other. Therefore two planes are also perpendicular.

Let the required plane passing through the points $$\left( {0,\, - 1,\,0} \right)$$   and $$\left( {0,\,0,\,1} \right)$$   be $$\frac{x}{\lambda } + \frac{y}{{ - 1}} + \frac{z}{1} = 1$$    and the given plane is $$y-z+5=0$$
$$\eqalign{ & \therefore \,\,\,\cos \frac{\pi }{4} = \frac{{ - 1 - 1}}{{\sqrt {\left( {\frac{1}{{{\lambda ^2}}} + 1 + 1} \right)} \sqrt 2 }} \cr & \Rightarrow {\lambda ^2} = \frac{1}{2} \cr & \Rightarrow \frac{1}{\lambda } = \pm \sqrt 2 \cr} $$
Then, the equation of plane is $$ \pm \sqrt 2 x - y + z = 1$$
Then the point $$\left( {\sqrt 2 ,\,1,\,4} \right)$$   satisfies the equation of plane $$ - \sqrt 2 x - y + z = 1$$

Given that $$P\left( {a,\,b,\,c} \right);\,Q\left( {a + 2,\,b + 2,\,c - 2} \right)$$       and $$R\left( {a + 6,\,b + 6,\,c - 6} \right)$$     are collinear, one point must divide, the other two points externally or internally.
Let $$R$$ divide $$P$$ and $$Q$$ in ratio $$k : 1$$
So, taking on $$x$$-co-ordinates
$$\eqalign{ & \frac{{k\left( {a + 2} \right) + a}}{{k + 1}} = a + 6 \cr & \Rightarrow k\left( {a + 2} \right) + a = \left( {k + 1} \right)\left( {a + 6} \right) \cr & \Rightarrow ka + 2k + a = ka + 6k + a + 6 \cr & \Rightarrow - 4k = 6{\text{ or }}k = - \frac{3}{2} \cr} $$
Negative sign shows that this is external division in ratio $$3 : 2.$$
So, $$R$$ is divided $$P$$ and $$Q$$ externally in $$3 : 2$$  ratio.
Putting this value for $$y$$ - and $$z$$ - coordinates satisfied :
for $$y$$ - co-rdinate :
$$\frac{{3\left( {b + 2} \right) - 2b}}{{3 - 2}} = \frac{{3b + 6 - 2b}}{1} = b + 6$$
and for $$z$$-co-rdinate :
$$\frac{{3\left( {c - 2} \right) - 2c}}{{3 - 2}} = \frac{{3c - 6 - 2c}}{1} = c - b$$
Statement $$\left( 2 \right)$$ is correct.
Also, let $$Q$$ divide $$P$$ and $$R$$ in ratio $$P : 1$$
Taking an $$x$$-co-ordinate :
$$\eqalign{ & \frac{{p\left( {a + 6} \right) + a}}{{p + 1}} = a + 2 \cr & \Rightarrow \frac{{p.a + 6p + a}}{{p + 1}} = a + 2 \cr & \Rightarrow pa + 6p + a = pa + a + 2p + 2 \cr & \Rightarrow 4p = 2 \cr & \Rightarrow p = \frac{1}{2} \cr} $$
Positive sign shows but the division is internal and in the ratio $$1 : 2$$
Verifying for $$y$$-and $$z$$-co-ordinates, satisfies this results.
For $$y$$- co-ordinate,
$$\frac{{\left( {b + 6} \right) \times 1 + 2b}}{3} = \frac{{3b + 6}}{3} = b + 2$$
and for $$z$$-co-ordinate,
$$\frac{{c - 6 + 2c}}{3} = \frac{{3c - 6}}{3} = c - 2$$
values are satisfied. So, statement $$\left( 3 \right)$$ is correct.

Given that $$P\left( {3,\,2,\,6} \right)$$   is a point in space and $$Q$$ is a point on line
$$\eqalign{ & \vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right) \cr & {\text{or, }}\frac{{x - 1}}{{ - 3}} = \frac{{y + 1}}{1} = \frac{{z - 2}}{5} = \mu \cr} $$
Let coordinates of $$Q$$ be $$\left( { - 3\mu + 1,\,\mu - 1,\,5\mu + 2} \right)$$
$$\therefore\, \,d.r'{\text{s }}\,{\text{of }}\overrightarrow {PQ} = - 3\mu - 2,\,\mu - 3,\,5\mu - 4$$
As $$\overrightarrow {PQ} $$  is parallel to the plane $$x-4y+3z=1$$
$$\eqalign{ & \therefore 1.\left( { - 3\mu - 2} \right) - 4.\left( {\mu - 3} \right) + 3.\left( {5\mu - 4} \right) = 0 \cr & \Rightarrow 8\mu = 2{\text{ or }}\mu = \frac{1}{4} \cr} $$

$$\eqalign{ & \frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4} = \lambda \cr & \Rightarrow x = 2\lambda + 1,\,\,y = 3\lambda - 1{\text{ and }}\,z = 4\lambda + 1 \cr & \frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1} = \mu \cr & \Rightarrow x = 3 + \mu ,\,\,y = k + 2\mu \,\,{\text{and }}z = \mu \cr} $$
Since above lines intersect
$$\eqalign{ & \Rightarrow 2\lambda + 1 = 3 + \mu .....(1) \cr & \,\,\,\,\,\,\,\,3\lambda - 1 = 2\mu + k.....(2) \cr & \,\,\,\,\,\,\,\,\mu = 4\lambda + 1.....(3) \cr} $$
Solving (1) and (3) and putting the value of $$\lambda $$ and $$\mu $$ in (2) we get, $$k = \frac{9}{2}$$

$$\eqalign{ & {\text{Let }}A = \left( {1,\,1,\,1} \right); \cr & B = \left( { - 2,\,4,\,1} \right); \cr & C = \left( { - 1,\,5,\,5} \right)\,\& \,D = \left( {2,\,2,\,5} \right) \cr & {\text{Then, }}AB = \sqrt {9 + 9 + 0} = 3\sqrt 2 , \cr & BC = \sqrt {1 + 1 + 16} = 3\sqrt 2 , \cr & CD = 3\sqrt 2 {\text{ and }}AD = 3\sqrt 2 \cr & {\text{Since, all sides are equal}}{\text{. Hence it is a square}}{\text{.}} \cr} $$

Any plane through $$\left( {1,\, - 2,\,3} \right)$$   is
$$A\left( {x - 1} \right) + B\left( {y + 2} \right) + C\left( {z - 3} \right) = 0......\left( 1 \right)$$
The point $$\left( { - 1,\,2,\, - 1} \right)$$   lies in this plane if $$ - 2A + 4B - 4C = 0$$
i.e. if $$A - 2B + 2C = 0......\left( 2 \right)$$
The plane $$\left( 1 \right)$$ is parallel to the given line with direction ratio, $$2,\,3,\,4$$
if $$2A + 3B + 4C = 0......\left( 3 \right)$$
From $$\left( 2 \right)$$ and $$\left( 3 \right)$$, we have
$$\eqalign{ & \frac{A}{{ - 8 - 6}} = \frac{B}{{4 - 4}} = \frac{C}{{3 + 4}} \cr & \Rightarrow \frac{A}{{ - 14}} = \frac{B}{0} = \frac{C}{7} \cr & \Rightarrow A:B:C = 2:0: - 1 \cr} $$

Let the direction cosines of line $$L$$ be $$l,\,m,\,n$$   then
$$2l + 3m + n = 0.....({\text{i}})$$
and $$l + 3m + 2n = 0.....({\text{ii}})$$
on solving equation (i) and (ii), we get
$$\frac{l}{{6 - 3}} = \frac{m}{{1 - 4}} = \frac{n}{{6 - 3}}\,\,\, \Rightarrow \frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3}$$
$$\eqalign{ & {\text{Now }}\frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} = \frac{{\sqrt {{l^2} + {m^2} + {n^2}} }}{{\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }} \cr & \because \,{l^2} + {m^2} + {n^2} = 1 \cr & \therefore \,\frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} = \frac{1}{{\sqrt {27} }} \cr & \Rightarrow l = \frac{3}{{\sqrt {27} }} = \frac{1}{{\sqrt 3 }},\,\,m = - \frac{1}{{\sqrt 3 }},\,\,n = \frac{1}{{\sqrt 3 }} \cr} $$
Line $$L,$$  makes an angle $$\alpha $$ with $$+ve\, x$$ -axis
$$\therefore l = \cos \,\alpha \,\,\,\, \Rightarrow \cos \,\alpha = \frac{1}{{\sqrt 3 }}$$