Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

$$A\left( {3,\,1,\,6} \right)\,;\,B\left( {1,\,3,\,4} \right)$$
Mid-point of $$AB = \left( {2,\,2,\,5} \right)$$     lies on the plane.
and d.r’s of $$AB = \left( {2,\, - 2,\,2} \right)$$
d.r’s of normal to plane $$= \left( {1,\, - 1,\,1} \right)$$.
Direction ratio of $$AB$$  and normal to the plane are proportional therefore,
$$AB$$  is perpendicular to the plane
$$\therefore \,A$$  is image of $$B$$
Statement-2 is correct but it is not correct explanation.

Let the point be $$P\left( {x,\,y,\,z} \right)$$   and two points, $$\left( {a,\,0,\,0} \right)$$   and $$\left( { - a,\,0,\,0} \right)$$   be a $$A$$ and $$B$$
As given in the problem,
$$\eqalign{ & P{A^2} + P{B^2} = 2{c^2} \cr & {\text{so, }}{\left( {x + a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {z - 0} \right)^2} + {\left( {x - a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {z - 0} \right)^2} = 2{c^2} \cr & {\text{or, }}{\left( {x + a} \right)^2} + {y^2} + {z^2} + {\left( {x - a} \right)^2} + {y^2} + {z^2} = 2{c^2} \cr & \Rightarrow {x^2} + 2a + {a^2} + {y^2} + {z^2} + {x^2} - 2a + {a^2} + {y^2} + {z^2} = 2{c^2} \cr & \Rightarrow 2\left( {{x^2} + {y^2} + {z^2} + {a^2}} \right) = 2{c^2} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + {a^2} = {c^2} \cr & \Rightarrow {x^2} + {a^2} = {c^2} - {y^2} - {z^2} \cr} $$

Given $$l+m+n=0$$    and an $${l^2} = {m^2} = {n^2}$$
$$\eqalign{ & {\text{Now }}{\left( { - m - n} \right)^2} = {m^2} + {n^2} \cr & \Rightarrow mn = 0 \cr & \Rightarrow m = 0{\text{ or }}n = 0 \cr & {\text{If }}m = 0{\text{ then }}l = - n \cr & {\text{We know }}{l^2} + {m^2} + {n^2} = 1\,\,\,\,\,\, \Rightarrow n = \pm \frac{1}{{\sqrt 2 }} \cr & {\text{i}}{\text{.e}}{\text{., }}\left( {{l_1},\,{m_1},\,{n_1}} \right) = \left( { - \frac{1}{{\sqrt 2 }},\,0,\,\frac{1}{{\sqrt 2 }}} \right) \cr & {\text{If }}n = 0{\text{ then }}l = - m \cr & {l^2} + {m^2} + {n^2} = 1 \cr & \Rightarrow 2{m^2} = 1 \cr & \Rightarrow m = \pm \frac{1}{{\sqrt 2 }} \cr & {\text{Let }}m = \frac{1}{{\sqrt 2 }} \Rightarrow l = - \frac{1}{{\sqrt 2 }}{\text{ and }}n = 0 \cr & \left( {{l_2},\,{m_2},\,{n_2}} \right) = \left( { - \frac{1}{{\sqrt 2 }},\,\frac{1}{{\sqrt 2 }},\,0} \right) \cr & \therefore \,\cos \,\theta = \frac{1}{2}\,\,\,\, \Rightarrow \theta = \frac{\pi }{3} \cr} $$

$$\eqalign{ & AB = 6,\,BC = \sqrt {13} ,\,CA = 3 \cr & \therefore \,AB:AC = 2:1 \cr} $$

Internal bisector of an angle divides the opposite side in the ratio of adjacent sides
$$\therefore \,\frac{{BD}}{{CD}} = \frac{{AB}}{{AC}} = \frac{2}{1}$$
$$\therefore $$  Coordinate of $$D$$ are $$\left( {2,\,\frac{{13}}{3},\,6} \right)$$
$$\therefore $$  Length $$AD = \frac{2}{3}\sqrt {34} $$

Equation of line through $$\left( {5,\,1,\,a} \right)$$   and $$\left( {3,\,b,\,1} \right)$$   is $$\frac{{x - 5}}{{ - 2}} = \frac{{y - 1}}{{b - 1}} = \frac{{z - a}}{{1 - a}} = \lambda $$
$$\therefore $$ Any point on this line is a $$\left[ { - 2\lambda + 5,\,\left( {b - 1} \right)\lambda + 1,\,\left( {1 - a} \right)\lambda + a} \right]$$
It crosses yz plane where $$ - 2\lambda + 5 = 0$$
$$\eqalign{ & \lambda = \frac{5}{2} \cr & \therefore \left( {0,\,\left( {b - 1} \right)\frac{5}{2} + 1,\,\left( {1 - a} \right)\frac{5}{2} + a} \right) = \left( {0,\,\frac{{17}}{2},\,\frac{{ - 13}}{2}} \right) \cr & \Rightarrow \left( {b - 1} \right)\frac{5}{2} + 1 = \frac{{17}}{2}\,{\text{ and }}\left( {1 - a} \right)\frac{5}{2} + a = - \frac{{13}}{2} \cr & \Rightarrow b = 4{\text{ and }}a = 6 \cr} $$

$$P,$$  the image of point (3, 1, 7) in the plane $$x-y+z=3$$   is given by
$$\eqalign{ & \frac{{x - 3}}{1} = \frac{{y - 1}}{{ - 1}} = \frac{{z - 7}}{1} = \frac{{ - 2\left( {3 - 1 + 7 - 3} \right)}}{{{1^2} + {1^2} + {1^2}}} \cr & \Rightarrow \frac{{x - 3}}{1} = \frac{{y - 1}}{{ - 1}} = \frac{{z - 7}}{1} = - 4 \cr & \Rightarrow x = - 1,\,\,y = 5,\,\,z = 3 \cr & \therefore P\left( { - 1,\,5,\,3} \right) \cr} $$
Now equation of plane through $$\left( { - 1,\,5,\,3} \right)$$   and containing the line $$\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$$   is
\[\begin{array}{l} \left| \begin{array}{l} \,\,\,\,\,\,\,x\,\,\,\,\,y\,\,\,\,\,z\\ - 1\,\,\,\,\,5\,\,\,\,\,3\\ \,\,\,\,\,\,\,1\,\,\,\,\,2\,\,\,\,\,1 \end{array} \right| = 0\\ \Rightarrow - x + 4y - 7z = 0\\ {\rm{or }}\,\,\,x - 4y + 7z = 0 \end{array}\]

The direction ratios of the line segment joining points $$A\left( {1,\,0,\,7} \right)$$   and $$B\left( {1,\,6,\,3} \right)$$   are $$0,\, 6,\,-4.$$
The direction ratios of the given line are $$1, \,2, \,3.$$
Clearly $$1 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0$$
So, the given line is perpendicular to line $$AB.$$
Also , the mid point of $$A$$ and $$B$$ is $$\left( {1,\,3,\,5} \right)$$  which lies on the given line.
So, the image of $$B$$ in the given line is $$A,$$  because the given line is the perpendicular bisector of line segment joining points $$A$$ and $$B,$$  But statement-2 is not a correct explanation for statement-1.

Equation of plane passing through the line of intersection of first two planes is :
$$\eqalign{ & \left( {2x - 2y + 3z - 2} \right) + \lambda \left( {x - y + z + 1} \right) = 0 \cr & {\text{or, }}x\left( {\lambda + 2} \right) - y\left( {2 + \lambda } \right) + z\left( {\lambda + 3} \right) + \left( {\lambda - 2} \right) = 0.....({\text{i}}) \cr} $$
is having infinite number of solution with $$x+2y-z-3=0$$     and $$3x-y+2z-1=0,$$     then
\[\left| \begin{array}{l} \left( {\lambda + 2} \right)\,\,\,\,\, - \left( {2 + \lambda } \right)\,\,\,\,\,\left( {\lambda + 3} \right)\\ \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\\ \,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \end{array} \right| = 0\,\,\,\,\, \Rightarrow \lambda = 5\]
Now put $$\lambda = 5$$  in (i), we get
$$7x-7y+8z+3=0$$
Now perpendicular distance from $$\left( {0,\,0,\,0} \right)$$  to the place containing $${L_1}$$ and $${L_2} = \frac{3}{{\sqrt {162} }} = \frac{1}{{3\sqrt 2 }}$$

$$\sin \,\left( {90 - \alpha } \right) = \frac{{\left| {\overrightarrow {AB} \times \overrightarrow {AD} } \right|}}{{\left| {\overrightarrow {AB} } \right|\left| {\overrightarrow {AD} } \right|}}$$
Where, \[\overrightarrow {AB} \times \overrightarrow {AD} = \left| \begin{array}{l} \,\,\,\,\,\,\hat i\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\hat k\\ \,\,\,\,\,\,2\,\,\,\,\,10\,\,\,\,\,\,11\\ - 1\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,2 \end{array} \right| = - 2\hat i - 15\hat j + 14\hat k\]
$$\eqalign{ & \therefore \left| {\overrightarrow {AB} \times \overrightarrow {AD} } \right| = \sqrt {4 + 225 + 196} = \sqrt {425} \cr & \left| {\overrightarrow {AB} } \right| = \sqrt {4 + 100 + 121} = \sqrt {225} = 15 \cr & \left| {\overrightarrow {AD} } \right| = \sqrt {1 + 4 + 4} = \sqrt 9 = 3 \cr & \therefore \sin \,\left( {90 - \alpha } \right) = \frac{{\sqrt {425} }}{{15 \times 3}} = \frac{{\sqrt {17} }}{9} \Rightarrow \cos \,\alpha = \frac{{\sqrt {17} }}{9} \cr} $$

Let $$A\left( {5,\,2,\,4} \right),\,B\left( {6,\, - 1,\,2} \right),\,C\left( {8,\, - 7,\,k} \right)$$       be the given points $$k = - 2.$$