Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

The given lines are $$2x=3y=-z$$
or $$\frac{x}{3} = \frac{y}{2} = \frac{z}{{ - 6}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Dividing by 6}}} \right]$$
and $$6x=-y=-4z$$
or $$\frac{x}{2} = \frac{y}{{ - 12}} = \frac{z}{{ - 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Dividing by 12}}} \right]$$
$$\therefore $$ Angle between two lines is
$$\eqalign{ & \cos \,\theta = \frac{{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)}}{{\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }} \cr & \Rightarrow \cos \,\theta = \frac{{6 - 24 + 18}}{{\sqrt {49} \sqrt {157} }} = 0 \cr & \Rightarrow \theta = {90^ \circ } \cr} $$

The variable point is $$P\left( {x,\,y,\,z} \right)$$
Its distance from the $$y$$-axis $$ = \sqrt {{x^2} + {z^2}} $$
Its distance from $$\left( {2,\,1,\, - 1} \right)$$
$$\eqalign{ & = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 1} \right)}^2} + {{\left( {z + 1} \right)}^2}} \cr & {\text{Given : }}\sqrt {{x^2} + {z^2}} \cr & = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 1} \right)}^2} + {{\left( {z + 1} \right)}^2}} \cr & = {y^2} - 2y - 4x + 2z + 6 = 0 \cr} $$

If $$\left( {{l_1},\,{m_1},\,{n_1}} \right)$$   and $$\left( {{l_2},\,{m_2},\,{n_2}} \right)$$   are the direction ratios then angle between the lines is
$$\eqalign{ & \cos \,q = \frac{{{l_1}{l_2} + {m_1}{m_2} + \,{n_1}{n_2}}}{{\sqrt {l_1^2 + m_1^2 + n_1^2} \,\sqrt {l_2^2 + m_2^2 + n_2^2} }} \cr & {\text{Here, }}{l_1} = 1,\,{m_1} = 1,\,{n_1} = 1\,{\text{and}} \cr & {l_2} = 1,\,{m_2} = - 1,\,{n_2} = n{\text{ and }}q = {60^ \circ } \cr & \therefore \,\cos \,{60^ \circ } = \frac{{1 \times 1 + 1 \times \left( { - 1} \right) + 1 \times n}}{{\sqrt {{1^2} + {1^2} + {1^2}} \times \sqrt {{1^2} + {1^2} + {n^2}} }} \cr & \Rightarrow \frac{1}{2} = \frac{n}{{\sqrt 3 \sqrt {2 + {n^2}} }} \cr & \Rightarrow {n^2} = 6 \cr & \Rightarrow n = \pm \sqrt 6 \cr} $$

The point $$A\left( {6,\,7,\,7} \right)$$   is on the line . Let the perpendicular from $$P$$ meet the line in $$L$$. Then
$$A{P^2} = {\left( {6 - 1} \right)^2} + {\left( {7 - 2} \right)^2} + {\left( {7 - 3} \right)^2} = 66$$

$$\eqalign{ & {\text{Also }}AL = {\text{ projection of }}AP\,{\text{on line}} \cr & \left( {{\text{actual d}}{\text{.c}}{\text{.'s }}\frac{3}{{\sqrt {17} }},\,\frac{2}{{\sqrt {17} }},\,\frac{{ - 2}}{{\sqrt {17} }}} \right) \cr & \Rightarrow \left( {6 - 1} \right).\frac{3}{{\sqrt {17} }} + \left( {7 - 2} \right).\frac{2}{{\sqrt {17} }} + \left( {7 - 3} \right).\frac{{ - 2}}{{\sqrt {17} }} = \sqrt {17} \cr & \therefore \, \bot \,{\text{distance }}d{\text{ of }}P{\text{ from the line is given by}} \cr & {d^2} = A{P^2} - A{L^2} \cr & \Rightarrow {d^2} = 66 - 17 \cr & \Rightarrow {d^2} = 49 \cr & {\text{So, that }}d = 7 \cr} $$

We know that sum of square of direction cosines $$ = 1$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{., }}{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \cr & \Rightarrow {\cos ^2}{45^ \circ } + {\cos ^2}\beta + {\cos ^2}\beta = 1 \cr & \left( {{\text{As given }}\alpha = {{45}^ \circ }{\text{ and }}\beta = \gamma } \right) \cr & \Rightarrow \frac{1}{2} + 2{\cos ^2}\beta = 1 \cr & \Rightarrow {\cos ^2}\beta = \frac{1}{4} \cr & \Rightarrow \cos \,\beta = \pm \frac{1}{2}, \cr} $$
Negative value is discarded,
Since the line makes angle with positive axes.
Hence, $$\cos \,\beta = \frac{1}{2} \Rightarrow \cos \,\beta = \cos \,{60^ \circ } \Rightarrow \beta = {60^ \circ }$$
$$\therefore $$  Required sum $$ = \alpha + \beta + \gamma = {45^ \circ } + {60^ \circ } + {60^ \circ } = {165^ \circ }$$

Since $$3\left( 1 \right) + 2\left( { - 2} \right) + \left( { - 1} \right)\left( { - 1} \right) = 3 - 4 + 1 = 0$$
$$\therefore $$  given line is $$ \bot $$ to the normal to the plane i.e., given line is parallel to the given plane.
Also $$\left( {1,\, - 1,\,3} \right)$$   lies on the plane $$x - 2y - z = 0$$    if $$1 - 2\left( { - 1} \right) - 3 = 0$$     i.e. $$1 + 2 - 3 = 0$$    which is true.
$$\therefore \,L$$  lies in plane $$\pi .$$

The equation of plane through the point $$\left( {1,\, - 2,\,1} \right)$$   and perpendicular to the planes $$2x-2y+z=0$$    and $$x-y+2z=4$$   is given by
\[\left| \begin{array}{l} x - 1\,\,\,\,\,y + 2\,\,\,\,\,z - 1\\ \,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,2 \end{array} \right| = 0\,\, \Rightarrow x + y + 1 = 0\]
It’s distance from the point (1, 2, 2) is
$$\left| {\frac{{1 + 2 + 1}}{{\sqrt 2 }}} \right| = 2\sqrt 2 .$$

Equation of lines
$$\eqalign{ & \frac{{x - b}}{a} = \frac{y}{1} = \frac{{z - d}}{c} \cr & \frac{{x - b'}}{{a'}} = \frac{y}{1} = \frac{{z - d'}}{{c'}} \cr} $$
Line are perpendicular
$$ \Rightarrow aa' + 1 + cc' = 0$$

Equation of straight line joining $$Q\left( {2,\,3,\,5} \right)$$   and $$R\left( {1,\, - 1,\,4} \right)$$   is
$$\eqalign{ & \frac{{x - 2}}{{ - 1}} = \frac{{y - 3}}{{ - 4}} = \frac{{z - 5}}{1} = \lambda \cr & {\text{Let }}P\left( { - \lambda + 2,\, - 4\lambda + 3,\, - \lambda + 5} \right) \cr} $$
As $$P$$ lies on $$5x-4y-z=1$$
$$\eqalign{ & \therefore - 5\lambda + 10 + 16\lambda - 12 + \lambda - 5 = 1 \cr & \Rightarrow 12\lambda = 8 \cr & \Rightarrow \lambda = \frac{2}{3} \cr & \therefore P = \left( {\frac{4}{3},\,\frac{1}{3},\,\frac{{13}}{3}} \right) \cr} $$
Now let point $$S$$ on $$QR$$  be
$$\left( { - \mu + 2,\, - 4\mu + 3,\, - \mu + 5} \right)$$
$$\because \,S$$  is the foot of perpendicular drawn from $$T\left( {2,\,1,\,4} \right)$$   to $$QR,$$  where dr’s of $$ST$$  are $$\mu ,\,4\mu - 2,\,\mu - 1$$   and dr’s of $$QR$$ are $$-1,\,-4,\,-1$$
$$\eqalign{ & \therefore - \mu - 16\mu + 8 - \mu + 1 = 0 \cr & \Rightarrow 18\mu = 9 \cr & \Rightarrow \mu = \frac{1}{2} \cr & \therefore S = \left( {\frac{3}{2},\,1,\,\frac{9}{2}} \right) \cr} $$
$$\therefore $$ Distance between $$P$$ and $$S$$
$$\eqalign{ & = \sqrt {{{\left( {\frac{4}{3} - \,\frac{3}{2}} \right)}^2} + {{\left( {\frac{1}{3} - 1} \right)}^2} + {{\left( {\frac{{13}}{3} - \frac{9}{2}} \right)}^2}} \cr & = \sqrt {\frac{1}{{36}} + \frac{4}{9} + \frac{1}{{36}}} \cr & = \frac{1}{{\sqrt 2 }} \cr} $$

The required equation of plane is given by
\[\begin{array}{l} \left| \begin{array}{l} x - 1\,\,\,\,\,y - 1\,\,\,\,\,z - 1\\ \,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\, - 2\\ \,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\, - 6\,\,\,\,\,\,\,\, - 2 \end{array} \right| = 0\\ \Rightarrow \left( {x - 1} \right)\left( { - 14} \right) - \left( {y - 1} \right)\left( 2 \right) + \left( {z - 1} \right)\left( { - 15} \right) = 0\\ \Rightarrow 14x - 14 + 2y - 2 + 15z - 15 = 0\\ \Rightarrow 14x + 2y + 15z = 31 \end{array}\]