Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

\[\begin{array}{l} \left| \begin{array}{l} {x_2} - {x_1}\,\,\,\,\,{y_2} - {y_1}\,\,\,\,\,{z_2} - {z_1}\\ \,\,\,\,\,\,\,\,\,{l_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_1}\\ \,\,\,\,\,\,\,\,{l_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{n_2} \end{array} \right| = 0\\ \therefore \left| \begin{array}{l} 1\,\,\,\,\, - 1\,\,\,\,\, - 1\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - k\\ k\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\,\,\,\,\, \Rightarrow \left| \begin{array}{l} \,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\, - 1\\ \,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 + k\,\,\,\,\, - k\\ k + 2\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\\ {k^2} + 3k = 0\\ \Rightarrow k\left( {k + 3} \right) = 0\\ \Rightarrow k = 0\,\,{\rm{ or }} - 3 \end{array}\]

Given lines will be coplanar
\[\begin{array}{l} {\rm{If }}\left| \begin{array}{l} - 1\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,1\,\,\,\,\,1\,\,\,\, - k\\ \,\,\,\,\,\,k\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\\ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0\\ \Rightarrow k = 0,\,\,- 3 \end{array}\]

Equation $${x^2} + {y^2} + {z^2} + 2ux + 2vy + 2wz = 0$$        represents a real sphere if $${u^2} + {v^2} + {w^2} > d$$

Since, coordinates of points $$O$$ and $$P$$ are $$\left( {0,\,0,\,0} \right)$$   and $$\left( {2,\,3,\, - 1} \right)$$   respectively.
Direction ratios of $$OP$$  are $$\left\langle {2,\,3,\, - 1} \right\rangle $$
The plane is perpendicular to $$OP.$$
So, its equation is $$2x + 3y - z + d = 0......\left( {\text{i}} \right)$$
Since, this plane passes through $$\left( {2,\,3,\, - 1} \right);$$
$$\eqalign{ & 2 \times 2 + 3 \times 3 - 1 \times - 1 + d = 0 \cr & \Rightarrow 4 + 9 + 1 + d = 0 \cr & \Rightarrow d = - 14 \cr} $$
On putting the value of $$d$$ in equation $$\left( {\text{i}} \right)$$
$$\eqalign{ & 2x + 3y - z - 14 = 0 \cr & \Rightarrow 2x + 3y - z = 14 \cr} $$
which is required equation of plane.

Suppose $$zx$$ -plane divides the join of $$\left( {1,\,2,\,3} \right)$$   and $$\left( {4,\,2,\,1} \right)$$   in the ratio $$\lambda :1.$$
Then, the co-ordinates of the point of division are
$$\left( {\frac{{4\lambda + 1}}{{\lambda + 1}},\,\frac{{2\lambda + 2}}{{\lambda + 1}},\,\frac{{\lambda + 3}}{{\lambda + 1}}} \right)$$
This point lies on $$zx$$ -plane. $$\therefore \,y$$ -coordinate $$ = 0 \Rightarrow \frac{{2\lambda + 2}}{{\lambda + 1}} = 0 \Rightarrow \lambda = - 1$$
Hence, $$zx$$ -plane divides the join of $$\left( {1,\,2,\,3} \right)$$   and $$\left( {4,\,2,\,1} \right)$$   externally in the ratio $$1 : 1.$$
Alternate solution :
We know that the $$zx$$ -plane divides the segment joining $$P\left( {{x_1},\,{y_1},\,{z_1}} \right)$$   and $$Q\left( {{x_2},\,{y_2},\,{z_2}} \right)$$   in the ratio $$ - {y_1}:{y_2}.$$
$$\therefore \,zx$$ -plane divides the join of $$\left( {1,\,2,\,3} \right)$$   and $$\left( {4,\,2,\,1} \right)$$   in the ratio $$ - 2:2{\text{ i}}{\text{.e}}{\text{., }}1:1$$   externally.

Let the ray of light comes along $$x$$-axis and strikes the mirror at the origin.
Direction cosines of normal are
$$\frac{1}{{\sqrt 3 }},\, - \frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}{\text{ so, }}\cos \frac{\theta }{2} = \frac{1}{{\sqrt 3 }}$$
Let the reflected ray has direction cosines $$l,\,m,\,n$$   then
$$\eqalign{ & \frac{{l + 1}}{{2\,\cos \frac{\theta }{2}}} = \frac{1}{{\sqrt 3 }} \Rightarrow l = \frac{2}{3} - 1 = - \frac{1}{3} \cr & \frac{{m + 0}}{{2\,\cos \frac{\theta }{2}}} = - \frac{1}{{\sqrt 3 }} \Rightarrow m = - \frac{2}{3} \cr & \frac{{n + 0}}{{2\,\cos \frac{\theta }{2}}} = \frac{1}{{\sqrt 3 }} \Rightarrow n = \frac{2}{3} \cr} $$

$${\text{From the figure}}$$

$$\eqalign{ & {x_1} + {x_2} = 2l,\,{y_1} + {y_2} = 0,\,{z_1} + {z_2} = 0 \cr & {x_2} + {x_3} = 0,\,{y_2} + {y_3} = 2m,\,{z_1} + {z_3} = 0 \cr & {\text{and }}{x_1} + {x_3} = 0,\,{y_1} + {y_3} = 0,\,{z_1} + {z_3} = 2n \cr & {\text{On solving, we get}} \cr & {x_1} = l,\,{x_2} = l,\,{x_3} = - l \cr & {y_1} = - m,\,{y_2} = m,\,{y_3} = m \cr & {\text{and }}{z_1} = n,\,{z_2} = - n,\,{z_3} = n \cr & \therefore \,{\text{Coordinates are}}\,: \cr & A\left( {l,\, - m,\,n} \right),\,B\left( {l,\,m,\, - n} \right){\text{and }}C\left( { - l,\,m,\,n} \right) \cr & \therefore \,\frac{{A{B^2} + B{C^2} + C{A^2}}}{{{l^2} + {m^2} + {n^2}}} \cr & = \frac{{\left( {4{m^2} + 4{n^2}} \right) + \left( {4{l^2} + 4{n^2}} \right) + \left( {4{l^2} + 4{m^2}} \right)}}{{{l^2} + {m^2} + {n^2}}} \cr & = 8 \cr} $$

$$\eqalign{ & 2x + y + 2z - 8 = 0.....\left( {{\text{Plane 1}}} \right) \cr & 2x + y + 2z + \frac{5}{2} = 0.....\left( {{\text{Plane 2}}} \right) \cr} $$
Distance between Plane 1 and 2
$$ = \left| {\frac{{ - 8 - \frac{5}{2}}}{{\sqrt {{2^2} + {1^2} + {2^2}} }}} \right| = \left| {\frac{{ - 21}}{6}} \right| = \frac{7}{2}$$

Let the variable point be $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   then according to question
$$\eqalign{ & {\left( {\frac{{\left| {\alpha + \beta + \gamma } \right|}}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{{\left| {\alpha - \gamma } \right|}}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\left| {\alpha - 2\beta + \gamma } \right|}}{{\sqrt 6 }}} \right)^2} = 9 \cr & \Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 9 \cr} $$
So, the locus of the point is $${x^2} + {y^2} + {z^2} = 9$$

Let the line be $$\frac{x}{a} = \frac{y}{b} = \frac{z}{c}......\left( {\text{i}} \right)$$
If line $$\left( {\text{i}} \right)$$ intersects with the line $$\frac{{x - 1}}{2} = \frac{{y + 3}}{4} = \frac{{z - 5}}{3},$$     then
\[\left| \begin{array}{l} a\,\,\,\,\,\,\,\,b\,\,\,\,\,c\\ 2\,\,\,\,\,\,\,\,4\,\,\,\,\,\,3\\ 4\,\,\, - 3\,\,\,\,\,14\, \end{array} \right| = 0 \Rightarrow 9a - 7b - 10c = 0......\left( {{\rm{ii}}} \right)\]
From $$\left( {\text{i}} \right)$$ and $$\left( {\text{i}} \right),$$  we have $$\frac{a}{1} = \frac{b}{{ - 3}} = \frac{c}{5}$$
$$\therefore $$  The line is $$\frac{x}{1} = \frac{y}{{ - 3}} = \frac{z}{5}$$