Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

Let $$Q$$ be the image of the point $$P\left( {2,\,3,\,4} \right)$$   in the plane $$x - 2y + 5z = 6,$$    then $$PQ$$  is normal to the plane.
$$\therefore $$  direction ratios of $$PQ$$  are $$\left\langle {1,\, - 2,\,5} \right\rangle $$
Since $$PQ$$  passes through $$P\left( {2,\,3,\,4} \right)$$   and has direction ratios $${1,\, - 2,\,5}$$
$$\therefore $$  Equation of $$PQ$$  is $$\frac{{x - 2}}{1} = \frac{{y - 3}}{{ - 2}} = \frac{{z - 4}}{5}$$

$$D$$ divides $$BC$$  in the ratio $$AB : AC$$   i.e. $$3 : 13.$$
Therefore, coordinates of $$D$$ are,
$$\eqalign{ & \left( {\frac{{3 \times - 9 + 13 \times 5}}{{3 + 13}},\,\frac{{3 \times 6 + 13 \times 5}}{{3 + 13}},\,\frac{{3 \times - 3 + 13 \times 2}}{{3 + 13}}} \right) \cr & {\text{or }}\left( {\frac{{19}}{8},\,\frac{{57}}{{16}},\,\frac{{17}}{{16}}} \right) \cr} $$

$$\eqalign{ & 2{x^2} - 2{y^2} + 4{z^2} + 6xz + 2yz + 3xy = 0 \cr & {\text{or, }}2{x^2} + x\left( {6z + 3y} \right) - 2{y^2} + 4{z^2} + 2yz = 0 \cr & \Rightarrow x = \frac{{ - \left( {6z + 3y} \right) \pm \sqrt {36{z^2} + 9{y^2} + 36yz - 8\left( { - 2{y^2} + 4{z^2} + 2yz} \right)} }}{4} \cr & \Rightarrow x = \frac{{ - \left( {6z + 3y} \right) \pm \sqrt {{{\left( {2z + 5y} \right)}^2}} }}{4} \cr & \Rightarrow x = \frac{{ - \left( {6z + 3y} \right) \pm \left( {2z + 5y} \right)}}{4} \cr & {\text{or, }}2x - y + 2z = 0,\,x + 2y + 2z = 0 \cr & \therefore {\text{ Angle between planes is }}{\cos ^{ - 1}}\left( {\frac{4}{9}} \right). \cr} $$

Suppose $$R$$ divides $$PQ$$ in the ration $$\lambda :1.$$
Then, the coordinates of $$R$$ are
$$\left( {\frac{{5\lambda + 3}}{{\lambda + 1}},\,\frac{{4\lambda + 2}}{{\lambda + 1}},\,\frac{{ - 6\lambda - 4}}{{\lambda + 1}}} \right)$$
But, the coordinates of $$R$$ are given as $$\left( {9,\,8,\, - 10} \right)$$
$$\eqalign{ & \therefore \,\frac{{5\lambda + 3}}{{\lambda + 1}} = 9,\,\frac{{4\lambda + 2}}{{\lambda + 1}} = 8{\text{ and }} \cr & \frac{{ - 6\lambda - 4}}{{\lambda + 1}} = - 10 \Rightarrow \lambda = - \frac{3}{2} \cr} $$
Hence, $$R$$ divides $$PQ$$  externally in the ratio $$3 : 2$$

If the line is parallel to the plane then $$a\ell + bm + cn = 0.$$

We know that $$z$$-co-ordinate of every point on $$xy$$ -plane is zero. So, let $$P\left( {x,\,y,\,0} \right)$$   be a point in $$xy$$ -plane such that
$$\eqalign{ & PA = PB = PC \cr & {\text{Now}},\,PA = PB \Rightarrow P{A^2} = P{B^2} \cr & \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {0 - 3} \right)^2} = {\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {0 - 2} \right)^2} \cr & \Rightarrow 4x - 6y = 0 \cr & \Rightarrow 2x - 3y = 0......\left( 1 \right) \cr & PB = PC \Rightarrow P{B^2} = P{C^2} \cr & \Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {0 - 2} \right)^2} = {\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {0 - 1} \right)^2} \cr & \Rightarrow - 6y + 12 = 0 \cr & \Rightarrow y = 2......\left( 2 \right) \cr & {\text{Putting }}y = 2{\text{ in }}\left( 1 \right){\text{, we obtain }}x = 3 \cr & {\text{Hence, the required point is }}\left( {3,\,2,\,0} \right) \cr} $$

Let a point on the line $$x=y+a=z$$    is $$\left( {\lambda ,\,\lambda - a,\,\lambda } \right)$$   and a point on the line $$x+a=2y=2z$$    is $$\left( {\mu - a,\,\frac{\mu }{2},\,\frac{\mu }{2}} \right),$$    then direction ratio of the line joining these points are $$\lambda - \mu + a,\,\lambda - a - \frac{\mu }{2},\,\lambda - \frac{\mu }{2}$$
If it respresents the required line, then
$$\frac{{\lambda - \mu + a}}{2} = \frac{{\lambda - a - \frac{\mu }{2}}}{1} = \frac{{\lambda - \frac{\mu }{2}}}{2}$$
on solving we get $$\lambda = 3a,\,\,\mu = 2a$$
$$\therefore $$ The required points of intersection are
$$\left( {3a,\,3a - a,\,3a} \right)$$    and $$\left( {2a - a,\,\frac{{2a}}{2},\,\frac{{2a}}{2}} \right)$$
or $$\left( {3a,\,2a,\,3a} \right)$$    and $$\left( {a,\,a,\,a} \right)$$

Let the point $$R$$ divides the line joining the points $$P\left( {2,\,4,\,5} \right)$$   and $$Q\left( {3,\,5,\, - 4} \right)$$   in the ratio $$m : n.$$

$$\therefore $$  The coordinate of $$R$$ is
$$\left( {\frac{{3m - 2n}}{{m + n}},\,\frac{{5m - 4n}}{{m + n}},\,\frac{{ - 4m + 5n}}{{m + n}}} \right)$$
Since, the point $$R$$ is on $$yz$$ -plane, therefore $$x$$-co-ordinate will be zero.
$$\eqalign{ & \therefore \,\frac{{3m - 2n}}{{m + n}} = 0 \cr & \Rightarrow 3m - 2n = 0 \cr & \Rightarrow 3m = 2n \cr & \Rightarrow \frac{m}{n} = \frac{2}{3} \cr} $$

$$\eqalign{ & \frac{{a - 1}}{2} = \frac{{b - 3}}{{ - 1}} = \frac{{c - 4}}{1} = \lambda \,\,\left( {{\text{let}}} \right) \cr & \Rightarrow a = 2\lambda + 1 \cr & b = 3 - \lambda \cr & c = 4 + \lambda \cr & P = \left( {\frac{{a + 1}}{2},\,\frac{{b + 3}}{2},\,\frac{{c + 4}}{2}} \right) \cr & = \left( {\lambda + 1,\,\frac{{6 - \lambda }}{2},\,\frac{{\lambda + 8}}{2}} \right) \cr & \therefore 2\left( {\lambda + 1} \right) - \frac{{6 - \lambda }}{2} + \frac{{\lambda + 8}}{2} + 3 = 0 \cr & \Rightarrow 3\lambda + 6 = 0\,\,\,\,\,\, \Rightarrow \lambda = - 2 \cr & a = - 3,\,\,\,b = 5,\,\,\,c = 2 \cr} $$
Required line is $$\frac{{x + 3}}{3} = \frac{{y - 5}}{1} = \frac{{z - 2}}{{ - 5}}$$

If the centre $$'P'$$ is with position vector $$\overrightarrow r ,$$
Then $$\overrightarrow a - \overrightarrow r = \overrightarrow {PA} ,\,\overrightarrow b - \overrightarrow r = \overrightarrow {PB} ,\,\overrightarrow c - \overrightarrow r = \overrightarrow {PC} ,$$
Where $$\left| {\overrightarrow {PA} } \right| = \left| {\overrightarrow {PB} } \right| = \left| {\overrightarrow {PC} } \right| = \left| {\overrightarrow {OP} } \right| = \left| {\overrightarrow r } \right|$$

$$\eqalign{ & {\text{Consider }}\left| {\overrightarrow a - \overrightarrow r } \right| = \left| {\overrightarrow r } \right| \cr & \Rightarrow \left( {\overrightarrow a - \overrightarrow r } \right).\left( {\overrightarrow a - \overrightarrow r } \right) = \overrightarrow r .\overrightarrow r \cr & \Rightarrow {a^2} - 2\overrightarrow a .\overrightarrow r + {r^2} = {r^2} \cr & \Rightarrow {a^2} = 2\overrightarrow a .\overrightarrow r \cr & {\text{Similarly, }}{b^2} = 2\overrightarrow b .\overrightarrow r {\text{ and }}{c^2} = 2\overrightarrow c .\overrightarrow r \cr & {\text{Since, }}\left( {\overrightarrow b \times \overrightarrow c } \right),\,\left( {\overrightarrow c \times \overrightarrow a } \right){\text{ and }}\left( {\overrightarrow a \times \overrightarrow b } \right){\text{ are non - coplanar,}} \cr & {\text{then }}\overrightarrow r = x\left( {\overrightarrow b \times \overrightarrow c } \right) + y\left( {\overrightarrow c \times \overrightarrow a } \right) + z\left( {\overrightarrow a \times \overrightarrow b } \right) \cr & \Rightarrow \overrightarrow a .\overrightarrow r = x\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right) + y.0 + z.0 \cr & \Rightarrow \overrightarrow a .\overrightarrow r = x\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] \cr & \Rightarrow x = \frac{{\overrightarrow a .\overrightarrow r }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr & \Rightarrow x = \frac{{{a^2}}}{{2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr & {\text{Similarly, }}y = \frac{{{b^2}}}{{2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{\text{ and }}z = \frac{{{c^2}}}{{2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr & {\text{Therefore, }}\overrightarrow r = \frac{{{a^2}\left( {\overrightarrow b \times \overrightarrow c } \right) + {b^2}\left( {\overrightarrow c \times \overrightarrow a } \right) + {c^2}\left( {\overrightarrow a \times \overrightarrow b } \right)}}{{2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr} $$