Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

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 $$ - 2\lambda + 5 = 0 \Rightarrow \lambda = \frac{5}{2}$$
$$\eqalign{ & \therefore \,\left( {0,\,\left( {b - 1} \right)\frac{5}{2} + 1,\,\left( {1 - a} \right)\frac{5}{2} + a} \right) = \left( {0,\,\frac{{17}}{2},\,\frac{{ - 13}}{2}} \right) \cr & \Rightarrow \left( {b - 1} \right)\frac{5}{2} + 1 = \frac{{17}}{2}{\text{ and }}\left( {1 - a} \right)\frac{5}{2} + a = - \frac{{13}}{2} \cr & \Rightarrow b = 4{\text{ and }}a = 6 \cr} $$

The equation of the sphere is
$$\left| {\overrightarrow r - 2\overrightarrow i - \overrightarrow j + 6\overrightarrow k } \right| = \sqrt {18} $$
$$ \Rightarrow $$  Its centre is at the point $$\left( {\overrightarrow r - 2\overrightarrow i - \overrightarrow j + 6\overrightarrow k } \right),$$      i.e., at $$\left( {2,\,1,\, - 6} \right).$$
Coordinates of $$A$$ are $$\left( {3,\,2,\, - 2} \right).$$
Let the coordinates of $$B$$ be $$\left( {\alpha ,\,\beta ,\,\gamma } \right).$$
$$\eqalign{ & {\text{Then, }}\frac{{3 + \alpha }}{2} = 2,\,\frac{{2 + \beta }}{2} = 1{\text{ and }}\frac{{ - 2 + \lambda }}{2} = - 6 \cr & \Rightarrow \alpha = 1,\,\beta = 0,\,\gamma = - 10 \cr} $$
Therefore, coordinates of $$B$$ are $$\left( {1,\,0,\, - 10} \right).$$

Let $$A\left( {a,\,b,\,c} \right)$$   be the fixed point on the variable plane

Now D.R ‘s of $$OM$$  are $$x - 0,\,y - 0,\,z - 0{\text{ i}}{\text{.e}}{\text{., }}x,\,y,\,z$$
D.R.’s of $$MA$$  are $$x - a,\,y - b,\,z - c$$
Since $$OM$$  perpendicular $$MA$$
$$\eqalign{ & x\left( {x - a} \right) + y\left( {y - b} \right) + z\left( {z - c} \right) = 0 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - ax - by - cz = 0 \cr} $$

$$\eqalign{ & {\text{Any plane through the line is}} \cr & \left( {x + 2y - z + 1} \right) + \lambda \left( {3x - y - 4z + 3} \right) = 0 \cr & {\text{It passes through }}\left( {1,\,1,\,1} \right). \cr & \therefore \,\left( {1 + 2 - 1 + 1} \right) + \lambda \left( {3 - 1 - 4 + 3} \right) = 0 \cr & \Rightarrow \lambda = - 3 \cr & {\text{Therefore, the required plane is}} \cr & 8x - 5y - 11z + 8 = 0 \cr} $$

Shortest distance $$=$$ perpendicular distance between the plane and sphere $$=$$ distance of plane from centre of sphere $$-$$ radius
$$\eqalign{ & = \left| {\frac{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327}}{{\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} \cr & = 26 - 13 \cr & = 13 \cr} $$

Let the equation of variable plane be $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$
Which meets the axes at $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$     and $$C\left( {0,\,0,\,c} \right)$$
The centroid of $$\Delta ABC$$   is $$\left( {\frac{a}{3},\,\frac{b}{3},\,\frac{c}{3}} \right)$$   and it satisfies the relation $$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k.$$
Thus,
$$\eqalign{ & \frac{9}{{{a^2}}} + \frac{9}{{{b^2}}} + \frac{9}{{{c^2}}} = k \cr & {\text{or, }}\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{k}{9}......\left( {\text{i}} \right) \cr} $$
Also it is given that the distance of the plane $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$   from $$\left( {0,\,0,\,0} \right)$$   is $$1$$ unit. Therefore,
$$\eqalign{ & \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }} = 1 \cr & {\text{or, }}\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 1......\left( {{\text{ii}}} \right) \cr} $$
From $$\left( {\text{i}} \right)$$ and $$\left( {\text{ii}} \right)$$, we get $$\frac{k}{9} = 1,\,{\text{i}}{\text{.e}}{\text{., }}k = 9$$

Let point on line be $$P\left( {2k + 1,\,3k - 1,\,4k + 2} \right)$$
Since, point $$P$$ lies on the plane $$x+2y+3z=15$$
$$2k+1+6k-2+12k+6=15$$
$$\eqalign{ & \Rightarrow k = \frac{1}{2} \cr & \therefore P \equiv \left( {2,\,\frac{1}{2},\,4} \right) \cr} $$
Then the distance of the point $$P$$ from the origin is
$$OP = \sqrt {4 + \frac{1}{4} + 16} = \frac{9}{2}$$

Both the lines pass through origin. Line $${L_1}$$ is parallel to the vector
$$\overrightarrow {{V_1}} = \left( {\cos \,\theta + \sqrt 3 } \right)\hat i + \left( {\sqrt 2 \,\sin \,\theta } \right)\hat j + \left( {\cos \,\theta - \sqrt 3 } \right)\hat k$$
and $${L_2}$$ is parallel to the vector
$$\eqalign{ & \overrightarrow {{V_2}} = a\hat i + b\hat j + c\hat k \cr & \therefore \,\cos \,\alpha = \frac{{\overrightarrow {{V_1}} .\overrightarrow {{V_2}} }}{{\left| {\overrightarrow {{V_1}} } \right|\left| {\overrightarrow {{V_2}} } \right|}} \cr & = \frac{{a\left( {\cos \,\theta + \sqrt 3 } \right) + \left( {b\sqrt 2 } \right)\sin \,\theta + c\left( {\cos \,\theta - \sqrt 3 } \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {\cos \,\theta + \sqrt 3 } \right)}^2} + 2\,{{\sin }^2}\theta + {{\left( {\cos \,\theta - \sqrt 3 } \right)}^2}} }} \cr & = \frac{{\left( {a + c} \right)\cos \,\theta + \left( {b\sqrt 3 } \right) + \sin \,\theta + \left( {a - c - \sqrt 3 } \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {2 + 6} }} \cr} $$
In order that $$\cos \,\alpha $$   is independent of $$\theta ,$$  we get $$a + c = 0$$   and $$b = 0$$
$$\therefore \,\cos \,\alpha = \frac{{2a\sqrt 3 }}{{a\sqrt 2 \,2\sqrt 2 }} = \frac{{\sqrt 3 }}{2}\, \Rightarrow \alpha = \frac{\pi }{6}$$

For $$n = - \left( {l + m} \right),$$    the second relation gives
$$\eqalign{ & a{l^2} + b{m^2} + c{\left( {l + m} \right)^2} = 0 \cr & {\text{or }}\left( {a + c} \right){l^2} + 2clm + \left( {b + c} \right){m^2} = 0 \cr} $$
For parallel lines, the two roots must be equal
$$\eqalign{ & \Rightarrow 4{c^2} - 4\left( {b + c} \right)\left( {a + c} \right) = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr} $$

Let $$Q\left( {\overrightarrow q } \right)$$  be the foot of altitude drawn from $$P\left( {\overrightarrow p } \right)$$  to the line $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b ,$$
$$\eqalign{ & \Rightarrow \left( {\overrightarrow q - \overrightarrow p } \right).\overrightarrow b = 0{\text{ and }}\overrightarrow q = \overrightarrow a + \lambda \overrightarrow b \cr & \Rightarrow \left( {\overrightarrow a + \lambda \overrightarrow b - \overrightarrow p } \right).\overrightarrow b = 0 \cr & {\text{or }}\left| {\left( {\overrightarrow a - \overrightarrow p } \right)} \right|.\overrightarrow b + \lambda {\left| {\overrightarrow b } \right|^2} = 0\,\,{\text{or }}\lambda = \frac{{\left( {\overrightarrow p - \overrightarrow a } \right).\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr & \Rightarrow \overrightarrow q - \overrightarrow p = \overrightarrow a + \frac{{\left( {\left( {\overrightarrow p - \overrightarrow a } \right).\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} - \overrightarrow p \cr & \Rightarrow \left| {\overrightarrow q - \overrightarrow p } \right| = \left| {\left( {\overrightarrow a - \overrightarrow p } \right) + \frac{{\left( {\left( {\overrightarrow p - \overrightarrow a } \right).\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}}} \right| \cr} $$