Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

$${\text{Let any point on the second line be}}\left( {\lambda ,\,2\lambda ,\,3\lambda } \right)$$

$$\eqalign{ & \cos \,\theta = \frac{6}{{\sqrt {42} }},\,\sin \,\theta = \frac{{\sqrt 6 }}{{\sqrt {42} }} \cr & {\Delta _{OAB}} = \frac{1}{2}\left( {OA} \right)OB\,\sin \,\theta \cr & = \frac{1}{2}\sqrt 3 \lambda \sqrt {14} \times \frac{{\sqrt 6 }}{{\sqrt {42} }} \cr & = \sqrt 6 {\text{ or }}\lambda = 2 \cr & {\text{So, }}B{\text{ is }}\left( {2,\,4,\,6} \right) \cr} $$

$$\eqalign{ & {\text{Equation of PO :}}\,\,\frac{{x - 1}}{1} = \frac{{y + 5}}{1} = \frac{{z - 9}}{1} = \lambda \cr & \Rightarrow x = \lambda + 1\,;\,\,y = \lambda - 5\,;\,\,z = \lambda + 9 \cr} $$
Putting these in equation of plane :-
$$\eqalign{ & \lambda + 1 - \lambda + 5 + \lambda + 9 = 5 \cr & \Rightarrow \lambda = - 10 \cr & \Rightarrow {\text{O is }}\left( { - 9,\, - 15,\, - 1} \right) \cr & \Rightarrow {\text{distance OP }} = 10\sqrt 3 \cr} $$

If a line makes angle $$\theta ,\,\phi $$  and $$\psi $$ with $$x,\, y,\, z$$   axes respectively, then
$$\eqalign{ & {\cos ^2}\theta + {\cos ^2}\phi + {\cos ^2}\psi = 1 \cr & \Rightarrow {\cos ^2}\theta + {\cos ^2}\phi = 1 - {\cos ^2}\psi \cr & \Rightarrow {\cos ^2}\theta + {\cos ^2}\phi = {\sin ^2}\psi \cr & \therefore {\text{ Statement }}\left( 2 \right){\text{ is correct}}{\text{.}} \cr} $$

Any line passing through the point $$\left( {1,\,1,\,1} \right)$$   is
$$\frac{{x - 1}}{a} = \frac{{y - 1}}{b} = \frac{{z - 1}}{c}......\left( {\text{i}} \right)$$
This line intersects the line
$$\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z - 3}}{4}$$
$${\text{If }}a:b:c \ne 2:3:4$$
\[{\rm{and }}\left| \begin{array}{l} 1 - 1\,\,\,\,\,2 - 1\,\,\,\,\, 3 - 1\\ \,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\ \,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \end{array} \right| = 0\]
$$ \Rightarrow a - 2b + c = 0.....\left( {{\text{ii}}} \right)$$
Again, line $$\left( {\text{i}} \right)$$ intersects line
$$\frac{{x - \left( { - 2} \right)}}{1} = \frac{{y - 3}}{2} = \frac{{z - \left( { - 1} \right)}}{4}$$
$${\text{If }}a:b:c \ne 2:3:4$$
\[{\rm{and }}\left| \begin{array}{l} - 2 - 1\,\,\,\,\,3 - 1\,\,\,\,\, - 1 - 1\\ \,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\ \,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \end{array} \right| = 0\]
$$ \Rightarrow 6a + 5b - 4c = 0.....\left( {{\text{iii}}} \right)$$
From $$\left( {{\text{ii}}} \right)$$ and $$\left( {{\text{iii}}} \right)$$ by cross multiplication, we have
$$\frac{a}{{8 - 5}} = \frac{b}{{6 + 4}} = \frac{c}{{5 + 12}}{\text{ or }}\frac{a}{3} = \frac{b}{{10}} = \frac{c}{{17}}$$
So, the required line is
$$\frac{{x - 1}}{3} = \frac{{y - 1}}{{10}} = \frac{{z - 1}}{{17}}.$$

$$\eqalign{ & {\text{Mid point of }}PQ{\text{ is}} = \left( {\frac{3}{2},\,\frac{7}{2},\,\frac{{ - 9}}{2}} \right) \cr & {\text{DR of the normal is}} \cr & = \left( {4 - \left( { - 1} \right)} \right),\,\left( {5 - 2} \right),\,\left( { - 10 - 1} \right) \cr & = 5,\,3,\, - 11 \cr & \therefore {\text{ Equation of plane is}} \cr & 5\left( {x - \frac{3}{2}} \right) + 3\left( {y - \frac{7}{2}} \right) - 11\left( {z + \frac{9}{2}} \right) = 0 \cr & \Rightarrow 5x + 3y - 11z = \frac{{135}}{2} \cr & \Rightarrow \overrightarrow r .\left( {5\hat i + 3\hat j - 11\hat k} \right) = \frac{{135}}{2} \cr} $$

At the point of intersection of the two given planes, we have
$$\eqalign{ & \overrightarrow b + {\lambda _1}\left( {\overrightarrow b - \overrightarrow a } \right) + {\mu _1}\left( {\overrightarrow a + \overrightarrow c } \right) = \overrightarrow c + {\lambda _2}\left( {\overrightarrow b - \overrightarrow c } \right) + {\mu _2}\left( {\overrightarrow a + \overrightarrow b } \right) \cr & \Rightarrow \left( { - {\lambda _1} + {\mu _1} - {\mu _2}} \right)\overrightarrow a + \left( {1 + {\lambda _1} - {\lambda _2} - {\mu _2}} \right)\overrightarrow b + \left( {{\mu _1} - 1 + {\lambda _2}} \right)\overrightarrow c = \overrightarrow 0 \cr & \Rightarrow - {\lambda _1} + {\mu _1} - {\mu _2} = 0,\,1 + {\lambda _1} - {\lambda _2} - {\mu _2} = 0{\text{ and}}\,{\mu _1} - 1 + {\lambda _2} = 0 \cr} $$
[$$\because \,\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$    are non-coplanar vectors]
From the last two equations, we get
$${\lambda _1} + {\mu _1} - {\mu _2} = 0$$
On solving this equation with $$ - {\lambda _1} + {\mu _1} - {\mu _2} = 0,$$    we get
$$\eqalign{ & {\mu _1} = {\mu _2}{\text{ and }}{\lambda _1} = 0 \cr & \therefore \,{\lambda _1} = 0,\,{\mu _1} = {\mu _2}{\text{ and }}{\lambda _2} = 1 - {\mu _1} \cr} $$
On substituting these values in either of the given equations, we obtain
$$\overrightarrow r = \overrightarrow b + {\mu _1}\left( {\overrightarrow a + \overrightarrow c } \right)$$
As the required line of intersection of the given planes.

The angle made by line $$AB$$  with $$x$$-axis is $${45^ \circ }$$
The angle made by line $$AB$$  with $$y$$-axis is $${120^ \circ }$$
Let the angle made by line $$AB$$  with $$z$$-axis is $$\theta $$
We have
$$\eqalign{ & {\cos ^2}\left( {45} \right) + {\cos ^2}\left( {120} \right) + {\cos ^2}\theta = 1 \cr & \Rightarrow {\cos ^2}\theta = 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \cr} $$
Given that $$\theta $$ is acute
So we get
$$\eqalign{ & \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \theta = {60^ \circ } \cr} $$

The intersection of given plane is
$$\eqalign{ & x - y + 2z - 1 + \lambda \left( {x + y - z - 3} \right) = 3 \cr & \Rightarrow x\left( {1 + \lambda } \right) + y\left( {\lambda - 1} \right) + z\left( {2 - \lambda } \right) - 3\lambda - 1 = 0 \cr} $$
DR’s of normal to the above plane is $$\left( {1 + \lambda ,\,\lambda - 1,\,2 - \lambda } \right)$$
By taking option (A)
$$\eqalign{ & - 1\left( {1 + \lambda } \right) + 3\left( {\lambda - 1} \right) + 2\left( {2 - \lambda } \right) = 0 \cr & \Rightarrow - 1 - \lambda + 3\lambda - 3 + 4 - 2\lambda = 0 \cr & \Rightarrow 0 = 0{\text{ which is true}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Ratio}} = - \left( {\frac{3}{{ - 2}}} \right) = \frac{3}{2} \cr & \therefore \,{\text{Required co - ordinates of the points are,}} \cr & \left[ {\frac{{6 - 6}}{5},\,\frac{{10 + 3}}{5},\,\frac{{ - 14 + 24}}{5}} \right] = \left( {0,\,\frac{{13}}{5},\,2} \right) \cr} $$

Plane $$2ax - 3ay + 4az + 6 = 0$$     passes through the mid point of the centre of spheres $${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$$       and $${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$$       respectively. Center of spheres are $$\left( { - 3,\,4,\,1} \right)$$   and $$\left( {5,\, - 2,\,1} \right).$$
Mid point of centres is $$\left( {1,\,1,\,1} \right).$$
Satisfying this in the equation of plane, we get
$$2a - 3a + 4a + 6 = 0 \Rightarrow a = - 2.$$