Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

$$\eqalign{ & 1 = \left| {\left( {\overrightarrow b - \overrightarrow a } \right).\frac{{\left( {\overrightarrow p \times \overrightarrow q } \right)}}{{\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right| \cr & \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right|\cos \,{60^ \circ } = 1 \cr & \Rightarrow AB = 2 \cr} $$

Let $$\left( {h,\,k,\,\ell } \right)$$   be the point which is equidistant from the points $$\left( {1,\,2,\,3} \right)$$   and $$\left( {3,\,2,\, - 1} \right)$$
$$\eqalign{ & \Rightarrow \sqrt {{{\left( {h - 1} \right)}^2} + {{\left( {k - 2} \right)}^2} + {{\left( {\ell - 3} \right)}^2}} = \sqrt {{{\left( {h - 3} \right)}^2} + {{\left( {k - 2} \right)}^2} + {{\left( {\ell + 1} \right)}^2}} \cr & \Rightarrow {\left( {h - 1} \right)^2} + {\left( {\ell - 3} \right)^2} = {\left( {h - 3} \right)^2} + {\left( {\ell + 1} \right)^2} \cr & \Rightarrow {h^2} + 1 - 2h + {\ell ^2} - 6\ell + 9 = {h^2} - 6h + 9 + {\ell ^2} + 2\ell + 1 \cr & \Rightarrow - 2h - 6\ell = - 6h + 2\ell \cr & \Rightarrow 6h - 2h - 6\ell - 2\ell = 0 \cr & \Rightarrow 4h - 8\ell = 0 \cr & \Rightarrow h - 2\ell = 0 \cr} $$
Putting $$h = x$$  and $$\ell = z$$
We get locus of points $$\left( {h,\,k,\,\ell } \right)$$   as, $$x - 2z = 0$$

$$\overrightarrow r .\overrightarrow {{n_1}} + \lambda \overrightarrow r .\overrightarrow {{n_2}} = {q_1} + \lambda {q_2}......\left( {\text{i}} \right)$$
where $$\lambda $$ is a parameter.
So, $$\overrightarrow {{n_1}} + \lambda \overrightarrow {{n_2}} $$   is normal to plane $$\left( {\text{i}} \right).$$  Now, any plane parallel to the line of intersection of the planes $$\overrightarrow r .\overrightarrow {{n_3}} = {q_3}$$   and $$\overrightarrow r .\overrightarrow {{n_4}} = {q_4}$$   is of the form $$\overrightarrow r .\left( {\overrightarrow {{n_3}} \times \overrightarrow {{n_4}} } \right) = d.$$
Hence, we must have
$$\eqalign{ & \left[ {\overrightarrow {{n_1}} + \lambda \overrightarrow {{n_2}} } \right].\left[ {\overrightarrow {{n_3}} \times \overrightarrow {{n_4}} } \right] = 0 \cr & {\text{or, }}\left[ {\overrightarrow {{n_1}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right] + \lambda \left[ {\overrightarrow {{n_2}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right] = 0 \cr & {\text{or, }}\lambda = \frac{{ - \left[ {\overrightarrow {{n_1}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right]}}{{\left[ {\overrightarrow {{n_2}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right]}} \cr} $$
On putting this value in Equation $$\left( {\text{i}} \right),$$  we have the equation of the required plane as
$$\eqalign{ & \overrightarrow r .\overrightarrow {{n_1}} - {q_1} = \frac{{\left[ {\overrightarrow {{n_1}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right]}}{{\left[ {\overrightarrow {{n_2}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right]}}\left( {\overrightarrow r .\overrightarrow {{n_2}} - {q_2}} \right) \cr & {\text{or, }}\left[ {\overrightarrow {{n_2}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right]\left( {\overrightarrow r .\overrightarrow {{n_1}} - {q_1}} \right) = \left[ {\overrightarrow {{n_1}} \overrightarrow {{n_3}} \overrightarrow {{n_4}} } \right]\left( {\overrightarrow r .\overrightarrow {{n_2}} - {q_2}} \right) \cr} $$

As $${\vec v}$$ lies in the plane of $${\vec a}$$ and $${\vec b}$$
$$\eqalign{ & \therefore \vec v = \lambda \vec a + \mu \vec b \cr & \Rightarrow \vec v = \left( {\lambda + \mu } \right)\hat i + \left( {\lambda - \mu } \right)\hat j + \left( {\lambda + \mu } \right)\hat k \cr} $$
$$\because $$ Projection of $${\vec v}$$ on $${\vec c}$$ is $$\frac{1}{{\sqrt 3 }}$$
$$\eqalign{ & \therefore \,\,\frac{{\vec v.\vec c}}{{\left| {\vec c} \right|}} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{{\left( {\lambda + \mu } \right) - \left( {\lambda - \mu } \right) - \left( {\lambda + \mu } \right)}}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \mu - \lambda = 1{\text{ or }}\mu = \lambda + 1 \cr & \Rightarrow \vec v = \left( {2\lambda + 1} \right)\hat i - \hat j + \left( {2\lambda + 1} \right)\hat k \cr & {\text{For }}\lambda = 1,\,\,\,\vec v = 3\hat i - \hat j + 3\hat k \cr} $$

$$\eqalign{ & {\text{Since, }}{l^2} + {m^2} + {n^2} = 1 \cr & \therefore \,{\cos ^2}\alpha + {\cos ^2}\alpha + {\cos ^2}\theta = 1......\left( {\text{i}} \right) \cr} $$
($$\because \,A$$  line makes the same angle $$\alpha $$ with $$x$$ and $$y$$-axes and $$\theta $$ with $$z$$-axis)
$$\eqalign{ & {\text{Also, }}{\sin ^2}\theta = 2\,{\sin ^2}\alpha \cr & \Rightarrow 1 - {\cos ^2}\theta = 2\left( {1 - {{\cos }^2}\alpha } \right)\,\,\,\left( {\because {{\sin }^2}A + {{\sin }^2}A = 1} \right) \cr & \Rightarrow {\cos ^2}\theta = 2\,{\cos ^2}\alpha - 1......\left( {{\text{ii}}} \right) \cr & \therefore \,{\text{From equations }}\left( {\text{i}} \right)\,{\text{and }}\left( {{\text{ii}}} \right), \cr & 2\,{\cos ^2}\alpha + 2{\cos ^2}\alpha - 1 = 1 \cr & \Rightarrow 4\,{\cos ^2}\alpha = 2 \cr & \Rightarrow \cos \,\alpha = \pm \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \alpha = \frac{\pi }{4},\,\frac{{3\pi }}{4} \cr} $$

The plane passing through the intersection line of given planes is
$$\eqalign{ & \left( {x + 2y + 3z - 2} \right) + \lambda \left( {x - y + z - 3} \right) = 0 \cr & {\text{or}}\,\,\,\left( {1 + \lambda } \right)x + \left( {2 - \lambda } \right)y + \left( {3 + \lambda } \right)z + \left( { - 2 - 3\lambda } \right) = 0 \cr} $$
Its distance from the point $$\left( {3,\,1,\, - 1} \right)$$   is $$\frac{2}{{\sqrt 3 }}$$
$$\eqalign{ & \therefore \left| {\frac{{3\left( {1 + \lambda } \right) + 1\left( {2 - \lambda } \right) - 1\left( {3 + \lambda } \right) + \left( { - 2 - 3\lambda } \right)}}{{\sqrt {{{\left( {1 + \lambda } \right)}^2} + {{\left( {2 - \lambda } \right)}^2} + {{\left( {3 + \lambda } \right)}^2}} }}} \right| = \frac{2}{{\sqrt 3 }} \cr & \Rightarrow \left| {\frac{{ - 2\lambda }}{{\sqrt {3{\lambda ^2} + 4\lambda + 14} }}} \right| = \frac{2}{{\sqrt 3 }} \cr & \Rightarrow 3{\lambda ^2} + 4\lambda + 14 = 3{\lambda ^2} \cr & \Rightarrow \lambda = - \frac{7}{2} \cr} $$
$$\therefore $$ Required equation of plane is
$$\eqalign{ & \left( {x + 2y + 3z - 2} \right) - \frac{7}{2}\left( {x - y + z - 3} \right) = 0 \cr & {\text{or }}5x - 11y + z = 17 \cr} $$

The equations of spheres are $${S_1} : {x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$$        and $${S_2} : {x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$$
Their plane of intersection is
$$\eqalign{ & {S_1} - {S_2} = 0 \cr & \Rightarrow 10x - 5y - 5z - 5 = 0 \cr & \Rightarrow 2x - y - z = 1 \cr} $$

Let coordinates of $$D$$ be $$\left( {x,\,y,\,z} \right)$$
Co-ordinates of centroid is $$\left( {1,\,1,\,1} \right),$$   and of $$A$$, is $$\left( {4,\,7,\, - 8} \right)$$
Centroid divides median is $$2 : 1$$  ratio

$$\eqalign{ & {\text{So, }}\frac{{AO}}{{OD}} = 2:1 \cr & {\text{For }}x\,:\,1 = \frac{{2 \times x + 1 \times 4}}{{1 + 2}} \Rightarrow x = - \frac{1}{2} \cr & {\text{For }}y\,:\,1 = \frac{{2 \times y + 1 \times 7}}{{1 + 2}} \Rightarrow y = - 2 \cr & {\text{For }}z\,:\,1 = \frac{{2 \times z + 1 \times - 8}}{{1 + 2}} \Rightarrow z = + \frac{{11}}{2} \cr & \therefore \,{\text{Co - ordinates of }}D{\text{ are}}\left( { - \frac{1}{2},\, - 2,\,\frac{{11}}{2}} \right) \cr} $$

The given equation of sphere is
$$a{x^2} + b{y^2} + c{z^2} - 2x + 4y + 2z - 3 = 0$$
This equation represents a equation of sphere, if coefficient of $${x^2},{y^2}$$  and $${z^2}$$ is same.
i.e., $$a = b = c$$
$$\therefore $$  Equation of sphere can be re-written as
$$\eqalign{ & b{x^2} + b{y^2} + b{z^2} - 2x + 4y + 2z - 3 = 0 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - \frac{{2x}}{b} + \frac{{4y}}{b} + \frac{{2z}}{b} - \frac{3}{b} = 0 \cr} $$
The centre of this sphere is $$\left( {\frac{1}{b},\,\frac{{ - 2}}{b},\,\frac{{ - 1}}{b}} \right)$$
Given that the centre of sphere is $$\left( {\frac{1}{2},\, - 1,\, - \frac{1}{2}} \right)$$
$$\frac{1}{b} = \frac{1}{2} \Rightarrow b = 2$$

Lines are $$\overrightarrow r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i - \hat j + \hat k} \right)$$       and $$\overrightarrow r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \mu \left( {\hat i + \hat j - \hat k} \right)$$
along vectors $$\left( {\hat i - \hat j + \hat k} \right)$$   and $$\left( {\hat i + \hat j - \hat k} \right)$$   respectively.
Angle between two lines
$$\eqalign{ & = {\cos ^{ - 1}}\left( {\frac{{\left( 1 \right) \times \left( 1 \right) + \left( { - 1} \right) \times \left( 1 \right) + \left( 1 \right) \times \left( { - 1} \right)}}{{\sqrt 3 \,\sqrt 3 }}} \right) \cr & = {\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right) \cr} $$
Which is an obtuse angle.
$$\therefore $$  Vector along acute angle bisector
$$\eqalign{ & = \lambda \left[ {\frac{{\hat i - \hat j + \hat k}}{{\sqrt 3 }} - \frac{{\hat i + \hat j - \hat k}}{{\sqrt 3 }}} \right] \cr & = \frac{{2\lambda }}{{\sqrt 3 }}\left( { - \hat j + \hat k} \right) \cr} $$
$$\therefore $$  Equation of acute angle bisector
$$ = \left( {\hat i + 2\hat j + 3\hat k} \right) + t\left( {\hat j - \hat k} \right)$$