3D Geometry and Vectors MCQ Questions & Answers in Geometry | Maths

$$\eqalign{ & \,\,\,\,\,\,\left\{ {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow b - \overrightarrow c } \right)} \right\}.\left( {\overrightarrow c - \overrightarrow a } \right) \cr & = \left( {\overrightarrow a \times \overrightarrow b - \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right).\left( {\overrightarrow c - \overrightarrow a } \right) \cr & = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] - \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right] \cr & = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr & = 0 \cr} $$

$$\eqalign{ & {\left\{ { - \left( {m + n} \right)} \right\}^2} = {m^2} + {n^2}\,\,\,\,\, \Rightarrow mn = 0 \cr & {\text{When }}m = 0,\,l = - n{\text{ direction ratios are }} - 1,\,0,\,1 \cr & {\text{When }}n = 0,\,l = - m{\text{ direction ratios are }} - 1,\,1,\,0 \cr & \therefore \,\cos \,\theta = \frac{{ - 1}}{{\sqrt 2 }}.\frac{{ - 1}}{{\sqrt 2 }} + \frac{0}{{\sqrt 2 }}.\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}.\frac{0}{{\sqrt 2 }} = \frac{1}{2} \cr} $$

$$\eqalign{ & \overrightarrow a \times \overrightarrow b = \left( {2\overrightarrow i + \overrightarrow j - 2\overrightarrow k } \right) \times \left( {\overrightarrow i + \overrightarrow j } \right) = 2\overrightarrow i - 2\overrightarrow j + \overrightarrow k \cr & \therefore \left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|\,\left| {\overrightarrow b } \right|\sin \,{30^ \circ } \cr & = \sqrt {{2^2} + {{\left( { - 2} \right)}^2} + {1^2}} .\left| {\overrightarrow c } \right|.\frac{1}{2} \cr & = \frac{3}{2}\left| {\overrightarrow c } \right| \cr & {\text{Now, }}\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 \,\,\,\,\,\, \Rightarrow {\left( {\overrightarrow c - \overrightarrow a } \right)^2} = 8 \cr & {\text{or }}{\left| {\overrightarrow c } \right|^2} + {\left| {\overrightarrow a } \right|^2} - 2\overrightarrow c .\overrightarrow a = 8 \cr & {\text{or }}{\left| {\overrightarrow c } \right|^2} + {\left( {\sqrt {{2^2} + {1^2} + {{\left( { - 2} \right)}^2}} } \right)^2} - 2\left| {\overrightarrow c } \right| = 8 \cr & {\text{or }}{\left| {\overrightarrow c } \right|^2} - 2\left| {\overrightarrow c } \right| + 1 = 0 \cr & \therefore \,\,\left| {\overrightarrow c } \right| = 1 \cr & \therefore \,\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|\,\left| {\overrightarrow c } \right|\sin \,{30^ \circ } = \frac{3}{2} \cr} $$

Projection of $${\vec v}$$ along $$\vec u = \frac{{\vec v.\vec u}}{{\left| {\vec u} \right|}} = \frac{{\vec v.\vec u}}{2}$$
projection of $${\vec w}$$ along $$\vec u = \frac{{\vec w.\vec u}}{{\left| {\vec u} \right|}} = \frac{{\vec w.\vec u}}{2}$$
$$\eqalign{ & {\text{Given }}\frac{{\vec v.\vec u}}{2} = \frac{{\vec w.\vec u}}{2}.....(1) \cr & {\text{Also,}}\,\,{\text{ }}\vec v.\,\vec w = 0.....(2) \cr & {\text{Now }}\,\,{\text{ }}{\left| {\vec u - \vec v + \vec w} \right|^2} \cr & = {\left| {\vec u} \right|^2} + {\left| {\vec v} \right|^2} + {\left| {\vec w} \right|^2} - 2\vec u.\vec v - 2\vec v.\vec w + 2\vec u.\vec w \cr & = 1 + 4 + 9 + 0\,\,\,\,\,\,\left[ {{\text{ From equation (1) and (2)}}} \right] \cr & = 14 \cr & \therefore \,\left| {\vec u - \vec v + \vec w} \right| = \sqrt {14} \, \cr} $$

Direction cosines of the line are $$\frac{{\sqrt 3 }}{4},\,\frac{2}{4},\,\frac{3}{4}.$$   So, $$P = \left( {3 \pm 1.\frac{{\sqrt 3 }}{4}, - 2 \pm 1.\,\frac{2}{4},4 \pm 1.\,\frac{3}{4}} \right)$$
$$\therefore \,\,P = \left( {3 \pm \frac{{\sqrt 3 }}{4}, - 2 \pm \,\frac{1}{2},4 \pm \,\frac{3}{4}} \right)$$

$$\eqalign{ & \left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow r \times \overrightarrow c } \right) = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow r } \right]\overrightarrow c ......\left( 1 \right) \cr & \left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow r \times \overrightarrow a } \right) = \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right]\overrightarrow r - \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow r } \right]\overrightarrow a \cr & \left( {\overrightarrow c \times \overrightarrow a } \right) \times \left( {\overrightarrow r \times \overrightarrow b } \right) = \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right]\overrightarrow r - \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow r } \right]\overrightarrow b \cr} $$
$$\therefore $$  the expression $$ = 3\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r - \left\{ {\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow r } \right]\overrightarrow c + \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow r } \right]\overrightarrow a + \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow r } \right]\overrightarrow b } \right\}......\left( 2 \right)$$
Again, $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow r \times \overrightarrow c } \right) = \left[ {\overrightarrow a \,\,\overrightarrow r \,\,\overrightarrow c } \right]\overrightarrow b - \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow r } \right]\overrightarrow a ......\left( 3 \right)$$
From (1) and (3), $$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow r } \right]\overrightarrow c = \left[ {\overrightarrow a \,\,\overrightarrow r \,\,\overrightarrow c } \right]\overrightarrow b - \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow r } \right]\overrightarrow a $$
$$\therefore \,\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow r } \right]\overrightarrow c + \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow r } \right]\overrightarrow b + \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow r } \right]\overrightarrow a $$
$$\therefore $$  from (2), the expression $$ = 3\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r = 2\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow r .$$

$$\eqalign{ & \theta = \alpha + \beta ,\,\beta = {\tan ^{ - 1}}\left( {\frac{3}{5}} \right){\text{ or }}\beta = \theta - \alpha \cr & \Rightarrow \tan \,\beta = \frac{{\tan \,\theta - \tan \,\alpha }}{{1 + \tan \,\theta .\tan \,\alpha }} \cr & {\text{or }}\frac{3}{5} = \frac{{\frac{h}{{40}} - \frac{h}{{160}}}}{{1 + \frac{h}{{40}}.\frac{h}{{160}}}} \cr & \Rightarrow {h^2} - 200h + 6400 = 0 \cr & \Rightarrow h = 40{\text{ or }}160{\text{ metre}} \cr & \therefore \,{\text{possible height}} = 40\,{\text{metre}} \cr} $$

$$\eqalign{ & \vec a + \vec b + \vec c = 0\,\, \Rightarrow \left( {\vec a + \vec b + \vec c} \right).\left( {\vec a + \vec b + \vec c} \right) = 0 \cr & {\left| {\vec a} \right|^2} + \,{\left| {\vec b} \right|^2} + \,{\left| {\vec c} \right|^2} + 2\left( {\vec a.\vec b + \vec b.\vec c + \vec c.\vec a} \right) = 0 \cr & \vec a.\vec b + \vec b.\vec c + \vec c.\vec a = \frac{{ - 1 - 4 - 9}}{2} = - 7 \cr} $$

The length of the perpendicular to the plane $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$    from the origin $$ = \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}.$$
The length of the perpendicular to the plane $$\frac{{x'}}{{a'}} + \frac{{y'}}{{b'}} + \frac{{z'}}{{c'}} = 1$$    from the origin (the same point) $$ = \frac{1}{{\sqrt {\frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}} }}.$$
These are equal. So $$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}.$$

$$\eqalign{ & \left( {\overrightarrow a + \overrightarrow b } \right).\overrightarrow a = \left| {\overrightarrow a + \overrightarrow b } \right|\left| {\overrightarrow a } \right|\cos \,{60^ \circ } \cr & \Rightarrow \cos \,{60^ \circ } = \frac{1}{2} = \frac{{\left( {\overrightarrow a + \overrightarrow b } \right).\overrightarrow a }}{{\left| {\overrightarrow a + \overrightarrow b } \right|\left| {\overrightarrow a } \right|}} \cr & \Rightarrow \frac{1}{2} = \frac{{{{\left| {\overrightarrow a } \right|}^2}}}{{\left| {\overrightarrow a + \overrightarrow b } \right|\left| {\overrightarrow a } \right|}} \cr & \Rightarrow \frac{1}{2} = \frac{{\left| {\overrightarrow a } \right|}}{{\left| {\overrightarrow a + \overrightarrow b } \right|}}......\left( 1 \right) \cr & \left\{ {{\text{as }}\overrightarrow a .\overrightarrow b = 0,\,\overrightarrow a \bot \overrightarrow b } \right\} \cr} $$

$$\eqalign{ & \left( {\overrightarrow a + \overrightarrow b } \right).\overrightarrow b = \left| {\overrightarrow a + \overrightarrow b } \right|\left| {\overrightarrow b } \right|\cos \,{30^ \circ } \cr & \Rightarrow \cos \,{30^ \circ } = \frac{{\sqrt 3 }}{2} = \frac{{\left( {\overrightarrow a + \overrightarrow b } \right).\left( {\overrightarrow b } \right)}}{{\left| {\overrightarrow a + \overrightarrow b } \right|\left| {\overrightarrow b } \right|}} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{{{\left| {\overrightarrow b } \right|}^2}}}{{\left| {\overrightarrow a + \overrightarrow b } \right|\left| {\overrightarrow b } \right|}} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{\left| {\overrightarrow b } \right|}}{{\left| {\overrightarrow a + \overrightarrow b } \right|}}......\left( 2 \right) \cr & {\text{Dividing}}\left( 2 \right){\text{ by}}\left( 1 \right)\,: \cr & \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}}} = \frac{{\left| {\overrightarrow b } \right|}}{{\left| {\overrightarrow a } \right|}} \cr & \Rightarrow \sqrt 3 = \frac{{\left| {\overrightarrow b } \right|}}{{\left| {\overrightarrow a } \right|}} \cr & \Rightarrow \left| {\overrightarrow b } \right| = \sqrt 3 \left| {\overrightarrow a } \right| \cr} $$