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

$$\because \,M$$  is mid point of $$BC$$
$$\eqalign{ & \therefore \,\overrightarrow {AM} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) \cr & = 4\hat i + \hat j + 4\hat k \cr} $$
Length of median $$AM$$
$$ = \sqrt {16 + 1 + 16} = \sqrt {33} $$

The positive vector of the centroid $$ = \frac{{\overrightarrow a + \overrightarrow b + \overrightarrow c }}{3},$$    where $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$   are the position vectors of the vertices.
$$\therefore \,\overrightarrow k = \frac{{\overrightarrow i + \overrightarrow j + 2\overrightarrow i - \overrightarrow j + \overrightarrow k + \overrightarrow a }}{3}\,\,\,\,\,\therefore \overrightarrow a = - 3\overrightarrow i + 2\overrightarrow k $$

Here, $$\overrightarrow a + \overrightarrow b = t\overrightarrow c ,\,\,\overrightarrow b + \overrightarrow c = s\overrightarrow a .$$      Subtracting, $$\overrightarrow a - \overrightarrow c = t\overrightarrow c - s\overrightarrow a $$
or $$\left( {1 + s} \right)\overrightarrow a = \left( {1 + t} \right)\overrightarrow c $$
But $$\overrightarrow a ,\,\overrightarrow c $$  are noncollinear.
$$\therefore \,1 + s = 0,\,\,1 + t = 0.$$     Hence, $$\overrightarrow a + \overrightarrow b = - \overrightarrow c $$

$$\eqalign{ & {\text{Let }}a = \frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j + \hat k \cr & {\text{and }}b = \frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat j + \hat k \cr & \therefore \,\cos \,\theta = \frac{{a.b}}{{\left| a \right|\left| b \right|}} \cr & \Rightarrow \cos \,\theta = \frac{{\left( {\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j + \hat k} \right).\left( {\frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat j + \hat k} \right)}}{{\sqrt {\frac{1}{2} + \frac{1}{2} + 1} .\sqrt {\frac{1}{2} + \frac{1}{2} + 1} }} \cr & \Rightarrow \cos \,\theta = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{2} + 1} \right] \cr & \Rightarrow \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \cos \,\theta = \cos \,{60^ \circ } \cr & \therefore \,\theta = {60^ \circ } \cr} $$

\[\begin{array}{l} \left| \begin{array}{l} a\,\,\,\,\,{a^2}\,\,\,\,\,1 + {a^3}\\ b\,\,\,\,\,{b^2}\,\,\,\,\,1 + {b^3}\\ c\,\,\,\,\,{c^2}\,\,\,\,\,1 + {c^3} \end{array} \right| = 0\\ \Rightarrow \left| \begin{array}{l} a\,\,\,\,{a^2}\,\,\,\,\,1\\ b\,\,\,\,{b^2}\,\,\,\,\,1\\ c\,\,\,\,\,{c^2}\,\,\,\,\,1 \end{array} \right| + \left| \begin{array}{l} a\,\,\,\,{a^2}\,\,\,\,\,{a^3}\\ b\,\,\,\,{b^2}\,\,\,\,\,{b^3}\\ c\,\,\,\,\,{c^2}\,\,\,\,\,{c^3} \end{array} \right| = 0\\ \Rightarrow \left( {1 + abc} \right)\left| \begin{array}{l} 1\,\,\,\,\,a\,\,\,\,{a^2}\\ 1\,\,\,\,\,b\,\,\,\,{b^2}\\ 1\,\,\,\,\,c\,\,\,\,\,{c^2} \end{array} \right| = 0\\ {\rm{As }}\left| \begin{array}{l} 1\,\,\,\,\,a\,\,\,\,{a^2}\\ 1\,\,\,\,\,b\,\,\,\,{b^2}\\ 1\,\,\,\,\,c\,\,\,\,\,{c^2} \end{array} \right| \ne 0\,\,\,\,\left( {{\rm{given \,\,condition}}} \right)\\ \therefore \,\,abc = - 1 \end{array}\]

$$\eqalign{ & \overrightarrow r = \left( {\overrightarrow a \times \overrightarrow b } \right)\sin \,x + \left( {\overrightarrow b \times \overrightarrow c } \right)\cos \,y + 2\left( {\overrightarrow c \times \overrightarrow a } \right) \cr & \overrightarrow r .\left( {\overrightarrow a + \,\overrightarrow b + \,\overrightarrow c } \right) = 0 \cr & \Rightarrow \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]\left( {\sin \,x + \cos \,y + 2} \right) = 0 \cr} $$
Since, $$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] \ne 0,$$    we have $$\sin \,x + \cos \,y = - 2$$
This is possible only when $$\sin \,x = - 1{\text{ and }}\cos \,y = - 1$$
For $${x^2} + {y^2}$$   to be minimum, $$x = \frac{\pi }{2}{\text{ and }}y = \pi $$
$$\therefore $$  Minimum value of $$\left( {{x^2} + {y^2}} \right) = \frac{{{\pi ^2}}}{2} + {\pi ^2} = \frac{{5{\pi ^2}}}{4}$$

Adding and simplifying, $$\overrightarrow \lambda + \overrightarrow \mu + \overrightarrow \nu = 0$$
$$\therefore $$  there is a linear relation between $$\overrightarrow \lambda ,\,\overrightarrow \mu ,\,\overrightarrow \nu $$   and, therefore, they are coplanar.

Moment of force, $$m = \overrightarrow r \times \overrightarrow F $$
\[m = \left| \begin{array}{l} \hat i\,\,\,\,\,\,\hat j\,\,\,\,\,\,\hat k\\ 2\,\,\, - 2\,\,\, - 3\\ 3\,\,\,\,\,\,4\,\,\,\, - 3 \end{array} \right|\]
$$\eqalign{ & = \hat i\left( {6 + 12} \right) - \hat j\left( { - 6 + 9} \right) + \hat k\left( {8 + 6} \right) \cr & = 18\hat i - 3\hat j + 14\hat k \cr & = \sqrt {{{\left( {18} \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( {14} \right)}^2}} \cr & = \sqrt {529} \cr & = 23{\text{ units}} \cr} $$

$$\eqalign{ & {\text{We have, }}\vec a.\vec b = 0,\,\,\,\vec a.\vec a = 1,\,\,\vec b.\vec b = 1 \cr & \left( {2\vec a - \vec b} \right).\left[ {\left( {\vec a \times \vec b} \right) \times \left( {\vec a + 2\vec b} \right)} \right] \cr & = \left( {2\vec a - \vec b} \right).\left[ {\left\{ {\vec a.\left( {\vec a + 2\vec b} \right)} \right\}\vec b - \left\{ {\vec b.\left( {\vec a + 2\vec b} \right)\vec a} \right\}} \right] \cr & = \left( {2\vec a - \vec b} \right).\left[ {\left( {\vec a.\vec a + 2\vec a.\vec b} \right)\vec b - \left( {\vec a.\vec b + 2\vec b.\vec b} \right)\vec a} \right] \cr & = \left( {2\vec a - \vec b} \right).\left[ {\left( {\vec b - 2\vec a} \right)} \right] \cr & = 4\vec a.\vec b - \vec b.\vec b - 4\vec a.\vec a \cr & = - 5 \cr} $$

$$\because \,\,\vec a$$  lies in the plane of $${\vec b}$$ and $${\vec c}$$
$$\eqalign{ & \therefore \,\,\vec a = \vec b + \lambda \vec c \cr & \Rightarrow \alpha \hat i + 2\hat j + \beta \hat k = \hat i + \hat j + \lambda \left( {\hat j + \hat k} \right) \cr & \Rightarrow \alpha = 1,\,2 = 1 + \lambda ,\,\,\beta = \lambda \cr & \Rightarrow \alpha = 1,\,\,\beta = 1 \cr} $$