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

$$\eqalign{ & \overrightarrow a .\overrightarrow b = \sqrt {1 + \frac{1}{3}} .\sqrt {1 + \frac{1}{3}} \cos \,\theta \cr & {\text{or }}\frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} = \frac{2}{{\sqrt 3 }}.\frac{2}{{\sqrt 3 }}\cos \,\theta \cr & {\text{or }}\cos \,\theta = \frac{{\sqrt 3 }}{2} \cr & \therefore \,\cos \,\theta = {30^ \circ } \cr} $$

The plane is $$x + 3y + z - 4 + \lambda \left( {2x - y} \right) = 0.$$
This always passes through the intersection of the planes $$x + 3y + z - 4 = 0$$     and $$2x - y = 0,$$   which is a line.
$$\eqalign{ & {\text{Now, }}2x - y = 0\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \frac{x}{1} = \frac{y}{2} \cr & \therefore \,\,x + 3y + z - 4 = 0 \cr & \Rightarrow x + 3.2x + z - 4 = 0 \cr & \Rightarrow \,7x + z - 4 = 0 \cr & \therefore \,7x = - \left( {z - 4} \right) \cr & \Rightarrow \frac{x}{1} = \frac{{z - 4}}{{ - 7}} \cr & \therefore \,\,{\text{ the line is }}\frac{x}{1} = \frac{y}{2} = \frac{{z - 4}}{{ - 7}}. \cr} $$

$$\overrightarrow a .\overrightarrow b $$  is a scalar and $$\overrightarrow a \times \overrightarrow b $$   is a vector. Dot or cross product is applicable for vectors only. Also, a vector may have a scalar multiple.

\[{\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = \left( {\overrightarrow a \times \overrightarrow i } \right).\left( {\overrightarrow a \times \overrightarrow i } \right) = \left| \begin{array}{l} \overrightarrow a .\overrightarrow a \,\,\,\,\,\overrightarrow a .\overrightarrow i \\ \overrightarrow i .\overrightarrow a \,\,\,\,\,\,\overrightarrow i .\overrightarrow i \end{array} \right| = {\overrightarrow a ^2} - {\left( {\overrightarrow a .\overrightarrow i } \right)^2}\]
$$\eqalign{ & {\text{Similarly, }}{\left( {\overrightarrow a \times \overrightarrow j } \right)^2} = {\overrightarrow a ^2} - {\left( {\overrightarrow a .\overrightarrow j } \right)^2},{\text{ and}}\,\,{\left( {\overrightarrow a \times \overrightarrow k } \right)^2} = {\overrightarrow a ^2} - {\left( {\overrightarrow a .k} \right)^2} \cr & \therefore {\text{ the expression}} = 3{\overrightarrow a ^2} - \left\{ {{{\left( {\overrightarrow a .\overrightarrow i } \right)}^2} + {{\left( {\overrightarrow a .\overrightarrow j } \right)}^2} + {{\left( {\overrightarrow a .\overrightarrow k } \right)}^2}} \right\} \cr & = 3{\overrightarrow a ^2} - {\overrightarrow a ^2} \cr & = 2{\overrightarrow a ^2} \cr} $$

Since $$\vec a,\,\vec c,\,\vec b$$  form a right handed system,
\[\therefore \,\,\,\vec c = \,\vec b \times \vec a = \left| \begin{array}{l} \hat i\,\,\,\,\,\hat j\,\,\,\,\,\hat k\\ 0\,\,\,\,\,1\,\,\,\,\,0\\ x\,\,\,\,\,y\,\,\,\,z \end{array} \right| = z\hat i - x\hat k\]

Given that $$\vec a,\,\vec b,\,\vec c,\,\vec d$$   are vectors such that
$$\left( {\vec a \times \vec b} \right) \times \left( {\vec c \times \vec d} \right) = 0.....(1)$$
$${P_1}$$  is the plane determined by vectors $${\vec a}$$ and $${\vec b}$$
$$\therefore $$ Normal vectors $$\overrightarrow {{n_1}} $$  to $${P_1}$$  will be given by $$\overrightarrow {{n_1}} = \vec a \times \vec b$$
Similarly, $${P_2}$$  is the plane determined by vectors $${\vec c}$$ and $${\vec d}$$
$$\therefore $$ Normal vectors $$\overrightarrow {{n_2}} $$ to $${P_2}$$  will be given by $$\overrightarrow {{n_2}} = \vec c \times \vec d$$
Substituting the values of $$\overrightarrow {{n_1}} $$  and $$\overrightarrow {{n_2}} $$  in equation (1)
We get, $$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} = 0\,\,\,\,\,\,\, \Rightarrow \overrightarrow {{n_1}} ||\,\overrightarrow {{n_2}} $$
and hence the planes will also be parallel to each other.
Thus angle between the planes $$= 0.$$

$$\eqalign{ & {\text{Vector moment}} = \overrightarrow {AB} \times \left( {\overrightarrow {{F_1}} + \overrightarrow {{F_2}} } \right) = \left( {\overrightarrow {OB} - \overrightarrow {OA} } \right) \times \left( {\overrightarrow {{F_1}} + \overrightarrow {{F_2}} } \right) \cr & = \left\{ {\left( { - 2\overrightarrow i + 3\overrightarrow j - \overrightarrow k } \right) - \left( {\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k } \right)} \right\} \times \left( {\overrightarrow i - 2\overrightarrow j + 5\overrightarrow k + 3\overrightarrow j - 4\overrightarrow k } \right) \cr & = \left( { - 3\overrightarrow i + \overrightarrow j - 4\overrightarrow k } \right) \times \left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) \cr & = 5\overrightarrow i - \overrightarrow j - 4\overrightarrow k \cr} $$

$$\eqalign{ & \because \,\left( {\vec a \times \vec b} \right) \times \vec a = \left( {\vec a.\vec a} \right)\vec b - \left( {\vec a.\vec b} \right)\vec a \cr & \therefore \left( {\hat j - \hat k} \right) \times \left( {\hat i + \hat j + \hat k} \right) = {\left( {\sqrt 3 } \right)^2}\left( {\vec b} \right) - \left( {\hat i + \hat j + \hat k} \right) \cr & \Rightarrow 3\hat b = 3\hat i\,\,\,\,\, \Rightarrow \hat b = \hat i \cr} $$

$$\eqalign{ & {\text{Let }}\overrightarrow d = x\hat i + y\hat j + z\hat k \cr & {\text{where }}{x^2} + {y^2} + {z^2} = 1......\left( {\text{i}} \right) \cr & \therefore \,\overrightarrow a .\overrightarrow d = 0 \Rightarrow x - y = 0{\text{ or }}x = y......\left( {{\text{ii}}} \right) \cr} $$
\[\left[ {\overrightarrow b \overrightarrow c \overrightarrow d } \right] = 0 \Rightarrow \left| \begin{array}{l} \,\,\,0\,\,\,\,\,\,1\,\, - 1\\ - 1\,\,\,\,0\,\,\,\,\,\,\,\,1\\ \,\,\,x\,\,\,\,\,\,y\,\,\,\,\,\,\,z \end{array} \right| = 0\]
$$\eqalign{ & {\text{or }}x + y + z = 0 \cr & {\text{or }}2x + z = 0\,\,\,\left[ {{\text{Using }}\left( {{\text{ii}}} \right)} \right] \cr & {\text{or }}z = - 2x......\left( {{\text{iii}}} \right) \cr} $$
From $$\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right)$$  and $$\left( {{\text{iii}}} \right),$$  we have
$$\eqalign{ & {x^2} + {x^2} + 4{x^2} = 1 \cr & \Rightarrow x = \pm \frac{1}{{\sqrt 6 }} \cr & \therefore \,\overrightarrow d = \pm \left( {\frac{1}{{\sqrt 6 }}\hat i + \frac{1}{{\sqrt 6 }}\hat j - \frac{2}{{\sqrt 6 }}\hat k} \right) = \pm \left( {\frac{{\hat i + \hat j - 2\hat k}}{{\sqrt 6 }}} \right) \cr} $$

Let $$OA = OB = OC = 1$$     and their direction cosines are $${l_1},\,{m_1},\,{n_1};\,{l_2},\,{m_2},\,{n_2}$$     and $${l_3},\,{m_3},\,{n_3}$$   respectively.
Then $$A = \left( {{l_1},\,{m_1},\,{n_1}} \right),\,B = \left( {{l_2},\,{m_2},\,{n_2}} \right),\,C = \left( {{l_3},\,{m_3},\,{n_3}} \right)$$
and $$OA \bot OB,\,OB \bot OC,\,OC \bot OA.$$
Clearly, $$ABC$$  is equilateral, and $$OG$$  is equally inclined with $$OA,\,OB,\,OC$$    where $$G$$ is the centroid (also circumcentre) of the $$\Delta ABC.$$
$$G = \left( {\frac{{{l_1} + {l_2} + {l_3}}}{3},\,\frac{{{m_1} + {m_2} + {m_3}}}{3},\,\frac{{{n_1} + {n_2} + {n_3}}}{3}} \right)$$
$$\therefore $$  direction ratios of $$OG = {l_1} + {l_2} + {l_3},\,{m_1} + {m_2} + {m_3},\,{n_1} + {n_2} + {n_3}$$
$$\eqalign{ & {\text{Also,}}\,{\left( {{l_1} + {l_2} + {l_3}} \right)^2} + {\left( {{m_1} + {m_2} + {m_3}} \right)^2} + {\left( {{n_1} + {n_2} + {n_3}} \right)^2} \cr & = \sum {l_1^2} + \sum {l_2^2} + \sum {l_3^2 + 2\sum {{l_1}{l_2}} + 2\sum {{l_2}{l_3} + 2\sum {{l_3}{l_1}} } } \cr & = 1 + 1 + 1 + 0 \cr & = 3, \cr} $$
from the conditions of the problem.
$$\therefore $$  direction cosines of $$OG$$  are $$\frac{{{l_1} + {l_2} + {l_3}}}{{\sqrt 3 }},\,\frac{{{m_1} + {m_2} + {m_3}}}{{\sqrt 3 }},\,\frac{{{n_1} + {n_2} + {n_3}}}{{\sqrt 3 }}.$$