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

Let $$\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow {k.} $$    Then $$\overrightarrow a .\overrightarrow i = x,\,\overrightarrow a .\overrightarrow j = y,\,\overrightarrow a .\overrightarrow k = z.$$

$$\eqalign{ & \left( {\vec u + \vec v - \vec w} \right).\left( {\vec u \times \vec v - \vec u \times \vec w - \vec v \times \vec v + \vec v \times \vec w} \right) \cr & = \left( {\vec u + \vec v - \vec w} \right).\left( {\vec u \times \vec v - \vec u \times \vec w + \vec v \times \vec w} \right) \cr & = \vec u.\left( {\vec u \times \vec v} \right) - \vec u.\left( {\vec u \times \vec w} \right) + \vec u.\left( {\vec v \times \vec w} \right) + \vec v.\left( {\vec u \times \vec v} \right) - \vec v.\left( {\vec u \times \vec w} \right) + \vec v.\left( {\vec v \times \vec w} \right) - \vec w.\left( {\vec u \times \vec v} \right) + \vec w.\left( {\vec u \times \vec w} \right) - \vec w.\left( {\vec u \times \vec w} \right) \cr & = \vec u.\left( {\vec v \times \vec w} \right) - \vec v.\left( {\vec u \times \vec w} \right) - \vec w.\left( {\vec u \times \vec v} \right) \cr & = \left[ {\vec u\,\vec v\,\vec w} \right] + \left[ {\vec v\,\vec w\,\vec u} \right] - \left[ {\vec w\,\vec u\,\vec v} \right] \cr & = \vec u.\left( {\vec v \times \vec w} \right) \cr} $$

$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \cr & = \left\{ {2\overrightarrow i + \left( {2 - y} \right)\overrightarrow j + 2\overrightarrow k } \right\} - \left\{ {\left( {2 - x} \right)\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k } \right\} \cr & = x\overrightarrow i - y\overrightarrow j \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} \cr & = \left\{ {2\overrightarrow i + 2\overrightarrow j + \left( {2 - z} \right)\overrightarrow k } \right\} - \left\{ {\left( {2 - x} \right)\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k } \right\} \cr & = x\overrightarrow i - z\overrightarrow k \cr & \overrightarrow {AD} = \overrightarrow {OD} - \overrightarrow {OA} \cr & = \left\{ {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right\} - \left\{ {\left( {2 - x} \right)\overrightarrow i + 2\overrightarrow j + 2\overrightarrow k } \right\} \cr & = \left( {x - 1} \right)\overrightarrow i - \overrightarrow j - \overrightarrow k \cr} $$
As these vectors are coplanar, \[\left| \begin{array}{l} \,\,\,\,x\,\,\,\,\,\,\,\,\,\, - y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\, - z\\ x - 1\,\,\,\,\, - 1\,\,\,\,\, - 1 \end{array} \right| = 0\]
On simplification, $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$$

Equations of the lines in symmetric form are $$\frac{x}{1} = \frac{y}{1} = \frac{z}{{ - 2}}{\text{ and }}\frac{{x - \frac{3}{4}}}{1} = \frac{{y - \frac{5}{4}}}{{ - 1}} = \frac{z}{{ - 4}}$$
$$\eqalign{ & {\text{Now,}}\,\,l + m - 2n = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,l - m - 4n = 0 \cr & {\text{Solving, }}\frac{l}{{ - 2}} = \frac{m}{2} = \frac{n}{{ - 2}}\,\,\,\,\,\, \Rightarrow l = \frac{{ - 1}}{{\sqrt 3 }},\,m = \frac{1}{{\sqrt 3 }},\,n = \frac{{ - 1}}{{\sqrt 3 }} \cr} $$
$$\therefore $$  shortest distance $$ = \left( {\frac{3}{4} - 0} \right)\frac{{ - 1}}{{\sqrt 3 }} + \left( {\frac{5}{4} - 0} \right)\frac{1}{{\sqrt 3 }} + \left( {0 - 0} \right)\frac{{ - 1}}{{\sqrt 3 }}.$$

\[\begin{array}{l} \vec a = \hat i - \hat k,\,\vec b = x\hat i + \hat j + \left( {1 - x} \right)\hat k\,\,{\rm{and}}\\ \vec c = y\hat i + x\hat j + \left( {1 + x - y} \right)\,\hat k\\ \left[ {\vec a,\,\vec b,\,\vec c} \right] = \vec a.\vec b \times \vec c = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\\ x\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1 - x\\ y\,\,\,\,\,\,\,\,x\,\,\,\,1 + x - y \end{array} \right|\\ = 1\left[ {1 + x - y - x + {x^2}} \right] - \left[ { - {x^2} - y} \right]\\ = 1 - y + {x^2} - {x^2} + y = 1 \end{array}\]
Hence $$\left[ {\vec a,\,\vec b,\,\vec c} \right]$$  is independent of $$x$$ and $$y$$ both.

Expression $$ = \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b + \left( {\overrightarrow b .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow b .\overrightarrow a } \right)\overrightarrow b = {\overrightarrow b ^2}\overrightarrow a = \overrightarrow a \,\left( {\because \,\overrightarrow b {\text{ is a unit vector}}} \right)$$

$$\eqalign{ & {\text{Expression}} = \frac{{\left( {\overrightarrow a + \overrightarrow b } \right).\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} + \frac{{\left( {\overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow c \times \overrightarrow a } \right)}}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} + \frac{{\left( {\overrightarrow c + \overrightarrow a } \right).\left( {\overrightarrow a \times \overrightarrow b } \right)}}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} + \frac{{\left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right]}}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} + \frac{{\left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right]}}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} \cr & = 1 + 1 + 1 \cr & = 3 \cr} $$

$$\eqalign{ & \overrightarrow {AB} .\overrightarrow {BC} = {a^2}\cos \,{120^ \circ } = - \frac{{{a^2}}}{2} \cr & \overrightarrow {BC} .\overrightarrow {CA} = - \frac{{{a^2}}}{2} \cr & \overrightarrow {CA} .\overrightarrow {AB} = - \frac{{{a^2}}}{2} \cr & \therefore {\text{ Answer}} = - \frac{{3{a^2}}}{2} \cr} $$

If the points be $$A,\,B$$  and $$C$$ respectively then
$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \overrightarrow a - 2\overrightarrow b + \lambda \overrightarrow c - \left( {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right) = - \overrightarrow a - \overrightarrow b + \left( {\lambda - 3} \right)\overrightarrow c \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = \mu \overrightarrow a - 5\overrightarrow b - \left( {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right) = \left( {\mu - 2} \right)\overrightarrow a - 4\overrightarrow b - 3\overrightarrow c \cr} $$
The points are collinear if $$\overrightarrow {AB} = t\overrightarrow {AC} $$
$$\eqalign{ & \Rightarrow - 1 = t\left( {\mu - 2} \right),\,\, - 1 = - 4t,\,\,\lambda - 3 = - 3t \cr & \therefore \,t = \frac{1}{4},{\text{ and }} - 1 = \frac{1}{4}\left( {\mu - 2} \right),\,\lambda - 3 = - \frac{3}{4} \cr} $$

Since $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c $$ are coplanar therefore
\[\begin{array}{l} \left| \begin{array}{l} \,1\,\,\,1\,\,\,1\\ 4\,\,\,3\,\,\,4\\ 1\,\,\,\alpha \,\,\beta \end{array} \right| = 0\\ \Rightarrow \beta = 1\,;\,\left| {\overrightarrow c } \right| = \sqrt {1 + {\alpha ^2} + {\beta ^2}} = \sqrt 3 \\ \Rightarrow {\alpha ^2} + {\beta ^2} = 2\\ \Rightarrow {\alpha ^2} = 1\\ \therefore \,\alpha = \pm 1 \end{array}\]