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

$$\eqalign{ & \overrightarrow u - \overrightarrow v = \overrightarrow w \cr & \left( {2x\overrightarrow \alpha + y\overrightarrow \beta } \right) - \left( {2y\overrightarrow \alpha + 3x\overrightarrow \beta } \right) = 2\overrightarrow \alpha - 5\overrightarrow \beta \cr & \left( {2x - 2y} \right)\overrightarrow \alpha + \left( {y - 3x} \right)\overrightarrow \beta = 2\overrightarrow \alpha - 5\overrightarrow \beta \cr & \therefore \,2x - 2y = 2......\left( {\text{i}} \right) \cr & {\text{and }}3x - y = - 5......\left( {{\text{ii}}} \right) \cr & {\text{Solving equations }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right){\text{, we get}} \cr & x = 2{\text{ and }}y = 1 \cr} $$

By definition of scalar triple product $${\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)}$$   can be written as $$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]$$
$$\eqalign{ & \frac{{\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left( {\overrightarrow c \times \overrightarrow a } \right).\overrightarrow b }} + \frac{{\overrightarrow b .\left( {\overrightarrow a \times \overrightarrow c } \right)}}{{\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right)}} \cr & = \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow c \overrightarrow a \overrightarrow b } \right]}} + \frac{{\left[ {\overrightarrow b \overrightarrow a \overrightarrow c } \right]}}{{\left[ {\overrightarrow c \overrightarrow a \overrightarrow b } \right]}} \cr & = \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} - \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr & = 1 - 1 \cr & = 0 \cr & \because \,\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left[ {\overrightarrow b \overrightarrow c \overrightarrow a } \right] = \left[ {\overrightarrow c \overrightarrow a \overrightarrow b } \right] \cr & {\text{but }}\left[ {\overrightarrow b \overrightarrow c \overrightarrow a } \right] = - \left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] \cr} $$

Any vector coplanar to $${\vec a}$$ and $${\vec b}$$ can be written as
$$\eqalign{ & \vec r = \vec a + \lambda \vec b \cr & \vec r = \left( {1 + 2\lambda } \right)\hat i + \left( { - 1 + \lambda } \right)\hat j + \left( {1 + \lambda } \right)\hat k \cr} $$
Since $${\vec r}$$ is orthogonal to $$5\hat i + 2\hat j + 6\hat k$$
$$\eqalign{ & \Rightarrow 5\left( {1 + 2\lambda } \right) + 2\left( { - 1 + \lambda } \right) + 6\left( {1 + \lambda } \right) = 0 \cr & \Rightarrow 9 + 18\lambda = 0 \cr & \Rightarrow \lambda = - \frac{1}{2} \cr & \therefore \,\vec r{\text{ is 3}}\hat j - \hat k \cr} $$
Since $${\hat r}$$ is a unit vector, $$\therefore \,\,\hat r = \frac{{{\text{3}}\hat j - \hat k}}{{\sqrt {10} }}$$

Let the given vectors be $$\overrightarrow {AB} = 3\hat i + 5\hat j + 4\hat k$$     and $$\overrightarrow {AC} = 5\hat i - 5\hat j + 2\hat k$$

Let $$AM$$  be the median through $$A$$
$$\eqalign{ & \therefore \,\overrightarrow {AM} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) \cr & = \frac{1}{2}\left[ {\left( {3\hat i + 5\hat j + 4\hat k} \right) + \left( {5\hat i - 5\hat j + 2\hat k} \right)} \right] \cr & = \frac{1}{2}\left( {8\hat i + 6\hat k} \right) \cr & = \left( {4\hat i + 3\hat k} \right) \cr} $$
$$\therefore $$   Length of the median $$AM = \sqrt {{4^2} + {3^2}} = 5{\text{ units}}$$

As $$c$$ is coplanar with $$a$$ and $$b,$$  we take,
$$c = \alpha a + \beta b.....(1)$$
where $$\alpha ,\beta $$  are scalars.
As $$c$$ is perpendicular to $$a,\,c.a=0$$
$$ \therefore $$  From (1) we get, $$0 = \alpha \,a.a + \beta \,b.a$$
$$\eqalign{ & \Rightarrow 0 = \alpha \left( 6 \right) + \beta \left( {2 + 2 - 1} \right) = 3\left( {2\alpha + \beta } \right) \Rightarrow \beta = - 2\alpha \cr & {\text{Thus, }}c = \alpha \left( {a - 2b} \right) = \alpha \left( { - 3j + 3k} \right) = 3\alpha \left( { - j + k} \right) \cr & \Rightarrow {\left| {\vec c} \right|^2} = 9{\alpha ^2}\left( {1 + 1} \right) = 18{\alpha ^2}\,\, \Rightarrow 1 = 18{\alpha ^2} \cr & \Rightarrow \alpha = \pm \frac{1}{{3\sqrt 2 }} \cr & \therefore c = \pm \frac{1}{{\sqrt 2 }}\left( { - j + k} \right) \cr} $$
Thus, we may take $$c = \frac{1}{{\sqrt 2 }}\left( { - j + k} \right).$$

$$\eqalign{ & \overrightarrow c ||{\text{ the plane of }}\overrightarrow a {\text{ and }}\overrightarrow b \Rightarrow \overrightarrow c \bot \overrightarrow a \times \overrightarrow b \cr & \therefore \,\overrightarrow a \times \overrightarrow b .\overrightarrow c = 0 \cr & {\text{or }}\left( {\overrightarrow i - 2\overrightarrow j + 3\overrightarrow k } \right) \times \left( {2\overrightarrow i + 3\overrightarrow j - \overrightarrow k } \right).\left\{ {\lambda \overrightarrow i + \overrightarrow j + \left( {2\lambda - 1} \right)\overrightarrow k } \right\} = 0 \cr & {\text{or }}\left( { - 7\overrightarrow i + 7\overrightarrow j + 7\overrightarrow k } \right).\left\{ {\lambda \overrightarrow i + \overrightarrow j + \left( {2\lambda - 1} \right)\overrightarrow k } \right\} = 0 \cr & {\text{or }} - 7\lambda + 7 + 7\left( {2\lambda - 1} \right) = 0 \cr & {\text{or }}\lambda = 0 \cr} $$

Let $$\vec a + 2\vec b = t\vec c$$   and $$\vec b + 3\vec c = s\vec a,$$   where $$t$$ and $$s$$ are scalars. Adding, we get
$$\eqalign{ & \vec a + 3\vec b + 3\vec c = t\vec c + s\vec a \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = t\vec c + s\vec a - \vec b + 3\vec c \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = t\vec c + \left( {\vec b + 3\vec c} \right) - \vec b + 3\vec c \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = \left( {t + 6} \right)\vec c\,\,\,\,\,\,\,\left[ {{\text{using }}s\vec a = \vec b + 3\vec c} \right] \cr & \Rightarrow \vec a + 2\vec b + 6\vec c = \lambda \vec c,{\text{ where }}\lambda = t + 6 \cr} $$

$$\eqalign{ & {\text{Given, }}\overrightarrow {{r_1}} = \lambda \hat i + 2\hat j + \hat k{\text{ and }}\,\overrightarrow {{r_2}} = \hat i + \left( {2 - \lambda } \right)\hat j + 2\hat k \cr & \therefore \,\left| {\overrightarrow {{r_1}} } \right| > \left| {\overrightarrow {{r_2}} } \right| \cr & \Rightarrow \sqrt {{\lambda ^2} + {{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}} > \sqrt {{{\left( 1 \right)}^2} + {{\left( {2 - \lambda } \right)}^2} + {{\left( 2 \right)}^2}} \cr & \Rightarrow {\lambda ^2} + 4 + 1 > 1 + 4 + {\lambda ^2} - 4\lambda + 4 \cr & \Rightarrow 5 > 9 - 4\lambda \cr & \Rightarrow 4\lambda > 4 \cr & \Rightarrow \lambda > 1 \cr} $$

$$\hat a,\,\hat b,\,\hat c$$   are units vectors.
$$\eqalign{ & \therefore \hat a.\hat a = \hat b.\hat b = \hat c.\hat c = 1 \cr & {\text{Now, }}x = {\left| {\hat a - \hat b} \right|^2} + {\left| {\hat b - \hat c} \right|^2} + {\left| {\hat c - \hat a} \right|^2} \cr & = \hat a.\hat a + \hat b.\hat b - 2\hat a.\hat b + \hat b.\hat b + \hat c.\hat c - 2\hat b.\hat c + c.\hat c + \hat a.\hat a - 2\hat c.\hat a \cr & = 6 - 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right).....(1) \cr & {\text{Also}}, \cr & \Rightarrow \left| {\hat a + \hat b + \hat c} \right| \geqslant 0\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\left| {\hat a + \hat b + \hat c} \right|^2} \geqslant 0 \cr & \Rightarrow \hat a.\hat a + \hat b.\hat b + \hat c.\hat c + 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \geqslant 0 \cr & \Rightarrow 3 + 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \geqslant 0 \cr & \Rightarrow 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \geqslant - 3 \cr & \Rightarrow - 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \leqslant 3 \cr & \Rightarrow 6 - 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \leqslant 9.....(2) \cr} $$
From (1) and (2), $$x \leqslant 9$$
$$\therefore \,x$$ does not exceed 9

$$\eqalign{ & {\text{Lines are}}\,\frac{{x - 4}}{2} = \frac{{y + 1}}{{ - 3}} = \frac{z}{6} \cr & \Rightarrow \frac{{x - 4}}{{ - 1}} = \frac{{y + 1}}{{\frac{3}{2}}} = \frac{z}{{ - 3}}\, \to \left( 1 \right) \cr & {\text{and }}\frac{x}{{ - 1}} = \frac{{y - 1}}{{\frac{3}{2}}} = \frac{{z + 1}}{{ - 3}}\, \to \left( 2 \right) \cr & {\text{Clearly both lines are parallel with}} \cr & \overrightarrow {{a_1}} = 4\hat i - \hat j,\,\overrightarrow {{a_2}} = \hat j - \hat k{\text{ and }}\overrightarrow b = - \hat i - \frac{3}{2}\hat j - 3\hat k \cr & {\text{Using shortest distance between parallel lines}} \cr} $$
\[\begin{array}{l} {\rm{distance}} = \frac{{\left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right|}}{{\left| {\overrightarrow b } \right|}}\\ = \frac{{\left| \begin{array}{l} \,\,\,\overrightarrow i \,\,\,\,\,\overrightarrow j \,\,\,\,\,\,\overrightarrow k \\ - 1\,\,\,\,\,\,\frac{3}{2}\,\,\,\, - 3\\ - 4\,\,\,\,\,\,\,2\,\,\,\,\, - 1 \end{array} \right|}}{{\sqrt {\frac{7}{2}} }}\\ = \frac{{\left| {\frac{9}{2}\hat i - 11\hat j + 4\hat k} \right|}}{{\frac{7}{2}}}\\ = \frac{{\sqrt {629} }}{7} \end{array}\]