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

Given that $$\vec a,\,\vec b,\,\vec c$$   are non coplanar
$$\therefore \left[ {\vec a\,\vec b\,\vec c} \right] \ne 0$$
Also $$\vec p = \frac{{\vec b \times \vec c}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}},\,\vec q = \frac{{\vec c \times \vec a}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}},\,\,\vec r = \frac{{\vec a \times \vec b}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}.....(1)$$
$$\eqalign{ & {\text{Now, }}\left( {\vec a + \vec b} \right).\vec p + \left( {\vec b + \vec c} \right).\vec q + \left( {\vec c + \vec a} \right).\vec r \cr & = \left( {\vec a + \vec b} \right).\frac{{\vec b \times \vec c}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} + \left( {\vec b + \vec c} \right).\frac{{\vec c \times \vec a}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} + \left( {\vec c + \vec a} \right).\frac{{\vec a \times \vec b}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} \cr & = \frac{{\vec a.\vec b \times \vec c}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} + \frac{{\vec b.\vec c \times \vec a}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} + \frac{{\vec c.\vec a \times \vec b}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} \cr & \left[ {{\text{Using }}\,\vec b.\vec b \times \vec c = \vec c.\vec c \times \vec a = \vec a.\vec a \times \vec b = 0} \right] \cr & = \frac{{\left[ {\vec a\,\vec b\,\vec c} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} + \frac{{\left[ {\vec a\,\vec b\,\vec c} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} + \frac{{\left[ {\vec a\,\vec b\,\vec c} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}} \cr & = 1 + 1 + 1 \cr & = 3 \cr} $$

$$a,\,b,\,c$$  are in HP $$ \Rightarrow \frac{1}{a},\,\frac{1}{b},\,\frac{1}{c}$$    are in A.P.
$$ \Rightarrow \frac{2}{b} = \,\frac{1}{a} + \,\frac{1}{c}\,\,\,\,\, \Rightarrow \frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$$
$$\therefore \frac{x}{a} + \frac{y}{a} + \frac{1}{c} = 0$$     passes through $$\left( {1,\, - 2} \right)$$

$$\eqalign{ & {\text{Here }}\left( {\overrightarrow p .\overrightarrow p } \right)\left( {\overrightarrow x - \overrightarrow q } \right) - \left\{ {\overrightarrow p .\left( {\overrightarrow x - \overrightarrow q } \right)} \right\}\overrightarrow p + \left( {\overrightarrow q .\overrightarrow q } \right)\left( {\overrightarrow x - \overrightarrow r } \right) - \left\{ {\overrightarrow q .\left( {\overrightarrow x - \overrightarrow r } \right)} \right\}\overrightarrow q + \left( {\overrightarrow r .\overrightarrow r } \right)\left( {\overrightarrow x - \overrightarrow p } \right) - \left\{ {\overrightarrow r .\left( {\overrightarrow x - \overrightarrow p } \right)} \right\}\overrightarrow r = 0 \cr & {\text{or }}{\lambda ^2}\left( {\overrightarrow x - \overrightarrow q + \overrightarrow x - \overrightarrow r + \overrightarrow x - \overrightarrow p } \right) = \left\{ {\overrightarrow p .\left( {\overrightarrow x - \overrightarrow q } \right)} \right\}\overrightarrow p + \left\{ {\overrightarrow q .\left( {\overrightarrow x - \overrightarrow r } \right)} \right\}\overrightarrow q + \left\{ {\overrightarrow r .\left( {\overrightarrow x - \overrightarrow p } \right)} \right\}\overrightarrow r , \cr & {\text{where }}\left| {\overrightarrow p } \right| = \left| {\overrightarrow q } \right| = \left| {\overrightarrow r } \right| = \lambda \cr & {\text{or }}{\lambda ^2}\left\{ {3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)} \right\} = \left( {\overrightarrow p .\overrightarrow x } \right)\overrightarrow p + \left( {\overrightarrow q .\overrightarrow x } \right)\overrightarrow q + \left( {\overrightarrow r .\overrightarrow x } \right)\overrightarrow r , \cr & {\text{because }}\overrightarrow p ,\,\overrightarrow q ,\,\overrightarrow r {\text{ are mutually perpendicular}}{\text{.}} \cr & {\text{Let }}\overrightarrow x = \alpha \overrightarrow p + \beta \overrightarrow q + \gamma \overrightarrow r {\text{. Then}} \cr & \,\,\,\,\,\,\,\,\,\,\,\overrightarrow p .\overrightarrow x = \alpha {\left| {\overrightarrow p } \right|^2} = \alpha {\lambda ^2},\,\overrightarrow q .\overrightarrow x = \beta {\left| {\overrightarrow q } \right|^2} = \beta {\lambda ^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\overrightarrow r .\overrightarrow x = \gamma {\left| {\overrightarrow r } \right|^2} = \gamma {\lambda ^2} \cr & \therefore \,{\lambda ^2}\left\{ {3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)} \right\} = {\lambda ^2}\overrightarrow x \,\,\,\,{\text{or }}3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right) = \overrightarrow x \cr} $$

$$\overrightarrow a + t\overrightarrow b = \left( {2 - t} \right)\hat i + \left( {2 + 2t} \right)\hat j + \left( {3 + t} \right)\hat k$$
$$\left( {\overrightarrow a + t\overrightarrow b } \right)$$   and $$\overrightarrow c $$ is perpendicular. Therefore,
$$\eqalign{ & \left( {\overrightarrow a + t\overrightarrow b } \right).\overrightarrow c = 0 \cr & \Rightarrow 3\left( {2 - t} \right) + 2 + 2t = 0 \cr & \Rightarrow 6 - 3t + 2t + 2 = 0 \cr & \Rightarrow t = 8 \cr} $$

Given that $${\vec a}$$ and $${\vec b}$$ are two unit vectors
$$\therefore \left| {\vec a} \right| = 1{\text{ and }}\left| {\vec b} \right| = 1$$
Also, given that ($$\left( {\vec a + 2\vec b} \right) \bot \left( {5\vec a - 4\vec b} \right)$$
$$\eqalign{ & \Rightarrow \left( {\vec a + 2\vec b} \right).\left( {5\vec a - 4\vec b} \right) = 0 \cr & \Rightarrow 5{\left| {\vec a} \right|^2} - 8{\left| {\vec b} \right|^2} - 4\vec a.\vec b + 10\vec b.\vec a = 0 \cr & \Rightarrow 5 - 8 + 6\vec a.\vec b = 0 \cr & \Rightarrow 6\left| {\vec a} \right|\,\left| {\vec b} \right|\,\cos \,\theta = 3 \cr} $$
[where $$\theta $$ is the angle between $${\vec a}$$ and $${\vec b}$$ ]
$$ \Rightarrow \cos \,\theta = \frac{1}{2}\,\,\,\,\, \Rightarrow \theta = {60^ \circ }$$

$$\eqalign{ & \because \,\overrightarrow {AB} = \overrightarrow {ED} {\text{ and }}\overrightarrow {AF} = \overrightarrow {CD} , \cr & {\text{So, }}\,\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} \cr & = \overrightarrow {ED} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {CD} \cr & = \left( {\overrightarrow {AC} + \overrightarrow {CD} } \right) + \left( {\overrightarrow {AE} + \overrightarrow {ED} } \right) + \overrightarrow {AD} \cr & = \overrightarrow {AD} + \overrightarrow {AD} + \overrightarrow {AD} \cr & = 3\overrightarrow {AD} \cr & \therefore \,k = 3 \cr} $$

$$\eqalign{ & {\text{As given :}} \cr & \overrightarrow a = \hat i + 2\hat j - 3\hat k{\text{ and }}\overrightarrow b = 3\hat i - \hat j + \lambda \hat k \cr & \overrightarrow a + \overrightarrow b = \hat i + 2\hat j - 3\hat k + 3\hat i - \hat j + \lambda \hat k \cr & \overrightarrow a + \overrightarrow b = 4\hat i + \hat j + \left( {\lambda - 3} \right)\hat k \cr & {\text{and}} \cr & \overrightarrow a - \overrightarrow b = \hat i + 2\hat j - 3\hat k - 3\hat i + \hat j - \lambda \hat k \cr & \overrightarrow a - \overrightarrow b = - 2\hat i + 3\hat j - \left( {3 + \lambda } \right)\hat k \cr & \left( {\overrightarrow a + \overrightarrow b } \right){\text{is perpendicular to}}\left( {\overrightarrow a - \overrightarrow b } \right) \cr & \Rightarrow \left( {\overrightarrow a + \overrightarrow b } \right).\left( {\overrightarrow a - \overrightarrow b } \right) = 0 \cr & \Rightarrow \left\{ {4\hat i + \hat j + \left( {\lambda - 3} \right)\hat k} \right\}\left\{ { - 2\hat i + 3\hat j - \left( {3 - \lambda } \right)\hat k} \right\} = 0 \cr & \Rightarrow - 8 + 3 + \left( {{3^2} - {\lambda ^2}} \right) = 0 \cr & \Rightarrow 4 - {\lambda ^2} = 0 \cr & \Rightarrow \lambda = \pm 2 \cr} $$

We have
$$\eqalign{ & \hat i \times \left( {\overrightarrow A \times \hat i} \right) + \hat j \times \left( {\overrightarrow A \times \hat j} \right) + \hat k \times \left( {\overrightarrow A \times \hat k} \right) \cr & \hat i \times \left( {\overrightarrow A \times \hat i} \right) = \left( {\hat i.\hat i} \right)\overrightarrow A - \left( {\hat i.\overrightarrow A } \right)\hat i = \overrightarrow A - \left( {\hat i.\overrightarrow A } \right)\hat i......\left( {\text{i}} \right) \cr & \hat j \times \left( {\overrightarrow A \times \hat j} \right) = \left( {\hat j.\hat j} \right)\overrightarrow A - \left( {\hat j.\overrightarrow A } \right)\hat j = \overrightarrow A - \left( {\hat j.\overrightarrow A } \right)\hat j......\left( {{\text{ii}}} \right) \cr & {\text{and }}\hat k \times \left( {\overrightarrow A \times \hat k} \right) = \left( {\hat k.\hat k} \right)\overrightarrow A - \left( {\hat k.\overrightarrow A } \right)\hat k = \overrightarrow A - \left( {\hat k.\overrightarrow A } \right)\hat k......\left( {{\text{iii}}} \right) \cr} $$
Now, equation $$\left( {\text{i}} \right) + $$  equation $$\left( {\text{ii}} \right) + $$  equation $$\left( {\text{iii}} \right) :$$
$$\eqalign{ & \hat i \times \left( {\overrightarrow A \times \hat i} \right) + \hat j \times \left( {\overrightarrow A \times \hat j} \right) + \hat k \times \left( {\overrightarrow A \times \hat k} \right) \cr & = 3\overrightarrow A - \left[ {\left( {\hat i.\overrightarrow A } \right)\hat i + \left( {\hat j.\overrightarrow A } \right)\hat j + \left( {\hat k.\overrightarrow A } \right)\hat k} \right] \cr & = 3\overrightarrow A - \overrightarrow A \cr & = 2\overrightarrow A \cr} $$

$$\vec a,\,\vec b,\,\vec c$$   are unit coplanar vectors, $$2\vec a - \vec b,\,2\vec b - \vec c$$   and $$2\vec c - \vec a$$  are also coplanar vectors being linear combination of.
Thus, $$\left[ {2\vec a - \vec b\,\,2\vec b - \vec c\,\,\,2\vec c - \vec a} \right] = 0$$

$$\eqalign{ & \left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = \left| {\vec a \times \vec b} \right|\left| {\vec c} \right|\,\sin \,{30^ \circ } \cr & = \frac{1}{2}\left| {\vec a \times \vec b} \right|\left| {\vec c} \right|.....(1) \cr & {\text{we have, }}\vec a = 2\hat i + \hat j - 2\hat k{\text{ and }}\vec b = \hat i + \hat j \cr & \Rightarrow \vec a \times \vec b = 2\hat i - 2\hat j + \hat k\,\,\,\,\, \Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt 9 = 3 \cr & {\text{Also given }}\left| {\vec c - \vec a} \right| = 2\sqrt 2 \cr & \Rightarrow {\left| {\vec c - \vec a} \right|^2} = 8 \cr & \Rightarrow \left( {\vec c - \vec a} \right).\left( {\vec c - \vec a} \right) = 8 \cr & \Rightarrow {\left| {\vec c} \right|^2} - \vec c.\vec a - \vec a.\vec c + {\left| {\vec a} \right|^2} = 8 \cr & {\text{As }}\left| {\vec a} \right| = 3{\text{ and }}\vec a.\vec c = \left| {\vec c} \right|,{\text{we get}} \cr & \Rightarrow {\left| {\vec c} \right|^2} - 2\left| {\vec c} \right| + 1 = 0 \cr & \Rightarrow \,{\left( {\left| {\vec c} \right| - 1} \right)^2} = 0 \cr & \Rightarrow \left| {\vec c} \right| = 1 \cr} $$
Substituting values of $$\left| {\vec a \times \vec b} \right|$$  and $$\left| {\vec c} \right|$$ in (1), we get
$$\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = \frac{1}{2} \times 3 \times 1 = \frac{3}{2}$$