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

$$\eqalign{ & {\text{Let}}\,\,\vec d = x\,\hat i\, + y\,\hat j + z\,\hat k \cr & {\text{where}}\,\,{x^2} + {y^2} + {z^2}.....(1) \cr & \left( {\vec d\,\,{\text{being}}\,{\text{unit}}\,{\text{vector}}} \right)\,\,\therefore \vec a.\vec d = 0 \cr & \Rightarrow x - y = 0\,\,\,\, \Rightarrow x = y.....(2) \cr} $$
\[\left[ {\vec b\,\vec c\,\vec d} \right] = 0 \Rightarrow \left| \begin{array}{l} \,\,\,\,\,\,0\,\,\,\,1\,\,\,\, - 1\\ - 1\,\,\,\,0\,\,\,\,1 = 0\\ \,\,\,\,\,\,x\,\,\,\,y\,\,\,\,z \end{array} \right|\]
$$\eqalign{ & \Rightarrow x + y + z = 0 \cr & \Rightarrow 2x + z = 0\,\,\,\,\,\,\,\,\left( {{\text{using }}\,(2)} \right) \cr & \Rightarrow z = - 2x.....(3) \cr & {\text{From (1),}}\,{\text{(2) and (3)}} \cr & {x^2} + {x^2} + 4{x^2} = 1 \Rightarrow x = \pm \frac{1}{{\sqrt 6 }} \cr & \therefore d = \pm \left( {\frac{1}{{\sqrt 6 }}\hat i + \frac{1}{{\sqrt 6 }}\hat j - \frac{2}{{\sqrt 6 }}\hat k} \right) = \pm \left( {\frac{{\hat i + \hat j - 2\hat k}}{{\sqrt 6 }}} \right) \cr} $$

Here,
$$\eqalign{ & 3\overrightarrow i + 2\overrightarrow j - 5\overrightarrow k = \lambda \left( {2\overrightarrow i - \overrightarrow j + \overrightarrow k } \right) + \mu \left( {\overrightarrow i + 3\overrightarrow j - 2\overrightarrow k } \right) + \nu \left( { - 2\overrightarrow i + \overrightarrow j - 3\overrightarrow k } \right) \cr & \Rightarrow 3 = 2\lambda + \mu - 2\nu ,\,\,2 = - \lambda + 3\mu + \nu ,\,\, - 5 = \lambda - 2\mu - 3\nu \cr} $$
Solving these, $$\mu = 1,\,\nu = 2,\,\lambda = 3.$$    Hence, $$\mu ,\,\frac{\lambda }{2},\,\nu $$   are in AP.

\[\overrightarrow b \times \overrightarrow c = \left| \begin{array}{l} \hat i\,\,\,\hat j\,\,\,\hat k\\ 1\,\,\,\,1\,\,\,\,2\\ 1\,\,\,\,3\,\,\,3 \end{array} \right| = - 3\hat i - \hat j + 2\hat k\]
If $$\theta $$ is the angle between $$\overrightarrow a $$ and the plane containing $$\overrightarrow b $$ and $$\overrightarrow c ,$$ then
$$\eqalign{ & \cos \left( {{{90}^ \circ } - \theta } \right) = \left| {\frac{{\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b \times \overrightarrow c } \right|}}} \right| \cr & \Rightarrow \cos \left( {{{90}^ \circ } - \theta } \right) = \frac{1}{{\sqrt {14} }}.\frac{1}{{\sqrt {14} }}\left| {\left( { - 9 - 2 + 2} \right)} \right| \cr & \Rightarrow \cos \left( {{{90}^ \circ } - \theta } \right) = \frac{9}{{14}} \cr & \Rightarrow \sin \,\theta = \frac{9}{{14}} \cr & \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{9}{{14}}} \right) \cr} $$

Work done $$ = \left| {\left( {\overrightarrow P + \overrightarrow Q + \overrightarrow R } \right).\overrightarrow {AB} } \right| = \left| {\left( {2\overrightarrow i - 3\overrightarrow j + 5\overrightarrow k } \right).\left( {2\overrightarrow i + 4\overrightarrow j - \overrightarrow k } \right)} \right| = 13$$

Clearly $$\vec a = - \frac{8}{7}\vec c$$
$$ \Rightarrow \vec a\,||\,\vec c$$   and are opposite in direction
$$\therefore $$ Angle between $${\vec a}$$ and $${\vec c}$$ is $$\pi $$

$$\eqalign{ & {\text{Let }}\overrightarrow b = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k \cr & \overrightarrow a .\overrightarrow b = 3\,\,\,\, \Rightarrow x + y + z = 3 \cr & \overrightarrow a \times \overrightarrow b = \overrightarrow c \,\,\, \Rightarrow \left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) \times \left( {x\overrightarrow i + y\overrightarrow j + z\overrightarrow k } \right) = \overrightarrow j - \overrightarrow k \cr & {\text{or }}\left( {z - y} \right)\overrightarrow i + \left( {x - z} \right)\overrightarrow j + \left( {y - x} \right)\overrightarrow k = \overrightarrow j - \overrightarrow k \cr & \Rightarrow \,z - y = 0,\,x - z = 1,\,y - x = - 1 \cr} $$
Solving the four equations in $$x,\,y,\,z,$$   we get $$x = \frac{5}{3},\,y = \frac{2}{3},\,z = \frac{2}{3}.$$

$$\eqalign{ & \left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow b + \overrightarrow c } \right) = \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c \cr & \therefore \,\left[ {\overrightarrow a + \overrightarrow b \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow c + \overrightarrow a } \right] = \left( {\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right).\left( {\overrightarrow c .\overrightarrow a } \right) \cr & = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] + \left[ {\overrightarrow b \,\overrightarrow c \,\overrightarrow a } \right]\left( {\because \,\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = 0{\text{ if two vectors are equal}}} \right) \cr & = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] + \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] \cr & = 2\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] \cr} $$

$$\eqalign{ & \vec c = \vec b \times \vec a\,\, \Rightarrow \vec b.\vec c = \vec b.\left( {\vec b \times \vec a} \right)\,\, \Rightarrow \vec b.\vec c = 0 \cr & \Rightarrow \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right).\left( {\hat i - \hat j - \hat k} \right) = 0, \cr & {\text{where }}\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k \cr & {b_1} + {b_2} + {b_3} = 0.....({\text{i}}) \cr & {\text{and }}\vec a.\vec b = 3\,\, \Rightarrow \left( {\hat j - \hat k} \right).\left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) = 3 \cr & \Rightarrow {b_2} - {b_3} = 3 \cr} $$
From equation (i)
$$\eqalign{ & {b_1} = {b_2} + {b_3} = \left( {3 + {b_3}} \right) + {b_3} = 3 + 2{b_3} \cr & \vec b = \left( {3 + 2{b_3}} \right)\hat i + \left( {3 + {b_3}} \right)\hat j + {b_3}\hat k \cr} $$
From the option given, it is clear that $${b_3}$$ equal to either 2 or $$-2$$
If $${b_3} = 2,$$   then $$\vec b = 7\hat i + 5\hat j + 2\hat k$$    which is not possible
If $${b_3} = - 2,$$   then $$\vec b = - \hat i + \hat j - 2\hat k$$

If three vectors are co-planar
\[ \Rightarrow \left| \begin{array}{l} \alpha \,\,\,\alpha \,\,\,\gamma \\ 1\,\,\,\,\,0\,\,\,\,\,1\\ \gamma \,\,\,\,\gamma \,\,\,\,\beta \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \alpha \left[ {0 - \gamma } \right] - \alpha \left[ {\beta - \gamma } \right] + \gamma \left[ {\gamma - 0} \right] = 0 \cr & \Rightarrow - \alpha \gamma - \alpha \beta + \alpha \gamma + {\gamma ^2} = 0 \cr & \Rightarrow {\gamma ^2} = \alpha \beta \cr & \Rightarrow {\text{So, }}\alpha ,\,\gamma ,\,\beta {\text{ are in G}}{\text{.P}}{\text{.}} \cr} $$

Since $$\overrightarrow {{r_1}} ,\,\overrightarrow {{r_2}} $$  and $$\overrightarrow {{r_3}} $$ are the position vector of three collinear points. Thus, $$\overrightarrow {{r_3}} $$ is the position vector of the point which divides the joining of points whose position vectors are $$\overrightarrow {{r_1}} $$ and $$\overrightarrow {{r_2}} $$ in the ratio $$m : n.$$
$$\eqalign{ & {\text{So, }}\overrightarrow {{r_3}} = \frac{{m\overrightarrow {{r_1}} + n\overrightarrow {{r_2}} }}{{m + n}} \cr & {\text{But as given, }}\overrightarrow {{r_3}} = m\overrightarrow {{r_1}} + n\overrightarrow {{r_2}} \cr & {\text{So, }}\frac{{\overrightarrow {m{r_1}} + \overrightarrow {n{r_2}} }}{{m + n}} = m{r_1} + n{r_2} \cr & \Rightarrow m + n = 1 \cr} $$