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

If the fourth vertex is $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   then centroid $$ = \left( {\frac{{0 + 6 - 4 + \alpha }}{4},\,\frac{{0 - 5 + 1 + \beta }}{4},\,\frac{{0 - 1 + 3 + \gamma }}{4}} \right) = \left( {1,\, - 2,\,5} \right).$$

$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {\vec a \times \vec b} \right).\left[ {\left( {\vec b \times \vec c} \right) \times \left( {\vec c \times \vec a} \right)} \right] \cr & = \left( {\vec a \times \vec b} \right).\left[ {\left( {\vec b \times \vec c.\vec a} \right)\vec c - \left( {\vec b \times \vec c.\vec c} \right)\vec a} \right] \cr & = \left( {\vec a \times \vec b} \right).\left[ {\left[ {\vec b\,\vec c\,\vec a} \right]\vec c} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\vec b \times \vec c.\vec c = 0} \right] \cr & = \left[ {\vec a\,\vec b\,\vec c} \right].\left( {\vec a \times \vec b.\vec c} \right) \cr & = {\left[ {\vec a\,\vec b\,\vec c} \right]^2} \cr & \left[ {\vec a \times \vec b\,\,\vec b \times \vec c\,\,\vec c \times \vec a} \right] = {\left[ {\vec a\,\vec b\,\vec c} \right]^2} \cr & {\text{So }}\lambda = 1 \cr} $$

$$\eqalign{ & \left( {\overrightarrow a + \overrightarrow b } \right).\overrightarrow c = \left( {\overrightarrow a - \overrightarrow b } \right).\overrightarrow c = 0\,\,\,\,\, \Rightarrow \overrightarrow a .\overrightarrow c = \overrightarrow b .\overrightarrow c = 0 \cr & \therefore \,\overrightarrow c \bot \overrightarrow a {\text{ as well as }}\overrightarrow b \cr & \therefore \,\overrightarrow c ||\overrightarrow a \times \overrightarrow b .{\text{ Hence, }}\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow a = \overrightarrow 0 . \cr} $$

$$\eqalign{ & \overrightarrow {AP} = \overrightarrow {OP} - \overrightarrow {OA} \cr & = \left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) - \left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) = - \overrightarrow i \cr & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \cr & = \left( {\overrightarrow i + 2\overrightarrow j + \overrightarrow k } \right) - \left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) = - \overrightarrow i + \overrightarrow j \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} \cr & = \left( {\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right) - \left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) = - \overrightarrow i + \overrightarrow k \cr & \overrightarrow {AC} \times \overrightarrow {AB} = \left( { - \overrightarrow i + \overrightarrow k } \right) \times \left( { - \overrightarrow i + \overrightarrow j } \right) = - \overrightarrow k - \overrightarrow j - \overrightarrow i \cr} $$
$$\therefore $$  unit vector perpendicular to the plane $$ABC = \frac{{ - \overrightarrow k - \overrightarrow j - \overrightarrow i }}{{\sqrt 3 }}$$
The required distance $$ = \overrightarrow {AP} .\frac{{ - \overrightarrow k - \overrightarrow j - \overrightarrow i }}{{\sqrt 3 }} = \left( { - \overrightarrow i } \right).\frac{{ - \overrightarrow k - \overrightarrow j - \overrightarrow i }}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$$

$$\eqalign{ & {\text{Expression}} = \left( {\overrightarrow i .\overrightarrow i } \right)\overrightarrow a - \left( {\overrightarrow i .\overrightarrow a } \right)\overrightarrow i + \left( {\overrightarrow j .\overrightarrow j } \right)\overrightarrow a - \left( {\overrightarrow j .\overrightarrow a } \right)\overrightarrow j + \left( {\overrightarrow k .\overrightarrow k } \right)\overrightarrow a - \left( {\overrightarrow k .\overrightarrow a } \right)\overrightarrow k \cr & = 3\overrightarrow a - \left\{ {\left( {\overrightarrow i .\overrightarrow a } \right)\overrightarrow i + \left( {\overrightarrow j .\overrightarrow a } \right)\overrightarrow j + \left( {\overrightarrow k .\overrightarrow a } \right)\overrightarrow k } \right\} \cr & = 3\overrightarrow a - \overrightarrow a \cr & = 2\overrightarrow a \cr} $$

The unit vector $$ = \pm \frac{{\overrightarrow a \times \overrightarrow b }}{{\left| {\overrightarrow a \times \overrightarrow b } \right|}} = \pm \frac{{4\overrightarrow i - \overrightarrow j - 3\overrightarrow k }}{{\sqrt {{4^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 3} \right)}^2}} }} = \pm \frac{1}{{\sqrt {26} }}\left( {4\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right).$$
Now, $$\overrightarrow k .\frac{{ \pm 1}}{{\sqrt {26} }}\left( {4\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right) = \frac{{ \pm 1}}{{\sqrt {26} }}\left( { - 3} \right).$$
$$\therefore $$  for acute angle the unit vector $$ = - \frac{1}{{\sqrt {26} }}\left( {4\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right).$$

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

Let $$O$$ be the origin and $$ABCD$$   be the parallelogram.
$$\eqalign{ & \ln \,\Delta ODC, \cr & \overrightarrow {OD} = \overrightarrow {OC} + \overrightarrow {CD} \cr & \overrightarrow {CD} = - \overrightarrow {AB} \cr & {\text{and, }}\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \,\,\left[ {\ln \,\Delta AOB} \right] \cr & = \overrightarrow b - \overrightarrow a \cr & {\text{Thus, }}\overrightarrow {CD} = - \overrightarrow {AB} = \overrightarrow a - \overrightarrow b \cr} $$

$$\eqalign{ & {\text{So, }}\overrightarrow {OD} = \overrightarrow c + \overrightarrow a - \overrightarrow b \cr & \left[ {{\text{since, }}\overrightarrow {OC} = \overrightarrow C {\text{ and }}\overrightarrow {CD} = \overrightarrow a - \overrightarrow b } \right] \cr} $$

$$\eqalign{ & \left| {\left( {\vec a \times \vec b} \right).\vec c} \right| = \left| {\vec a} \right|\left| {\vec b} \right|\left| {\vec c} \right| \cr & \Rightarrow \left| {\left| {\hat a} \right|\left| {\hat b} \right|\sin \,\theta \,\hat n.\,\vec c} \right|\, = \left| {\hat a} \right|\left| {\hat b} \right|\left| {\hat c} \right| \cr & {\text{where }}\theta {\text{ is angle between }}\vec a{\text{ and }}\vec b \cr & \Rightarrow \left| {\hat a} \right|\left| {\hat b} \right|\left| {\hat c} \right|\,\left| {\sin \,\theta \,\cos \,\alpha } \right| = \left| {\hat a} \right|\left| {\hat b} \right|\left| {\hat c} \right| \cr & {\text{where }}\alpha \,{\text{is angle betwen }}\vec c{\text{ and }}\hat n \cr & \Rightarrow \left| {\sin \,\theta \,} \right|\,\left| {\cos \,\alpha } \right| = 1 \cr & \Rightarrow \theta = \frac{\pi }{2}{\text{ and }}\alpha = 0 \cr & \Rightarrow \vec a \bot \vec b{\text{ and }}\vec c\left\| {\hat n} \right. \cr & \Rightarrow \vec a.\vec b = \vec b.\vec c = \vec c.\vec a = 0 \cr} $$

$$\overrightarrow a ||\left( {\overrightarrow b \times \overrightarrow c } \right)\,\,\, \Rightarrow \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = 0\,\,\, \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b = \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c $$
But $${\overrightarrow b }$$ is not collinear with $${\overrightarrow c }.$$ So, $$\overrightarrow a .\overrightarrow c = 0{\text{ and }}\overrightarrow a .\overrightarrow b = 0$$
Now, \[\left( {\overrightarrow a \times \overrightarrow b } \right).\left( {\overrightarrow a \times \overrightarrow c } \right) = \left| \begin{array}{l} \overrightarrow a .\overrightarrow a \,\,\,\,\,\overrightarrow a .\overrightarrow c \\ \overrightarrow b .\overrightarrow a \,\,\,\,\,\overrightarrow b .\overrightarrow c \end{array} \right| = \left| \begin{array}{l} \overrightarrow a .\overrightarrow a \,\,\,\,\,0\\ 0\,\,\,\,\,\,\overrightarrow b .\overrightarrow c \end{array} \right| = {\overrightarrow a ^2}\left( {\overrightarrow b .\overrightarrow c } \right).\]