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

$$\eqalign{ & {\text{If }}\overrightarrow a = 2\hat i - 2\hat j + \hat k{\text{ and }}\overrightarrow c = - \hat i + 2\hat k \cr & \left| {\overrightarrow c } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {2^2}} = \sqrt {1 + 4} = \sqrt 5 \cr & \left| {\overrightarrow c } \right|.\overrightarrow a = \sqrt 5 .\left( {2\hat i - 2\hat j + \hat k} \right) \cr & \therefore \,\left| {\overrightarrow c } \right|.\overrightarrow a = 2\sqrt 5 \hat i - 2\sqrt 5 \hat j + \sqrt 5 \hat k \cr} $$

$$\eqalign{ & \left[ {\lambda \left( {\vec a + \vec b} \right){\lambda ^2}\vec b\,\lambda \vec c} \right] = \left[ {\vec a\,\vec b + \vec c\,\vec b} \right] \cr & \Rightarrow {\lambda ^4}\left[ {\vec a + \vec b\,\,\vec b\vec c} \right] = \left[ {\vec a\,\vec b + \vec c\,\vec b} \right] \cr & \Rightarrow {\lambda ^4}\left\{ {\left[ {\vec a\,\vec b\,\vec c} \right] + \left[ {\vec b\,\vec b\,\vec c} \right]} \right\} = \left[ {\vec a\,\vec b\,\vec b} \right] + \left[ {\vec a\,\vec c\,\vec b} \right] \cr & \Rightarrow {\lambda ^4}\left[ {\vec a\,\vec b\,\vec c} \right] = - \left[ {\vec a\,\vec b\,\vec c} \right] \cr & \Rightarrow {\lambda ^4} = - 1 \cr & \Rightarrow \lambda \,\,\,{\text{has no real values}} \cr} $$

The velocity vector $$ = \overrightarrow i + \frac{5}{4}\overrightarrow j + \frac{1}{2}\overrightarrow k $$
$$\therefore $$  velocity $$ = \left| {\overrightarrow i + \frac{5}{4}\overrightarrow j + \frac{1}{2}\overrightarrow k } \right| = \sqrt {{1^2} + {{\left( {\frac{5}{4}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \frac{{3\sqrt 5 }}{4}$$

$$\eqalign{ & \overrightarrow {AP} + \overrightarrow {PH} = \overrightarrow {AH} ,\,\overrightarrow {BP} + \overrightarrow {CP} = 2\overrightarrow {MP} \cr & \therefore \,\overrightarrow {AP} + \overrightarrow {BP} + \overrightarrow {CP} + \overrightarrow {PH} = \overrightarrow {AH} + 2\left( {\overrightarrow {MO} + \overrightarrow {OP} } \right) \cr & = 2\overrightarrow {OM} + 2\overrightarrow {MO} + 2\overrightarrow {OP} \,\,\,\,\left( {\because \,AH = 2OM} \right) \cr & = 2\overrightarrow {OP} \cr & \therefore {\text{ the required force}} = 2\overrightarrow {PO} . \cr} $$

Let the given position vectors be of point $$A,\,B$$  and $$C$$ respectively, then
$$\eqalign{ & \left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( {\beta - \alpha } \right)}^2} + {{\left( {\,\gamma - \,\beta } \right)}^2} + {{\left( {\alpha - \,\gamma } \right)}^2}} \cr & \left| {\overrightarrow {BC} } \right| = \sqrt {{{\left( {\gamma - \beta } \right)}^2} + {{\left( {\alpha - \,\gamma } \right)}^2} + {{\left( {\,\alpha - \,\beta } \right)}^2}} \cr & \left| {\overrightarrow {CA} } \right| = \sqrt {{{\left( {\alpha - \,\gamma } \right)}^2} + {{\left( {\beta - \alpha } \right)}^2} + {{\left( {\,\gamma - \,\beta } \right)}^2}} \cr & \left| {\overrightarrow {AB} } \right| = \left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {CA} } \right| \cr} $$
$$ \Rightarrow \,\,\Delta ABC$$   is an equilateral $$\Delta .$$

Let $$\overrightarrow c $$ is the unit vector perpendicular to both the vectors $$\overrightarrow a $$ and $$\overrightarrow b .$$
So, A unit vector which is perpendicular to both the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$\frac{{\left( {\overrightarrow a \times \overrightarrow b } \right)}}{{\left| {\overrightarrow a \times \overrightarrow b } \right|}}$$

From the question, the lines are not parallel.
Any point on the first line is $$\left( {\alpha ' + \alpha r,\,\beta ' + \beta r,\,\gamma ' + \gamma r} \right).$$       It is on the other line if
$$\eqalign{ & \frac{{\alpha ' + \alpha r - \alpha }}{{\alpha '}} = \frac{{\beta ' + \beta r - \beta }}{{\beta '}} = \frac{{\gamma ' + \gamma r - \gamma }}{{\gamma '}} \cr & \Rightarrow 1 + \frac{{\alpha \left( {r - 1} \right)}}{{\alpha '}} = 1 + \frac{{\beta \left( {r - 1} \right)}}{{\beta '}} = 1 + \frac{{\gamma \left( {r - 1} \right)}}{{\gamma '}} \cr & \Rightarrow \frac{{\alpha \left( {r - 1} \right)}}{{\alpha '}} = \frac{{\beta \left( {r - 1} \right)}}{{\beta '}} = \frac{{\gamma \left( {r - 1} \right)}}{{\gamma '}} \cr & \Rightarrow r - 1 = 0\,\,\,\,\,\,\left( {\because \frac{\alpha }{{\alpha '}} = \frac{\beta }{{\beta '}} = \frac{\gamma }{{\gamma '}}{\text{ is not true}}} \right) \cr} $$
$$\therefore $$  the point of intersection is $$\left( {\alpha ' + \alpha .1,\,\beta ' + \beta .1,\,\gamma ' + \gamma .1} \right).$$

Let $$\vec a + \vec b + \vec c = \vec r.$$   Then
$$\eqalign{ & \vec a \times \left( {\vec a + \vec b + \vec c} \right) = \vec a \times \vec r \cr & \Rightarrow 0 + \vec a \times \vec b + \vec a \times \vec c = \vec a \times \vec r \cr & \Rightarrow \vec a \times \vec b - \vec c \times \vec a = \vec a \times \vec r \cr & \Rightarrow \vec a \times \vec r = 0 \cr & {\text{Similarly }}\vec b \times \vec r = \vec 0{\text{ & }}\vec c \times \vec r = \vec 0 \cr} $$
Above three conditions will be satisfied for non-zero vectors if and only if $$\vec r = \vec 0$$

As $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$   are noncoplanar, $$\overrightarrow b \times \overrightarrow c ,\,\overrightarrow c \times \overrightarrow a ,\,\overrightarrow a \times \overrightarrow b $$      are also noncoplanar.
So, any vector can be expressed as a linear combination of these vectors.
$$\eqalign{ & {\text{Let }}\overrightarrow a = \lambda \overrightarrow b \times \overrightarrow c + \mu \overrightarrow c \times \overrightarrow a + \nu \overrightarrow a \times \overrightarrow b \cr & \therefore \overrightarrow a .\overrightarrow a = \lambda \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right],\,\,\overrightarrow a .\overrightarrow b = \mu \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right],\,\,\overrightarrow a .\overrightarrow c = \nu \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr & \therefore \,\overrightarrow a = \frac{{\left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow b \times \overrightarrow c }}{{\left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right]}} + \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c \times \overrightarrow a }}{{\left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right]}} + \frac{{\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow a \times \overrightarrow b }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} \cr} $$

$$\eqalign{ & \overrightarrow {CA} = \left( {2 - a} \right)\hat i + 2\hat j\,; \cr & \overrightarrow {CB} = \left( {1 - a} \right)\hat i - 6\hat k \cr & \overrightarrow {CA} .\overrightarrow {CB} = 0 \cr & \Rightarrow \left( {2 - a} \right)\left( {1 - a} \right) = 0 \cr & \Rightarrow a = 2,\,1 \cr} $$