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

A vector in the plane of $${\vec a}$$ and $${\vec b}$$ is
$$\vec u = \vec a + \lambda \vec b = \left( {1 + \lambda } \right)\hat i + \left( {2 - \lambda } \right)\hat j + \left( {1 + \lambda } \right)\hat k$$
Projection of $${\vec u}$$ on $$\vec c = \frac{1}{{\sqrt 3 }}\,\,\, \Rightarrow \frac{{\vec u.\vec c}}{{\left| {\vec c} \right|}} = \frac{1}{{\sqrt 3 }}$$
$$\eqalign{ & \Rightarrow \vec u.\vec c = 1\,\, \Rightarrow \left| {1 + \lambda + 2 - \lambda - 1 - \lambda } \right| = 1 \cr & \Rightarrow \left| {2 - \lambda } \right| = 1\,\, \Rightarrow \lambda = 1{\text{ or }}3 \cr & \Rightarrow \vec u = 2\hat i + \hat j + 2\hat k{\text{ or 4}}\hat i - \hat j + 4\hat k \cr} $$

Equation of plane passing through given line is given by
$$a\left( {x - 1} \right) + b\left( {y + 1} \right) + cz = 0$$
Also direction ratio of line are $$2,\,-1,\,3$$
$$ \Rightarrow 2a - b + 3c = 0......\left( 1 \right)$$
and plane is parallel to direction with direction $$3,\,4,\,2$$
$$ \Rightarrow 3a + 4b + 2c = 0......\left( 2 \right)$$
Solving equation (1) and (2), we get $$a = \frac{{ - 14c}}{{11}},\,\,b = \frac{{5c}}{{11}}$$
Hence, required place is $$14x - 5y - 11z = 19$$

Any vector $$\overrightarrow r $$ can be represented in terms of three non-coplanar vectors $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c $$ as
$$\overrightarrow r = x\left( {\overrightarrow a \times \overrightarrow b } \right) + y\left( {\overrightarrow b \times \overrightarrow c } \right) + z\left( {\overrightarrow c \times \overrightarrow a } \right)......\left( {\text{i}} \right)$$
Taking dot product with $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c ,$$ respectively, we have
$$x = \frac{{\overrightarrow r .\overrightarrow c }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}},\,y = \frac{{\overrightarrow r .\overrightarrow a }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{\text{ and}}\,z = \frac{{\overrightarrow r .\overrightarrow b }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}$$
From $$\left( {\text{i}} \right),$$ we have
$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]\overrightarrow r = \frac{1}{2}\left( {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right)$$
$$\therefore $$ Area of $$\Delta ABC = \frac{1}{2}\left| {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right| = \left| {\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]\overrightarrow r } \right|$$

$$\eqalign{ & \vec a.\vec b \ne 0,\,\,\,\vec a.\vec d = 0 \cr & {\text{Now, }}\vec b \times \vec c = \vec b \times \vec d \cr & \Rightarrow \vec a \times \left( {\vec b \times \vec c} \right) = \vec a \times \left( {\vec b \times \vec d} \right) \cr & \Rightarrow \left( {\vec a.\vec c} \right)\vec b - \left( {\vec a.\vec b} \right)\vec c = \left( {\vec a.\vec d} \right)\vec b - \left( {\vec a.\vec b} \right)\vec d \cr & \Rightarrow \left( {\vec a.\vec b} \right)\vec d = - \left( {\vec a.\vec c} \right)\vec b + \left( {\vec a.\vec b} \right)\vec c \cr & \Rightarrow \vec d = \vec c - \left( {\frac{{\vec a.\vec c}}{{\vec a.\vec b}}} \right)\vec b \cr} $$

Vectors $$\vec a + 2\vec b + 3\vec c,\,\lambda \vec b + 4\vec c$$     and $$\left( {2\lambda - 1} \right)\vec c$$   are coplanar if \[\left| \begin{array}{l} 1\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\\ 0\,\,\,\,\lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\\ 0\,\,\,\,0\,\,\,\,\,\,2\lambda - 1 \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \lambda \left( {2\lambda - 1} \right) = 0 \cr & \Rightarrow \lambda = 0\,\,{\text{or}}\,\,\frac{1}{2} \cr} $$
$$\therefore $$ Forces are noncoplanar for all $$\lambda ,$$ except $$\lambda = 0,\,\frac{1}{2}$$

$${\text{ar}}\left( {\Delta ABC} \right) = \frac{1}{2}\left| {\overrightarrow b \times \overrightarrow c } \right| = \frac{1}{2}\left| {\overrightarrow b - \overrightarrow c } \right|h,$$         where $$h = $$  the length of the perpendicular.
$$\therefore \,\,h = \frac{{\left| {\overrightarrow b \times \overrightarrow c } \right|}}{{\left| {\overrightarrow b - \overrightarrow c } \right|}}.$$

$$\eqalign{ & \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \frac{1}{2}\overrightarrow b \,\,\,\,\,\, \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c = \frac{1}{2}\overrightarrow b \cr & \therefore \,\overrightarrow a .\overrightarrow c = \frac{1}{2},\,\,\,\,\overrightarrow a .\overrightarrow b = 0 \cr & \therefore \,\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \,\beta = \frac{1}{2},\,\,\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \,\alpha = 0 \cr & {\text{or }}\cos \,\beta = \frac{1}{2},\,\,\cos \,\alpha = 0\,\,\, \Rightarrow \,\beta = \frac{\pi }{3},\,\alpha = \frac{\pi }{2} \cr} $$

Position vectors of vertices $$A,\,B$$  and $$C$$ are $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c .$$

$$\because $$  triangle is equilateral.
$$\therefore $$  Centroid and orthocenter will coincide.
Centroid $$ \equiv $$ orthocenter position vector $$ = \frac{1}{3}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$$
$$\because $$  given in question orthocenter is at origin.
Hence, $$\frac{1}{3}\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0{\text{ or }}\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$

$$\eqalign{ & \,\,\,\,\,\,\,\left\{ {\overrightarrow a \times \left( {\overrightarrow b + \overrightarrow c } \right)} \right\}.\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \cr & = \left( {\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \cr & = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] + \left[ {\overrightarrow a \,\,\overrightarrow c \,\,\overrightarrow b } \right] \cr & = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr & = 0 \cr} $$

Volume of parallelepiped $$ = \left[ {\vec a\,\vec b\,\vec c} \right]$$
\[ = \left| \begin{array}{l} 2\,\,\, - 2\,\,\,\,\,0\\ 1\,\,\,\,\,\,1\,\,\, - 1\\ 3\,\,\,\,\,0\,\,\, - 1 \end{array} \right| = 2\left( { - 1} \right) + 2\left( { - 1 + 3} \right) = 2\]