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

$$\eqalign{ & \overrightarrow d .\overrightarrow a = \lambda \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right] + \mu \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow a } \right] + \nu \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = \lambda \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr & {\text{Similarly, }}\overrightarrow d .\overrightarrow b = \mu \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right]{\text{ and }}\overrightarrow d .\overrightarrow c = \nu \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] \cr & \therefore \,\lambda = \frac{{\overrightarrow d .\overrightarrow a }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}},\,\,\mu = \frac{{\overrightarrow d .\overrightarrow b }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}},\,\,\nu = \frac{{\overrightarrow d .\overrightarrow c }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}} \cr & \therefore \,\overrightarrow d = \frac{{\overrightarrow d .\overrightarrow a }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}}\overrightarrow b \times \overrightarrow c + \frac{{\overrightarrow d .\overrightarrow b }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}}\overrightarrow c \times \overrightarrow a + \frac{{\overrightarrow d .\overrightarrow c }}{{\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]}}\overrightarrow a \times \overrightarrow b \cr & \therefore {\text{ the expression}} = \left| {\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow d } \right| = \left| {\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]} \right|\,\left| {\overrightarrow d } \right| = \left| {\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]} \right| \cr} $$

$$\eqalign{ & 1 = {\left| {\overrightarrow a - \overrightarrow b } \right|^2} = {\left( {\overrightarrow a - \overrightarrow b } \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} - 2\overrightarrow a .\overrightarrow b = 1 + 1 - 2\overrightarrow a .\overrightarrow b \cr & \therefore \,\,\,\overrightarrow a .\overrightarrow b = \frac{1}{2} \cr & {\text{or }}\left| {\overrightarrow a } \right|\,\left| {\overrightarrow b } \right|\cos \,\theta = \frac{1}{2} \cr & {\text{or }}\cos \,\theta = \frac{1}{2} \cr & {\text{or }}\,\theta = \frac{\pi }{3} \cr} $$

$$\eqalign{ & {\text{Here, }}\overrightarrow c = t\overrightarrow a + s\overrightarrow b {\text{ and }}\left| {\overrightarrow c } \right| = \sqrt 3 = \sqrt {1 + {\alpha ^2} + {\beta ^2}} \cr & {\text{Now, }}\overrightarrow i + \alpha \overrightarrow j + \beta \overrightarrow k = t\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) + s\left( {4\overrightarrow i + 3\overrightarrow j + 4\overrightarrow k } \right) \cr & \Rightarrow 1 = t + 4s,\,\,\alpha = t + 3s,\,\,\beta = t + 4s \cr & \therefore {\text{ we get }}3 = 1 + {\alpha ^2} + {\beta ^2}\,\, \Rightarrow 2 = {\left( {t + 3s} \right)^2} + {\left( {t + 4s} \right)^2} \cr & {\text{or }}2 = {\left( {1 - 4s + 3s} \right)^2} + {\left( {1 - 4s + 4s} \right)^2} \cr & {\text{or }}1 = {\left( {1 - s} \right)^2} \cr & \therefore \,1 - s = \pm 1\,\,\,\therefore \,s = 1 \mp 1 = 0,\,2 \cr & \therefore \,t = 1 - 4s = 1,\, - 7 \cr & \therefore \,\,\alpha = t + 3s = 1 + 3.0,\, - 7 + 3.2 = 1,\, - 1\,\,{\text{and}} \cr & \,\,\,\,\,\,\,\,\beta = t + 4s = 1 + 4.0,\, - 7 + 4.2 = 1,\,1 \cr} $$

Diagonal $${d_1},\,\,\overrightarrow {AC} = 3\hat i + \hat j - 2\hat k$$
Diagonal $${d_2},\,\,\overrightarrow {BD} = \hat i - 3\hat j + 4\hat k$$

Area of parallelogram is $$\frac{1}{2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$
Hence area \[ = \frac{1}{2}\left| \begin{array}{l} \hat i\,\,\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\,\hat k\\ 3\,\,\,\,\,\,\,\,\,\,1\,\, - 2\\ 1\,\, - 3\,\,\,\,\,\,\,\,\,4 \end{array} \right|\]
$$\eqalign{ & = \frac{1}{2}\left| {\left[ {\hat i\left( {4 - 6} \right) - \hat j\left( {12 + 2} \right) + \hat k\left( { - 9 - 1} \right)} \right]} \right| \cr & = \frac{1}{2}\left| { - 2\hat i - 14\hat j - 10\hat k} \right| \cr & = \frac{1}{2}\sqrt {4 + 196 + 100} \cr & = \frac{{10\sqrt 3 }}{2} \cr & = 5\sqrt 3 {\text{ square units}} \cr} $$

$$\eqalign{ & \vec a + \vec b + \vec c = 0 \cr & \Rightarrow \vec b + \vec c = - \vec a \cr & \Rightarrow {\left( {\vec b + \vec c} \right)^2} = {\left( {\vec a} \right)^2} \cr & \Rightarrow {5^2} + {3^2} + 2\,\vec b.\vec c = {7^2} \cr & \Rightarrow 2\left| {\vec b} \right|\,\left| {\vec c} \right|\,\cos \,\theta = 49 - 34 = 15 \cr & \Rightarrow 2 \times 5 \times 3\,\cos \,\theta = 15 \cr & \Rightarrow \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \theta = \frac{\pi }{3} = {60^ \circ } \cr} $$

We can consider the two velocities as $$\overrightarrow {{v_1}} = u\hat i{\text{ and }}\overrightarrow {{v_2}} = \left( {ft\,\cos \,\alpha } \right)\hat i + \left( {ft\,\sin \,\alpha } \right)\hat j$$

$$\therefore $$  Relative velocity of second with respect to first
$$\eqalign{ & \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}} = \left( {ft\,\cos \,\alpha - u} \right)\hat i + \left( {ft\,\sin \,\alpha } \right)\hat j \cr & \Rightarrow {\left| {\overrightarrow v } \right|^2} = {\left( {ft\,\cos \,\alpha - u} \right)^2} + {\left( {ft\,\sin \,\alpha } \right)^2} \cr & \Rightarrow {\left| {\overrightarrow v } \right|^2} = {f^2}{t^2} + {u^2} - 2uft\,\cos \,\alpha \cr} $$
For $$\left| {\overrightarrow v } \right|$$ to be min we should have
$$\eqalign{ & \frac{{d{{\left| v \right|}^2}}}{{dt}} = 0 \cr & \Rightarrow 2{f^2}t - 2uf\,\cos \,\alpha = 0 \cr & \Rightarrow t = \frac{{u\,\cos \,\alpha }}{f} \cr} $$
Also, $$\frac{{{d^2}{{\left| v \right|}^2}}}{{d{t^2}}} = 2{f^2} = + ve$$
$$\therefore \,{\left| v \right|^2}$$  and hence $$\left| v \right|$$ is least at the time $$\frac{{u\,\cos \,\alpha }}{f}$$

Any point on the first line is $$\left( {4r + k,\,2r + 1,\,r - 1} \right),$$     and any point on the second line is $$\left( {r' + k + 1,\, - r',\,2r' + 1} \right).$$      The lines are intersecting if $$4r + k = r' + k + 1,\,2r + 1 = - r',\,r - 1 = 2r' + 1$$          for some $$r$$ and $$r' \Rightarrow 4r - r' = 1,\,2r + r' = - 1,\,r - 2r' = 2.$$
Now, $$4r - r' = 1,\,2r + r' = - 1\,\, \Rightarrow r = 0,\,r' = - 1$$         which satisfy $$r - 2r' = 2.$$
This is true for all $$k.$$

Centroid $$G = \left( {\frac{{2 - 1 + \lambda }}{3},\,\frac{{3 + 3 + 5}}{3},\,\frac{{5 + 2 + \mu }}{3}} \right).$$
Direction ratios of $$AG$$  are $$2 - \frac{{1 + \lambda }}{3},\,3 - \frac{{11}}{3},\,5 - \frac{{7 + \mu }}{3},{\text{ i}}{\text{.e}}{\text{., }}\frac{{5 - \lambda }}{3},\,\frac{{ - 2}}{3},\,\frac{{8 - \mu }}{3}.$$
As $$AG$$  is equally inclined with the axes, the direction ratios are $$1,\,1,\,1$$   also.
$$\therefore \frac{{\frac{{5 - \lambda }}{3}}}{1} = \frac{{\frac{{ - 2}}{3}}}{1} = \frac{{\frac{{8 - \mu }}{3}}}{1} \Rightarrow 5 - \lambda = - 2 = 8 - \mu $$

For the first line, $$x - b = ay,\,z - d = cy\,\,\,\,\, \Rightarrow \frac{{x - b}}{a} = \frac{y}{1} = \frac{{z - d}}{c}.$$
For the second line, $$x - b' = a'y,\,z - d' = c'y\,\,\, \Rightarrow \frac{{x - b'}}{{a'}} = \frac{y}{1} = \frac{{z - d'}}{{c'}}.$$
These lines are perpendicular $$ \Rightarrow \,a.a' + 1.1 + c.c' = 0.$$

Given image of the point $$P\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   by the plane $$lx + my + nz = 0$$    is the point $$Q\left( {\alpha ',\,\beta ',\,\gamma '} \right)$$
But $$lx + my + nz = 0$$    passes through origin.
Let $$O$$ be the origin.
Therefore, $$O$$ is equidistant from $$P$$ and $$Q$$
$$\eqalign{ & \Rightarrow OP = OQ \cr & \Rightarrow O{P^2} = O{Q^2} \cr & \therefore \,{\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} \cr} $$
Here, option B is correct.