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

$$\eqalign{ & \overrightarrow p = \lambda \left( {\overrightarrow u \times \overrightarrow v } \right) + \mu \left( {\overrightarrow v \times \overrightarrow w } \right) + \nu \left( {\overrightarrow w \times \overrightarrow u } \right) \cr & \Rightarrow \overrightarrow p .\overrightarrow w = \lambda \left( {\overrightarrow u \times \overrightarrow v } \right).\overrightarrow w + \mu \left( {\overrightarrow v \times \overrightarrow w } \right).\overrightarrow w + \nu \left( {\overrightarrow w \times \overrightarrow u } \right).\overrightarrow w \cr & \Rightarrow \lambda \left[ {\overrightarrow u \overrightarrow v \overrightarrow w } \right] + 0 + 0 = \frac{\lambda }{5} \cr & \Rightarrow \lambda = 5\left( {\overrightarrow p .\overrightarrow w } \right) \cr & {\text{Similarly, }}\mu = 5\left( {\overrightarrow p .\overrightarrow u } \right){\text{ and }}\nu = 5\left( {\overrightarrow p .\overrightarrow v } \right) \cr & \therefore \,\lambda + \mu + \nu \cr & = 5\left( {\overrightarrow p .\overrightarrow w } \right) + 5\left( {\overrightarrow p .\overrightarrow u } \right) + 5\left( {\overrightarrow p .\overrightarrow v } \right) \cr & = 5\overrightarrow p \left( {\overrightarrow u + \overrightarrow v + \overrightarrow w } \right) \cr & {\text{Hence, }}\lambda + \mu + \nu {\text{ depends on the vectors}}{\text{.}} \cr} $$

$$\eqalign{ & \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow b \times \overrightarrow c } \right] = \overrightarrow b .\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow c } \right) \cr & = \overrightarrow b .\left\{ {\left( {\overrightarrow c .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow c } \right\} \cr & = \left( {\overrightarrow c .\overrightarrow c } \right)\left( {\overrightarrow b .\overrightarrow b } \right) - {\left( {\overrightarrow b + \overrightarrow c } \right)^2} \cr & \therefore \,\left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow b \times \overrightarrow c } \right] + {\left( {\overrightarrow b + \overrightarrow c } \right)^2} = {\left| {\overrightarrow c } \right|^2}\,{\left| {\overrightarrow b } \right|^2} \cr} $$

$$\eqalign{ & \left| {\overrightarrow b \times \overrightarrow c } \right| = \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\,\sin \,\alpha = \sin \,\alpha \cr & {\text{Now, }}\frac{{\overrightarrow b \times \overrightarrow c }}{{\left| {\overrightarrow b \times \overrightarrow c } \right|}} = \overrightarrow a \,\,\left( {{\text{given}}} \right) \cr & \therefore \,\,\overrightarrow b \times \overrightarrow c = \sin \,\alpha \,\overrightarrow a . \cr} $$

Let the equation of variable plane be $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$    which meets the axes at $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$     and $$C\left( {0,\,0,\,c} \right).$$
$$\therefore $$ Centroid of $$\Delta ABC$$   is $$\left( {\frac{a}{3},\,\frac{b}{3},\,\frac{c}{3}} \right)$$
and it satisfies the relation
$$\eqalign{ & \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k \cr & \Rightarrow \frac{9}{{{a^2}}} + \frac{9}{{{b^2}}} + \frac{9}{{{c^2}}} = k \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{k}{9}.....(1) \cr} $$
Also given that the distance of plane $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$    from (0, 0, 0) is 1 unit.
$$\eqalign{ & \Rightarrow \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }} = 1 \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 1.....(2) \cr} $$
From (1) and (2), we get $$\frac{k}{9} = 1\,\,\,\,\,i.e.,\,\,k = 9$$

$$\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow b = \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow b } \right)\overrightarrow c .$$           Taking dot product with $$\overrightarrow b \times \overrightarrow c ,\,\left( {\overrightarrow a .\overrightarrow b } \right)\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0\,\,\,\,\,\,\,\, \Rightarrow \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0.$$

The position vector of points $$D,\,E,\,F$$   are respectively
$$\frac{{\hat i + \hat j}}{2} + \hat k,\,\hat i + \frac{{\hat k + \hat j}}{2}{\text{ and }}\frac{{\hat i + \hat k}}{2} + \hat j$$
So, position vector of centre of $$\Delta DEF$$
$$\eqalign{ & = \frac{1}{3}\left[ {\frac{{\hat i + \hat j}}{2} + \hat k + \hat i + \frac{{\hat k + \hat j}}{2}{\text{ + }}\frac{{\hat i + \hat k}}{2} + \hat j} \right] \cr & = \frac{2}{3}\left[ {\hat i + \hat j + \hat k} \right] \cr} $$

$$\eqalign{ & \frac{1}{a} = A + \left( {p - 1} \right)D\,; \cr & \frac{1}{b} = A + \left( {q - 1} \right)D\,; \cr & \frac{1}{c} = A + \left( {r - 1} \right)D \cr & \therefore \,q - r = \frac{{c - b}}{{bcD}},\,r - p = \frac{{a - c}}{{acD}},\,p - q = \frac{{b - a}}{{abD}} \cr & \Rightarrow \frac{{q - r}}{a} + \frac{{r - p}}{b} + \frac{{p - q}}{c} = 0 \cr & \Rightarrow \vec u.\vec v = 0 \cr} $$

Area of $$\Delta ABC\,{\bf{:}}$$
\[\begin{array}{l} = \frac{1}{2}\left( {\overrightarrow {AB} \times \overrightarrow {AC} } \right)\\ = \frac{1}{2}\left| \begin{array}{l} \,\,\,\hat i\,\,\,\,\,\hat j\,\,\,\,\hat k\\ - 2\,\,\,\,3\,\,\,\,2\\ - 4\,\,\,\,5\,\,\,\,2 \end{array} \right| \end{array}\]
$$\eqalign{ & = \frac{1}{2}\left[ {\hat i\left( {6 - 10} \right) - \hat j\left( { - 4 + 8} \right) + \hat k\left( { - 10 + 12} \right)} \right] \cr & = \frac{1}{2}\left[ { - 4\hat i - 4\hat j + 2\hat k} \right] \cr & = \frac{1}{2}\sqrt {16 + 16 + 4} \cr & = \frac{1}{2}\sqrt {36} \cr & = \frac{1}{2} \times 6 \cr & = 3{\text{ square units}} \cr} $$

Let $$\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow {k.} $$    Then $$\overrightarrow a .\overrightarrow i = x,\,\overrightarrow a .\overrightarrow j = y,\,\overrightarrow a .\overrightarrow k = z.$$
As $$\overrightarrow a = \left( {\overrightarrow a .\overrightarrow i } \right)\overrightarrow i + \left( {\overrightarrow a .\overrightarrow j } \right)\overrightarrow j + \left( {\overrightarrow a .\overrightarrow k } \right)\overrightarrow k ,$$
$${\left| {\overrightarrow a } \right|^2} = {\left( {\overrightarrow a .\overrightarrow i } \right)^2} + {\left( {\overrightarrow a .\overrightarrow j } \right)^2} + {\left( {\overrightarrow a .\overrightarrow k } \right)^2}{\text{ and }}{\left| {\overrightarrow a } \right|^2} = {\overrightarrow a ^2}$$

Required value $$ = \frac{{\frac{{b.a}}{{\left| b \right|}}}}{{\frac{{a.b}}{{\left| b \right|}}}} = \frac{{\left| a \right|}}{{\left| b \right|}} = \frac{7}{3}.$$