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

$$\eqalign{ & \overrightarrow {BD} = \overrightarrow a - 3\overrightarrow b ,\,\overrightarrow {AC} = \overrightarrow a + 3\overrightarrow b \cr & \overrightarrow {BD} \times \overrightarrow {AC} = \left( {\overrightarrow a - 3\overrightarrow b } \right) \times \left( {\overrightarrow a + 3\overrightarrow b } \right) = 6\overrightarrow a \times \overrightarrow b \cr & \overrightarrow {OD} \times \overrightarrow {OC} = \left( {\overrightarrow a - 2\overrightarrow b } \right) \times \left( {2\overrightarrow a + 3\overrightarrow b } \right) = 7\overrightarrow a \times \overrightarrow b \cr & \left( {\overrightarrow {BD} \times \overrightarrow {AC} } \right).\left( {\overrightarrow {OD} \times \overrightarrow {OC} } \right) = 42{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} \cr} $$

$$\eqalign{ & \left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow b = \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow b } \right)\overrightarrow a \cr & \therefore \,\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b = \left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow b + \overrightarrow a \cr & \therefore \,\overrightarrow u = \overrightarrow a - \left\{ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow b + \overrightarrow a } \right\} = \overrightarrow b \times \left( {\overrightarrow a \times \overrightarrow b } \right) \cr & \therefore \,\overrightarrow u = \overrightarrow b \times \overrightarrow v \cr & \Rightarrow \left| {\overrightarrow u } \right| = \left| {\overrightarrow b } \right|\left| {\overrightarrow v } \right|{\text{ because }}\overrightarrow b .\overrightarrow v = 0,{\text{ i}}{\text{.e}}{\text{., }}\overrightarrow b \bot \overrightarrow v \cr & \Rightarrow \left| {\overrightarrow u } \right| = \left| {\overrightarrow v } \right|{\text{ because }}\left| {\overrightarrow b } \right| = 1. \cr} $$

$$\eqalign{ & {\text{Let }}\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k .{\text{ Then }}\left| {\overrightarrow a } \right| = 4 = \sqrt {{x^2} + {y^2} + {z^2}} \cr & {\text{Now, }}\frac{{\overrightarrow a .\left( {\overrightarrow i + \overrightarrow j } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow i + \overrightarrow j } \right|}} = \frac{{\overrightarrow a .\left( {\overrightarrow j + \overrightarrow k } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow j + \overrightarrow k } \right|}} = \frac{{\overrightarrow a .\left( {\overrightarrow k + \overrightarrow i } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow k + \overrightarrow i } \right|}}\left( {{\text{from the question}}} \right) \cr & {\text{or }}x + y = y + z = z + x = t\,\,\left( {{\text{say}}} \right) \cr & {\text{Adding, }}2\left( {x + y + z} \right) = 3t\,\,\,\,{\text{or}}\,\,x + y + z = \frac{{3t}}{2} \cr & \therefore \,x = y = z = \frac{t}{2} \cr & \therefore \,4 = \sqrt {{x^2} + {y^2} + {z^2}} \, \Rightarrow 16 = 3.{\left( {\frac{t}{2}} \right)^2} \cr & \therefore t = \pm \frac{8}{{\sqrt 3 }}.{\text{ So, }}\overrightarrow a = \pm \frac{8}{{2\sqrt 3 }}\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) \cr} $$

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

$$\eqalign{ & {\text{Given }}\vec a + \vec b + \vec c = 0\,\,\,\,\,\,\,\,\,\,\left( {{\text{by triangle law}}} \right) \cr & \therefore \vec a \times \left( {\vec a + \vec b + \vec c} \right) = \vec a \times \vec 0 = \vec 0 \cr & \Rightarrow \vec a \times \vec a + \vec a \times \vec b + \vec a \times \vec c = \vec 0 \cr & \Rightarrow \vec a \times \vec b = \vec c \times \vec a\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\vec a \times \vec a = 0} \right] \cr & {\text{Similarly, }}\vec a \times \vec b = \vec b \times \vec c\,; \cr & {\text{Therefore }}\vec a \times \vec b = \vec b \times \vec c = \vec c \times \vec a \cr} $$

$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \overrightarrow i + 2\overrightarrow j - \left( { - \overrightarrow i + 2\overrightarrow k } \right) = 2\overrightarrow i + 2\overrightarrow j - 2\overrightarrow k \cr & \therefore {\text{ vector moment about }}C = \overrightarrow {CA} \times \frac{{10\left( {2\overrightarrow i + 2\overrightarrow j - 2\overrightarrow k } \right)}}{{\sqrt {12} }} \cr & = - \left( {3\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right) \times \frac{{10}}{{\sqrt 3 }}\left( {\overrightarrow i + \overrightarrow j - \overrightarrow k } \right) \cr & = \frac{{10}}{{\sqrt 3 }}\left( {5\overrightarrow i - 5\overrightarrow j } \right) \cr & \therefore {\text{moment}} = \frac{{10}}{{\sqrt 3 }}.\sqrt {{5^2} + {5^2}} = \frac{{50\sqrt 2 }}{{\sqrt 3 }} = \frac{{50\sqrt 6 }}{3}. \cr} $$

Point $$P$$ lies on $${x^2} + 3{y^2} = 3......\left( {\text{i}} \right)$$

Now from the diagram, according to the given conditions,
$$\eqalign{ & AP = AB{\text{ or }}{\left( {x + \sqrt 3 } \right)^2} + {\left( {y - 0} \right)^2} = 4 \cr & {\text{or }}{\left( {x + \sqrt 3 } \right)^2} + {y^2} = 4......\left( {{\text{ii}}} \right) \cr} $$
Solving $$\left( {\text{i}} \right)$$ and $$\left( {\text{ii}} \right),$$ we get $$x = 0$$  and $$y = \pm 1$$
Hence, point $$P$$ has position vector $$ \pm \hat j.$$

$$\eqalign{ & P.V.{\text{ of }}\overrightarrow {AD} = \frac{{\left( {3 + 5} \right)i + \left( {0 - 2} \right)j + \left( {4 + 4} \right)k}}{2}\, \cr & = 4i - j + 4k{\text{ or }}\left| {\overrightarrow {AD} } \right| = \sqrt {16 + 16 + 1} = \sqrt {33} \cr} $$

$$\eqalign{ & {\text{Since }}\vec a \times \left( {\vec b \times \vec c} \right) = \frac{{\vec b + \vec c}}{{\sqrt 2 }} \cr & \therefore \left( {\vec a.\vec c} \right)\vec b - \left( {\vec a.\vec b} \right)\vec c = \frac{1}{{\sqrt 2 }}\vec b + \frac{1}{{\sqrt 2 }}\vec c \cr & \Rightarrow \vec a.\vec c = \frac{1}{{\sqrt 2 }}\,\,\,\,\,\,\,\,\left[ {\because \vec b\,{\text{and }}\vec c\,{\text{are non - coplanar}}} \right] \cr & {\text{and }}\vec a.\vec b = - \frac{1}{{\sqrt 2 }} \Rightarrow \cos \,\theta = - \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \cos \,\frac{{3\pi }}{4} = \cos \,\theta \cr & \Rightarrow \theta = \frac{{3\pi }}{4} \cr} $$

$$\eqalign{ & {\text{Here }}\overrightarrow r .\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0 \cr & \therefore \,\overrightarrow r .\overrightarrow a + \overrightarrow r .\overrightarrow b + \overrightarrow r .\overrightarrow c = 0 \cr & {\text{Now }}\overrightarrow r .\overrightarrow a = l\left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right]\,\,\overrightarrow r .\overrightarrow b = m\left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right]\,\,\overrightarrow r .\overrightarrow c = n\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr & \therefore l\left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right] + m\left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right] + n\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0 \cr & \Rightarrow l + m + n = 0\,\,\,\,\,\,\,\left( {\because \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 2 \ne 0} \right) \cr} $$