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

$$\eqalign{ & \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \frac{{\overrightarrow b + \overrightarrow c }}{{\sqrt 2 }}\,\,\,\,\, \Rightarrow \,\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c = \frac{{\overrightarrow b + \overrightarrow c }}{{\sqrt 2 }}\, \cr & \Rightarrow \overrightarrow a .\overrightarrow c = \frac{1}{{\sqrt 2 }},\,\,\,\,\,\overrightarrow a .\overrightarrow b = - \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \,\alpha = \frac{1}{{\sqrt 2 }},\,\,\,\,\,\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \,\beta = - \frac{1}{{\sqrt 2 }} \cr & \therefore \,\,\cos \,\alpha = \frac{1}{{\sqrt 2 }},\,\,\,\,\,\cos \,\beta = - \frac{1}{{\sqrt 2 }} \cr & \therefore \alpha = \frac{\pi }{4},\,\,\,\,\,\beta = \frac{{3\pi }}{4} \cr} $$

$$\eqalign{ & \overrightarrow a + \overrightarrow b + 2\overrightarrow a - \overrightarrow b = 2\overrightarrow i + \overrightarrow i - \overrightarrow j {\text{ or }}3\overrightarrow a = 3\overrightarrow i - \overrightarrow j \cr & \therefore \,\overrightarrow a = \overrightarrow i - \frac{1}{3}\overrightarrow j \cr & \therefore \overrightarrow b = 2\overrightarrow i - \overrightarrow a = \overrightarrow i + \frac{1}{3}\overrightarrow j \cr & {\text{So, }}\cos \,\theta = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}} = \frac{{1 - \frac{1}{9}}}{{\sqrt {1 + \frac{1}{9}} \sqrt {1 + \frac{1}{9}} }} = \frac{8}{{10}} = \frac{4}{5} \cr} $$

Force $$\left( {\vec F} \right) = 2i + j - k$$    and its position vector of $$A = 2i - j.$$   We know that the position vector of a force about origin $$\left( r \right) = \left( {2i - j} \right) - \left( {0i + 0j + 0k} \right)$$       or $$r = 2i - j.$$
Therefore, moment of the force about origin
\[ = r \times \overrightarrow F = \left| \begin{array}{l} i\,\,\,\,\,\,\,\,\,j\,\,\,\,\,\,\,\,\,\,\,k\\ 2\,\,\, - 1\,\,\,\,\,\,\,\,0\\ 2\,\,\,\,\,\,\,\,1\,\,\,\, - 1 \end{array} \right| = i + 2j + 4k.\]

Any point on $$\frac{x}{1} = \frac{y}{{ - 3}} = \frac{z}{2}$$   is $$\left( {r,\, - 3r,\,2r} \right).$$   It is on the first plane if
$$\eqalign{ & r + 2\left( { - 3r} \right) - 3\left( {2r} \right) - 4 = 0 \cr & {\text{or }} - 11r - 4 = 0 \cr & {\text{or }}r = - \frac{4}{{11}} \cr & \therefore {\text{ the point is }}\left( { - \frac{4}{{11}},\,\frac{{12}}{{11}},\, - \frac{8}{{11}}} \right) \cr & \left( {r,\, - 3r,\,2r} \right){\text{ is on the second plane if }} \cr & 2r + 4\left( { - 3r} \right) - 6\left( {2r} \right) = 1 \cr & {\text{or }} - 22r = 1 \cr & {\text{or }}r = - \frac{1}{{22}} \cr & \therefore {\text{ the point is }}\left( { - \frac{1}{{22}},\,\frac{3}{{22}},\, - \frac{1}{{11}}} \right) \cr & \therefore {\text{ The distance }} = \sqrt {{{\left( { - \frac{4}{{11}} + \frac{1}{{22}}} \right)}^2} + {{\left( {\frac{{12}}{{11}} - \frac{3}{{22}}} \right)}^2} + {{\left( { - \frac{8}{{11}} + \frac{1}{{11}}} \right)}^2}} \cr & = \sqrt {\frac{{49}}{{{{\left( {22} \right)}^2}}} + \frac{{441}}{{{{\left( {22} \right)}^2}}} + \frac{{49}}{{{{\left( {11} \right)}^2}}}} \cr & = \frac{1}{{22}}\sqrt {49 + 441 + 196} \cr} $$

$$\left| {\overrightarrow a \times \overrightarrow b .\overrightarrow c } \right| = $$    volume of the parallelepiped $$ = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$$
$$ \Rightarrow $$  it is a rectangular parallelepiped, i.e., concurrent edges are perpendicular to each other. So, $$\overrightarrow a .\overrightarrow b = 0,\,\overrightarrow b .\overrightarrow c = 0,\,\overrightarrow c .\overrightarrow a = 0.$$

$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \cr & = - \overrightarrow i - \overrightarrow j + 8\overrightarrow k - \left( {\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k } \right) \cr & = - 2\overrightarrow i - 3\overrightarrow j + 5\overrightarrow k \cr} $$
Similarly, find $$\overrightarrow {BC} $$  and $$\overrightarrow {CA} .$$  Obtain $$\left| {\overrightarrow {AB} } \right| = \left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {CA} } \right|.$$

$$\overrightarrow a = 2\hat i - m\hat j + 3m\hat k\,\,\& \,\,\overrightarrow b = \left( {1 + m} \right)\hat i - 2m\hat j + \hat k$$
and if angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is an acute, then $$\overrightarrow a .\overrightarrow b > 0$$
$$\eqalign{ & \Rightarrow 2\left( {1 + m} \right) + 2{m^2} + 3m > 0 \cr & \Rightarrow 2{m^2} + 5m + 2 > 0 \cr & \Rightarrow 2{m^2} + 4m + m + 2 > 0 \cr & \Rightarrow \left( {2m + 1} \right)\left( {m + 2} \right) > 0 \cr & \Rightarrow m < - 2{\text{ or }}m > - \frac{1}{2} \cr} $$

$${\left( {\vec a \times \vec b} \right)^2} = {\left| {\vec a} \right|^2}{\left| {\vec b} \right|^2}{\sin ^2}\frac{\pi }{6} = 16 \times 4 \times \frac{1}{4} = 16$$

Since, $$\vec a + \vec b + \vec c = \vec 0$$    and $$\vec a,\,\vec b,\,\vec c$$   are unit vectors, therefore $$\vec a,\,\vec b,\,\vec c$$   form an equilateral triangle.
$$\eqalign{ & \Rightarrow \vec a \times \left( {\vec a + \vec b + \vec c} \right) = \vec 0 \cr & \Rightarrow \vec a \times \vec a + \vec a \times \vec b + \vec a \times \vec c = \vec 0 \cr & \Rightarrow \vec a \times \vec b = \vec c \times \vec a \cr & {\text{Similarly, }}\vec b \times \vec c = \vec c \times \vec a \cr & \therefore \vec a \times \vec b = \vec b \times \vec c = \vec c \times \vec a \cr} $$
Also since $$\vec a,\,\vec b,\,\vec c$$   are non parallel (these form an equilateral $$\Delta $$ ).
$$\therefore \vec a \times \vec b = \vec b \times \vec c = \vec c \times \vec a \ne \vec 0$$

$$\because \,\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w $$   are non-coplanar vectors
$$\eqalign{ & \therefore \left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] \ne 0 \cr & {\text{Now, }}\left[ {3\overrightarrow u ,\,p\overrightarrow v ,\,p\overrightarrow w } \right] - \left[ {p\overrightarrow v ,\,p\overrightarrow w ,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow w ,\,q\overrightarrow v ,\,q\overrightarrow u } \right] = 0 \cr & \Rightarrow 3{p^2}\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] - pq\left[ {\overrightarrow v ,\,\overrightarrow w ,\,\overrightarrow u } \right] - 2{q^2}\left[ {\overrightarrow w ,\,\overrightarrow v ,\,\overrightarrow u } \right] = 0 \cr & \Rightarrow 3{p^2}\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] - pq\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] + 2{q^2}\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] = 0 \cr & \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\overrightarrow u ,\,\overrightarrow v ,\,\overrightarrow w } \right] = 0 \cr & \Rightarrow 3{p^2} - pq + 2{q^2} = 0 \cr & \Rightarrow 2{p^2} + {p^2} - pq + \frac{{{q^2}}}{4} + \frac{{7{q^2}}}{4} = 0 \cr & \Rightarrow 2{p^2} + {\left( {p - \frac{q}{2}} \right)^2} + \frac{7}{4}{q^2} = 0 \cr & \Rightarrow p = 0,\,q = 0,\,p = \frac{q}{2} \cr} $$
This is possible only when $$p = 0,\,q = 0$$
$$\therefore $$  There is exactly one value of $$\left( {p,\,q} \right)$$