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

Vector perpendicular to the face $$OAB$$
\[ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| \begin{array}{l} \hat i\,\,\,\,\,\hat j\,\,\,\,\,\hat k\\ 1\,\,\,\,\,2\,\,\,\,\,1\\ 2\,\,\,\,\,1\,\,\,\,\,3 \end{array} \right| = 5\hat i - \hat j - 3\hat k\]
Vector perpendicular to the face $$ABC$$
\[ = \overrightarrow {AB} \times \overrightarrow {AC} = \left| \begin{array}{l} \,\,\,\,\,\hat i\,\,\,\,\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\,\,\,\,\,\,\hat k\\ \,\,\,\,1\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ - 2\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\, 1 \end{array} \right| = \hat i - 5\hat j - 3\hat k\]
Angle between the faces $$=$$ angle between their normals
$$\cos \,\theta = \left| {\frac{{5 + 5 + 9}}{{\sqrt {35} .\sqrt {35} }}} \right| = \frac{{19}}{{35}}{\text{ or }}\theta = {\cos ^{ - 1}}\left( {\frac{{19}}{{35}}} \right)$$

$$\vec a,\,\vec b,\,\vec c$$   and $${\vec d}$$ are unit vectors,
$$\eqalign{ & {\text{Let }}\vec a \times \vec b = \left( {\sin \,\alpha } \right)\overrightarrow {{n_1}} {\text{ and }}\vec c \times \vec d = \left( {\sin \,\beta } \right)\overrightarrow {{n_2}} \cr & {\text{then }}\left( {\vec a \times \vec b} \right).\left( {\vec c \times \vec d} \right) = 1 \cr & \Rightarrow \left( {\sin \,\alpha } \right)\overrightarrow {{n_1}} .\left( {\sin \,\beta } \right)\overrightarrow {{n_2}} = 1 \cr & \Rightarrow \sin \,\alpha \,\sin \,\beta \,\overrightarrow {{n_1}} .\overrightarrow {{n_2}} = 1 \cr & \Rightarrow \sin \,\alpha \,\sin \,\beta \,\cos \,\gamma = 1 \cr} $$
where $$\gamma $$ is the angle between $$\overrightarrow {{n_1}} $$  and $$\overrightarrow {{n_2}} .$$
$$\eqalign{ & \alpha = \frac{\pi }{2},\,\,\beta = \frac{\pi }{2},\,\,\gamma = {0^ \circ } \cr & {\text{Now }}\gamma = {0^ \circ }\,\, \Rightarrow \vec a \times \vec b\,||\,\vec c \times \vec d \cr & {\text{Let }}\vec a \times \vec b = \lambda \left( {\vec c \times \vec d} \right)\,\, \Rightarrow \left( {\vec a \times \vec b} \right).\vec c = \lambda \left( {\vec c \times \vec d} \right).\vec c = 0 \cr & {\text{and }}\left( {\vec a \times \vec b} \right).\vec d = \lambda \left( {\vec c \times \vec d} \right).\vec d = 0 \cr} $$
$$\therefore {\text{ }}\vec a,\,\vec b,\,\vec c$$   are coplanar and $$\vec a,\,\vec b,\,\vec d$$  are coplanar
$$ \Rightarrow \vec a,\,\vec b,\,\vec c,\,\vec d$$   are coplanar
Also $$\alpha = {90^ \circ }\,\, \Rightarrow \vec a\, \bot \,\vec b{\text{ and }}\beta = {90^ \circ }\,\, \Rightarrow \,\vec c\, \bot \,\vec d$$
But angle between $${\vec a}$$ and $${\vec c}$$ is $$\frac{\pi }{3}\,\,\,\,\,\left( {\because \vec a.\vec c = \frac{1}{2}} \right)$$
So, angle between $${\vec b}$$ and $${\vec d}$$ should also be $$\frac{\pi }{3}.$$
Hence $${\vec b}$$ and $${\vec d}$$ are non parallel.

Let $$A$$ be the first term and $$x$$ the common ratio of G.P.
So, $$a = A{x^{p - 1}} \Rightarrow \log \,a = \log \,A + \left( {p - 1} \right)\log \,x$$
Similarly, $$\log \,b = \log \,A + \left( {q - 1} \right)\log \,x$$
and $$\log \,c = \log \,A + \left( {r - 1} \right)\log \,x$$
If $$\overrightarrow \alpha = \log \,{a^2}\hat i + \log \,{b^2}\hat j + \log \,{c^2}\hat k$$
and $$\overrightarrow \beta = \left( {q - r} \right)\hat i + \left( {r - p} \right)\hat j + \left( {p - q} \right)\hat k\,{\text{then}}$$
$$\eqalign{ & \overrightarrow \alpha .\overrightarrow \beta = 2\left[ {\log \,a\left( {q - r} \right) + \log \,b\left( {r - p} \right) + \log \,c\left( {p - q} \right)} \right] \cr & = 2\left[ {\left( {q - r} \right)\left\{ {\log \,A + \left( {p - 1} \right)\log \,x} \right\} + \left( {r - p} \right)\left\{ {\log \,A + \left( {q - 1} \right)\log \,x} \right\} + \left( {p - q} \right)\left\{ {\log \,A + \left( {r - 1} \right)\log \,x} \right\}} \right] \cr & = 2\left[ {\left( {q - r + r - p + p - q} \right)\log \,A + \left( {qp - pr - p + r + pr - pq - r + p + pr - qr - p + q} \right)\log \,x} \right] \cr & = 0 \cr} $$
Hence, the angle between $$\overrightarrow \alpha $$ and $$\overrightarrow \beta $$ is $$\frac{\pi }{2}.$$

Volume of the tetrahedron $$ = V = \left| {\frac{1}{6}\left[ {\overrightarrow {AB} \,\,\overrightarrow {AC} \,\,\overrightarrow {AO} } \right]} \right|$$
Now,
$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = 2\overrightarrow j - \overrightarrow k - \left( {\overrightarrow i + \overrightarrow j } \right) = - \overrightarrow i + \overrightarrow j - \overrightarrow k \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = \overrightarrow i + \overrightarrow k - \left( {\overrightarrow i + \overrightarrow j } \right) = \overrightarrow k - \overrightarrow j \cr} $$
\[\therefore \,V = \left| {\frac{1}{6}\left[ { - \overrightarrow i + \overrightarrow j - \overrightarrow k \,\,\,\overrightarrow k - \overrightarrow j \,\,\, - \overrightarrow i - \overrightarrow j } \right]} \right| = \left| {\frac{1}{6}\left| \begin{array}{l} - 1\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\, - 1\\ \,\,\,\,\,\,0\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,1\\ - 1\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,0\, \end{array} \right|} \right| = \frac{1}{6}\]

Let point $$P$$ is $$\left( {1,\, - 1,\,2} \right)$$
and point $$Q$$ is $$\left( {2,\, - 1,\,3} \right)$$
$$ \Rightarrow $$  Position vector of $$P$$ w.r.t. $$Q$$ is
$$\eqalign{ & \overrightarrow r = \left( {1 - 2} \right)\hat i + \left( { - 1 + 1} \right)\hat j + \left( {2 - 3} \right)\hat k \cr & \Rightarrow \overrightarrow r = - \hat i + 0\hat j - \hat k{\text{ and }}\overrightarrow F = 3\hat i + 2\hat j - 4\hat k \cr} $$
\[ \Rightarrow {\rm{Moment}} = \overrightarrow r \times \overrightarrow F = \left| \begin{array}{l} \,\,\,\,\,\hat i\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\hat k\\ - 1\,\,\,\,0\,\,\, - 1\\ \,\,\,\,\,3\,\,\,\,2\,\,\, - 4 \end{array} \right|\]
$$\eqalign{ & = \hat i\left( {0 + 2} \right) - \hat j\left( {4 + 3} \right) + \hat k\left( { - 2 + 0} \right) \cr & = 2\hat i - 7\hat j - 2\hat k \cr} $$

$$\overrightarrow {OM} = $$  component of $$\overrightarrow a $$ along $$\overrightarrow b $$
$$\overrightarrow {MA} = $$  component of $$\overrightarrow a $$ perpendicular to $$\overrightarrow b $$

$$\eqalign{ & \Delta OMA \Rightarrow \cos \,\theta = \frac{{OM}}{{OA}} \cr & \Rightarrow OM = \left| {\overrightarrow {OM} } \right| = \left| {\overrightarrow {OA} } \right|\cos \,\theta = \left| {\overrightarrow a } \right|\cos \,\theta \cr & \because \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \,\theta = \left| {\overrightarrow b } \right|\left( {OM} \right) \cr & \therefore \,\overrightarrow {OM} = \left| {\overrightarrow {OM} } \right|\hat b = \left( {\frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\vec b} \right|}}} \right)\frac{{\overrightarrow b }}{{\left| {\vec b} \right|}} = \left( {\frac{{\overrightarrow a .\overrightarrow b }}{{{{\left| {\vec b} \right|}^2}}}} \right)\overrightarrow b \cr & \overrightarrow {OM} + \overrightarrow {MA} = \overrightarrow {OA} \cr & \therefore \,\overrightarrow {MA} = \overrightarrow {OA} - \overrightarrow {OM} = \overrightarrow a - \left( {\frac{{\overrightarrow a .\overrightarrow b }}{{{{\left| {\vec b} \right|}^2}}}} \right)\overrightarrow b \cr} $$

Here, $$\overrightarrow a = 2p\overrightarrow i + \overrightarrow j .$$   After rotation, let the vector be $$\overrightarrow b $$ and let the unit vectors along the new axes be $$\overrightarrow {i'} ,\,\overrightarrow {j'} .$$
Then $$\overrightarrow b = \left( {p + 1} \right)\overrightarrow {i'} + \overrightarrow {j'} .$$     But the magnitude of a vector does not change with rotation of axes.
$$\eqalign{ & \therefore \,\,\,\overrightarrow {\left| a \right|} = \left| {\overrightarrow b } \right| \cr & \Rightarrow \sqrt {{{\left( {2p} \right)}^2} + {1^2}} = \sqrt {{{\left( {p + 1} \right)}^2} + {1^2}} \cr & \Rightarrow 4{p^2} + 1 = {p^2} + 2p + 2{\text{ or }}3{p^2} - 2p - 1 = 0 \cr & \therefore \,\,p = 1,\, - \frac{1}{3} \cr} $$

Given $$\left( {\vec a \times \vec b} \right) \times \vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\,\vec a$$
Clearly $${\vec a}$$ and $${\vec b}$$ are non-collinear
$$\eqalign{ & \Rightarrow \left( {\vec a.\vec c} \right)\vec b - \left( {\vec b.\vec c} \right)\vec a = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\,\vec a \cr & \therefore \vec a.\vec c = 0{\text{ and }} - \vec b.\vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right| \Rightarrow \cos \,\theta = \frac{{ - 1}}{3} \cr & \therefore \sin \,\theta = \sqrt {1 - \frac{1}{9}} = \frac{{2\sqrt 2 }}{3}\,\,\,\left[ {\theta {\text{ is acute angle between }}\vec b{\text{ and }}\vec c} \right] \cr} $$

$$\eqalign{ & {\left( {\overrightarrow a \times \overrightarrow b } \right)^2} = {\left( {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \,\theta \,\widehat n} \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}{\sin ^2}\theta \cr & {\left( {\overrightarrow a .\overrightarrow b } \right)^2} = {\left( {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \,\theta } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}{\cos ^2}\theta .{\text{ Add these}}{\text{.}} \cr} $$

From the location of the points it is clear that they are not on the same plane. Through four noncoplanar points one and only one sphere passes.