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

$$\eqalign{ & \vec a'\left( {\vec b'\vec c} \right) = \frac{{\sqrt 3 }}{2}\left( {\vec b + \vec c} \right) \cr & \left( {\vec a \times \vec c} \right)\vec b - \left( {\vec a \times \vec b} \right)\vec c = \frac{{\sqrt 3 }}{2}\vec b + \frac{{\sqrt 3 }}{2}\vec c \cr} $$
On comparing both sides
$$\vec a \times \vec b = - \frac{{\sqrt 3 }}{2}\,\,\cos \,q = - \frac{{\sqrt 3 }}{2}$$
[$$\because \,\vec a$$  and $${\vec b}$$ are unit vectors ]
where $$\theta $$ is the angle between $${\vec a}$$ and $${\vec b}$$
$$\theta = \frac{{5\pi }}{6}$$

Let $$\left| {\overrightarrow p } \right| = \left| {\overrightarrow q } \right| = \left| {\overrightarrow r } \right| = k$$
Let $$\hat p,\,\hat q,\,\hat r$$   be unit vectors along $$\overrightarrow p ,\,\overrightarrow q ,\,\overrightarrow r $$   respectively. Clearly $$\hat p,\,\hat q,\,\hat r$$   are mutually perpendicular vectors, so any vector $$\overrightarrow x $$ can be written as $${a_1}\hat p + {a_2}\hat q + {a_3}\hat r.$$
$$\eqalign{ & \therefore \,\overrightarrow p \times \left\{ {\left( {\overrightarrow x - \overrightarrow q } \right) \times \overrightarrow p } \right\} = \left( {\overrightarrow p .\overrightarrow p } \right)\left( {\overrightarrow x - \overrightarrow q } \right) - \left\{ {\overrightarrow {p.} \left( {\overrightarrow x - \overrightarrow q } \right)} \right\}\overrightarrow p \cr & = {k^2}\left( {\overrightarrow x - \overrightarrow q } \right) - \left( {\overrightarrow p .\overrightarrow x } \right)\overrightarrow p \,\,\,\,\,\,\,\,\,\left[ {\because \,\overrightarrow p .\overrightarrow q = 0} \right] \cr & = {k^2}\left( {\overrightarrow x - \overrightarrow q } \right) - k\hat p.\left( {{a_1}\hat p + {a_2}\hat q + {a_3}\hat r} \right)k\hat p \cr & = {k^2}\left( {\overrightarrow x - \overrightarrow q - {a_1}\hat p} \right) \cr & {\text{Similarly, }}\overrightarrow q \times \left\{ {\left( {\overrightarrow x - \overrightarrow r } \right) \times \overrightarrow q } \right\} = {k^2}\left( {\overrightarrow x - \overrightarrow r - {a_2}\hat q} \right) \cr & {\text{and }}\overrightarrow r \times \left\{ {\left( {\overrightarrow x - \overrightarrow p } \right) \times \overrightarrow r } \right\} = {k^2}\left( {\overrightarrow x - \overrightarrow p - {a_3}\hat r} \right) \cr} $$
According to the given condition
$$\eqalign{ & {k^2}\left( {\overrightarrow x - \overrightarrow q - {a_1}\hat p + \overrightarrow x - \overrightarrow r - {a_2}\hat q + \overrightarrow x - \overrightarrow p - {a_3}\hat r} \right) = 0 \cr & \Rightarrow {k^2}\left\{ {3\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right) - \left( {{a_1}\hat p + {a_2}\hat q + {a_3}\hat r} \right)} \right\} = 0 \cr & \Rightarrow {k^2}\left[ {2\overrightarrow x - \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)} \right] = \overrightarrow 0 \cr & \Rightarrow \overrightarrow x = \frac{1}{2}\overrightarrow {} \left( {\overrightarrow p + \overrightarrow q + \overrightarrow r } \right)\,\,\,\,\,\left[ {\because \,k \ne 0} \right] \cr} $$

A point equidistant from $$AB$$  and $$AC$$  is on the bisectors of the angle $$BAC.$$
A vector along the internal bisector of the angle $$BAC$$
$$\eqalign{ & = \frac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \frac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}} \cr & = \frac{{3\overrightarrow i + \overrightarrow j - \overrightarrow k }}{{\sqrt {{3^2} + {1^2} + {{\left( { - 1} \right)}^2}} }} + \frac{{\overrightarrow i - \overrightarrow j + 3\overrightarrow k }}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {3^2}} }} \cr & = \frac{1}{{\sqrt {11} }}\left( {4\overrightarrow i + 2\overrightarrow k } \right) \cr & \therefore \,\overrightarrow {AP} = t\left( {2\overrightarrow i + \overrightarrow k } \right) \cr & \therefore \,\overrightarrow {BP} = \overrightarrow {AP} - \overrightarrow {AB} = t\left( {2\overrightarrow i + \overrightarrow k } \right) - \left( {3\overrightarrow i + \overrightarrow j - \overrightarrow k } \right) \cr & \therefore \,\overrightarrow {BP} = \left( {2t - 3} \right)\overrightarrow i - \overrightarrow j + \left( {t + 1} \right)\overrightarrow k \cr & {\text{Also,}}\,\overrightarrow {BC} = \overrightarrow {AC} - \overrightarrow {AB} = \left( {\overrightarrow i - \overrightarrow j + 3\overrightarrow k } \right) - \left( {3\overrightarrow i + \overrightarrow j - \overrightarrow k } \right) = - 2\overrightarrow i - 2\overrightarrow j + 4\overrightarrow k \cr & {\text{But }}\overrightarrow {BP} = s\overrightarrow {BC} \cr & \therefore \,\left( {2t - 3} \right)\overrightarrow i - \overrightarrow j + \left( {t + 1} \right)\overrightarrow k = s\left( { - 2\overrightarrow i - 2\overrightarrow j + 4\overrightarrow k } \right) \cr & {\text{or }}2t - 3 = - 2s,\,\,\, - 1 = - 2s,\,\,\,t + 1 = 4s \cr & \therefore \,s = \frac{1}{2}{\text{ and }}t = 1 \cr & \therefore \,\overrightarrow {AP} = 2\overrightarrow i + \overrightarrow k . \cr} $$

$$\eqalign{ & \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \left( {2\overrightarrow i + \overrightarrow j - 2\overrightarrow k } \right) - \left( {\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right) = \overrightarrow i + 2\overrightarrow j + \overrightarrow k \cr & \overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = \left( { - 5\overrightarrow i + 2\overrightarrow j - 6\overrightarrow k } \right) - \left( {\overrightarrow i - \overrightarrow j - 3\overrightarrow k } \right) = - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k \cr} $$
A vector along the bisector of the angle $$BAC$$
$$\eqalign{ & = \frac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \frac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}} \cr & = \frac{{\overrightarrow i + 2\overrightarrow j + \overrightarrow k }}{{\sqrt {{1^2} + {2^2} + {1^2}} }} + \frac{{ - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k }}{{\sqrt {{{\left( { - 6} \right)}^2} + {3^2} + {{\left( { - 3} \right)}^2}} }} \cr & = \frac{1}{{\sqrt 6 }}\left( {\overrightarrow i + 2\overrightarrow j + \overrightarrow k } \right) + \frac{1}{{3\sqrt 6 }}\left( { - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k } \right) \cr & = \frac{1}{{3\sqrt 6 }}\left( { - 3\overrightarrow i + 9\overrightarrow j } \right) \cr & = \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{\sqrt 6 }} \cr} $$
$$\therefore $$  the unit vector along $$AD = \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{\sqrt {10} }}$$
$$\eqalign{ & \therefore \,\overrightarrow {AD} = \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{10}}AD \cr & {\text{As }}D{\text{ is on }}BC,\,\overrightarrow {BD} = t\overrightarrow {BC} \cr & \therefore \,\overrightarrow {BA} + \overrightarrow {AD} = t\left( {\overrightarrow {BA} + \overrightarrow {AC} } \right) \cr & {\text{or }} - \overrightarrow i - 2\overrightarrow j - \overrightarrow k + \frac{{ - \overrightarrow i + 3\overrightarrow j }}{{10}}AD = t\left( { - \overrightarrow i - 2\overrightarrow j - \overrightarrow k - 6\overrightarrow i + 3\overrightarrow j - 3\overrightarrow k } \right) \cr & = t\left( { - 7\overrightarrow i + \overrightarrow j - 4\overrightarrow k } \right) \cr & \Rightarrow \,\, - 1 - \frac{{AD}}{{10}} = - 7t,\,\,\, - 2 + \frac{3}{{10}}AD = t,\,\,\, - 1 = - 4t \cr & \therefore \,t = \frac{1}{4} \cr & \therefore \, - 1 - \frac{{AD}}{{10}} = - \frac{7}{4}{\text{ or }}\frac{{AD}}{{10}} = \frac{3}{4} \cr & \therefore \,AD = \frac{{15}}{2}. \cr} $$

Volume of parallelepiped formed by
$$\vec u = \hat i + a\hat j + \hat k,\,\,\vec v = \hat j + a\hat k,\,\,\vec w = a\hat i + \hat k$$             is
\[\begin{array}{l} V = \left[ {\vec u\,\vec v\,\vec w} \right] = \left| \begin{array}{l} 1\,\,\,\,a\,\,\,\,1\\ 0\,\,\,\,1\,\,\,\,a\\ a\,\,\,\,0\,\,\,\,1 \end{array} \right|\\ = 1\left( {1 - 0} \right) - a\left( {0 - {a^2}} \right) + 1\left( {0 - a} \right)\\ = 1 + {a^3} - a \end{array}\]
For $$V$$ to be min. $$\frac{{dV}}{{da}} = 0$$
$$ \Rightarrow 3{a^2} - 1 = 0\,\,\,\,\,\,\, \Rightarrow a = \pm \frac{1}{{\sqrt 3 }}$$

$$\eqalign{ & {\text{Let}}\,\,\left( {\alpha ,\,\beta ,\,\gamma } \right) = \left( {1 + \frac{r}{4},\, - 1 + \frac{r}{3},\,2} \right) \cr & {\text{Then }}\alpha = 1 + \frac{r}{4},\,\beta = - 1 + \frac{r}{3},\,\gamma = 2 \cr & \therefore \,4\left( {\alpha - 1} \right) = 3\left( {\beta + 1} \right) = r{\text{ and }}\gamma = 2 \cr & \Rightarrow \,\,\frac{{\alpha - 1}}{3} = \frac{{\beta + 1}}{4} = \frac{{\gamma - 2}}{0} \cr} $$

$$\eqalign{ & {\text{Suppose, }}\overrightarrow p = {p_1}\hat i + {p_2}\hat j + {p_3}\hat k \cr & \overrightarrow p \times \hat i = {p_2}\hat j \times \hat i + {p_3}\hat k \times \hat i = - {p_2}\hat k + {p_3}\hat j \cr & {\left| {\overrightarrow p \times \hat i} \right|^2} = p_2^2 + p_3^2 \cr & {\text{Similarly, }}{\left| {\overrightarrow p \times \hat j} \right|^2} = p_3^2 + p_1^2 \cr & {\text{and }}{\left| {\overrightarrow p \times \hat k} \right|^2} = p_1^2 + p_2^2 \cr & \therefore \,\frac{3}{2}\left\{ {{{\left| {\overrightarrow p \times \hat i} \right|}^2} + {{\left| {\overrightarrow p \times \hat j} \right|}^2} + {{\left| {\overrightarrow p \times \hat k} \right|}^2}} \right\} \cr & = 3\left( {p_1^2 + p_2^2 + p_3^2} \right) \cr & = 3{\overrightarrow p ^2} \cr} $$

The lines are coplanar if they intersect.
Any point on the first line is $$\left( {1 + 4r,\, - 1 + 3r,\,kr} \right),$$     and any point on the second line is $$\left( {r',\,k + 3r',\,1 - 2r'} \right).$$     They intersect if
$$\eqalign{ & 1 + 4r = r',\, - 1 + 3r = k + 3r',\,1 - 2r' = kr \cr & \Rightarrow 9r + 4 + k = 0{\text{ and }}8r + 1 + kr = 0\,\, \Rightarrow r = \frac{{ - 4 - k}}{9} = \frac{{ - 1}}{{8 + k}} \cr} $$
$$ \Rightarrow {k^2} + 12k + 23 = 0$$     which gives two real values of $$k.$$

The given lines are
$$\eqalign{ & x - 1 = \frac{{y + 3}}{{ - \lambda }} = \frac{{z - 1}}{\lambda } = s.....(1) \cr & {\text{and }}2x = y - 1 = \frac{{z - 2}}{{ - 1}} = t.....(2) \cr} $$
The lines are coplanar, if
\[\begin{array}{l} \left| \begin{array}{l} 0 - \left( { - 1} \right)\,\,\,\,\, - 1 - 3\,\,\,\,\, - 2 - 1\\ \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda \\ \,\,\,\,\,\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 \end{array} \right| = 0\\ {c_2} \to {c_2} + {c_3}\,;\left| \begin{array}{l} 1\,\,\,\,\, - 5\,\,\,\,\, - 1\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\lambda \\ \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\, - 1 \end{array} \right| = 0\\ \Rightarrow 5\left( { - 1 - \frac{\lambda }{2}} \right) = 0\\ \Rightarrow \lambda = - 2 \end{array}\]

$$\eqalign{ & {\text{Here,}} \cr & \cos \,\alpha = \overrightarrow a .\overrightarrow b {\text{ or }}2{\cos ^2}\frac{\alpha }{2} - 1 = \overrightarrow a .\overrightarrow b \,\,\,{\text{or }}4{\cos ^2}\frac{\alpha }{2} = 2 + 2\overrightarrow a .\overrightarrow b \cr & \therefore \,{\left( {2{{\cos }^2}\frac{\alpha }{2}} \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b = {\left( {\overrightarrow a + \overrightarrow b } \right)^2} = {\left| {\overrightarrow a + \overrightarrow b } \right|^2} \cr & {\text{or }}2\cos \frac{\alpha }{2} = \left| {\overrightarrow a + \overrightarrow b } \right| \cr & \therefore \,\cos \frac{\alpha }{2} = \frac{1}{2}\left| {\overrightarrow a + \overrightarrow b } \right|. \cr} $$