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

If the direction cosines be $$\cos \,\alpha ,\,\cos \,\beta ,\,\cos \,\gamma $$    and the length is $$l$$ then $$l\cos \,\alpha = 3,\,l\cos \,\beta = 4,\,l\cos \,\gamma = 12.$$
Squaring and adding, $${l^2}\left( {{{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma } \right) = {3^2} + {4^2} + {12^2} = {13^2}.$$
$$\therefore \,{l^2} = {13^2}.$$

Given that $$\vec a \times \left( {2\hat i + 3\hat j + 4\hat k} \right) = \left( {2\hat i + 3\hat j + 4\hat k} \right) \times \vec b$$
$$ \Rightarrow \left( {\vec a + \vec b} \right) \times \left( {2\hat i + 3\hat j + 4\hat k} \right) = \vec 0$$
But neither $${\vec a + \vec b}$$  nor $${2\hat i + 3\hat j + 4\hat k}$$   is a null vector
$$\therefore \left( {\vec a + \vec b} \right)||\,\left( {2\hat i + 3\hat j + 4\hat k} \right) \Rightarrow \vec a + \vec b = \lambda \,\left( {2\hat i + 3\hat j + 4\hat k} \right)$$
Also given $$\left| {\vec a + \vec b} \right| = \sqrt {29} \,\,\, \Rightarrow \lambda = \pm 1$$
$$\eqalign{ & \therefore \vec a + \vec b = \pm \,\left( {2\hat i + 3\hat j + 4\hat k} \right) \cr & \therefore \left( {\vec a + \vec b} \right).\,\left( { - 7\hat i + 2\hat j + 3\hat k} \right) = \pm 4 \cr} $$

Given $$\vec a = \hat i + \hat j + \hat k,\,\vec b = \hat i - \hat j + 2\hat k$$       and $$\vec c = x\hat i + \left( {x - 2} \right)\hat j - \hat k$$
If $${\vec c}$$ lies in the plane of $${\vec a}$$ and $${\vec b},$$  then $$\left[ {\vec a\,\vec b\,\vec c} \right] = 0$$
\[\begin{array}{l} {\rm{i}}{\rm{.e}}{\rm{.,}}\,\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\\ x\,\,\,\,\left( {x - 2} \right)\,\,\,\, - 1 \end{array} \right| = 0\\ \Rightarrow 1\left[ {1 - 2\left( {x - 2} \right) - 1\left[ { - 1 - 2x} \right]} \right] + 1\left[ {x - 2 + x} \right] = 0\\ \Rightarrow 1 - 2x + 4 + 1 + 2x + 2x - 2 = 0\\ \Rightarrow 2x = - 4\\ \Rightarrow x = - 2 \end{array}\]

$$\eqalign{ & {\text{Let the coordinates of }}B{\text{ be }}\left( {x,{\text{ }}y} \right) \cr & \overrightarrow a = i - 3j \cr & {\text{P}}{\text{.V}}{\text{. of }}A{\text{ is}}\left( { - 1,\,5} \right) \cr & {\text{So,}}\,\overrightarrow {OA} = i + 5j,\,\overrightarrow {OB} = xi + yj \cr & \therefore \,\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \overrightarrow a \cr} $$
$$\Rightarrow \left(x+1\right)\stackrel{⌢}{i}+\left(y-5\right)\stackrel{⌢}{j}=\stackrel{⌢}{i}-3\stackrel{⌢}{j}$$
$$\eqalign{ & \Rightarrow x + 1 = 1{\text{ and }}y - 5 = - 3 \cr & \Rightarrow x = 0{\text{ and }}y = 2 \cr & \therefore \,{\text{Coordinates of }}B{\text{ are }}\left( {0,\,2} \right) \cr} $$

$$\eqalign{ & \vec A.\left( {\vec B + \vec C} \right) \times \left( {\vec A + \vec B + \vec C} \right) \cr & = \vec A.\left[ {\vec B \times \vec A + \vec B \times \vec B + \vec B \times \vec C + \vec C \times \vec A + \vec C \times \vec B + \vec C \times \vec C} \right] \cr & = \vec A.\vec B \times \vec A + \vec A.\vec B \times \vec C + \vec A.\vec C \times \vec A + \vec A.\vec C \times \vec B\,\,\,\,\left[ {{\text{Using }}\vec a \times \vec a = 0} \right] \cr & = 0 + \left[ {\vec A\,\vec B\,\vec C} \right] + 0 + \left[ {\vec A\,\vec C\,\vec B} \right] \cr} $$
(as $$\left[ {\vec a\vec b\vec c} \right] = 0$$   if any two vector are equal out of $$\vec a,\,\vec b,\,\vec c$$   )
$$\eqalign{ & = \left[ {\vec A\,\vec B\,\vec C} \right] - \left[ {\vec A\,\vec B\,\vec C} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{using }}\left[ {\vec a\vec b\vec c} \right] = - \left[ {\vec a\vec c\vec b} \right]} \right] \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Let }}\vec a = x\vec i + y\vec j + z\vec k \cr & \vec a \times \vec i = z\vec j - y\vec k\,\,\,\,\, \Rightarrow {\left( {\vec a \times \hat i} \right)^2} = {y^2} + {z^2} \cr & {\text{Similarly, }}{\left( {\vec a \times \hat j} \right)^2} = {x^2} + {z^2}{\text{ and }}{\left( {\vec a \times \hat k} \right)^2} = {x^2} + {y^2} \cr & \Rightarrow {\left( {\vec a \times \hat i} \right)^2} + {\left( {\vec a \times \hat j} \right)^2} + {\left( {\vec a \times \hat k} \right)^2} = 2\left( {{x^2} + {y^2} + {z^2}} \right) \cr & \Rightarrow {\left( {\vec a \times \hat i} \right)^2} + {\left( {\vec a \times \hat j} \right)^2} + {\left( {\vec a \times \hat k} \right)^2} = 2{{\vec a}^2} \cr} $$

If $$\overrightarrow F $$ be the resultant force, then $$\overrightarrow F = 2\hat i + 4\hat j + 2\hat k$$
Also, $$\overrightarrow r = \overrightarrow {AP} = - \hat i + 3\hat j + 2\hat k$$
$$\therefore $$  Required moment
\[ = \overrightarrow r \times \overrightarrow F = \left| \begin{array}{l} \,\,\,\hat i\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\hat k\\ - 1\,\,\,\,\,3\,\,\,\,\,2\\ \,\,\,2\,\,\,\,\,\,\,4\,\,\,\,\,2 \end{array} \right| = - 2\hat i + 6\hat j - 10\hat k\]

$$\eqalign{ & {\text{Here }}\left| {\overrightarrow a } \right| = 1,\,\,\left| {\overrightarrow b } \right| = 1,\,\,\left| {\overrightarrow c } \right| = 2 \cr & {\text{Now, }}\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow b = \overrightarrow 0 \cr & {\text{or }}\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow c + \overrightarrow b = \overrightarrow 0 \cr & {\text{or }}2\cos \,\theta \,\overrightarrow a - \overrightarrow c + \overrightarrow b = 0 \cr & {\text{or }}{\left( {\overrightarrow c - 2\cos \,\theta \,\overrightarrow a } \right)^2} = {\overrightarrow b ^2} \cr & {\text{or }}{\left| {\overrightarrow c } \right|^2} + 4{\cos ^2}\theta \,{\left| {\overrightarrow a } \right|^2} - 4\cos \,\theta \,\overrightarrow c .\overrightarrow a = {\left| {\overrightarrow b } \right|^2} \cr & {\text{or }}4 + 4{\cos ^2}\theta - 4\cos \,\theta .2\cos \,\theta = 1 \cr & {\text{or }}4{\cos ^2}\theta = 3 \cr & \therefore \,\,\theta = \frac{\pi }{6}. \cr} $$

$$\eqalign{ & \overrightarrow c = t\overrightarrow a + s\overrightarrow b .{\text{ Also }}\left| {\overrightarrow c } \right| = 1\left( {{\text{given}}} \right) \cr & {\text{Now, }}\overrightarrow c = t\left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) + s\left( {\overrightarrow i + 2\overrightarrow j - \overrightarrow k } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2t + s} \right)\overrightarrow i + \left( {t + 2s} \right)\overrightarrow j + \left( {t - s} \right)\overrightarrow k \cr & \therefore \,\,\left| {\overrightarrow c } \right| = 1\,\,\,\therefore \,{\left( {2t + s} \right)^2} + {\left( {t + 2s} \right)^2} + {\left( {t - s} \right)^2} = 1.....\left( 1 \right) \cr & {\text{and }}\overrightarrow a .\overrightarrow c = 0\,\, \Rightarrow 2\left( {2t + s} \right) + 1\left( {t + 2s} \right) + 1\left( {t - s} \right) = 0 \cr & {\text{or }}6t + 3s = 0{\text{ or }}2t + s = 0 \cr & \therefore \,{\text{from}}\left( 1 \right),\,{0^2} + {\left( {t - 4t} \right)^2} + {\left( {t + 2t} \right)^2} = 1 \cr & {\text{or }}18{t^2} = 1\,\,\,\,\,\therefore t = \pm \frac{1}{{3\sqrt 2 }} \cr & \therefore \,s = - 2t = \mp \frac{{\sqrt 2 }}{3} \cr & \therefore \,\overrightarrow c = \left( {2t + s} \right)\overrightarrow i + \left( {t + 2s} \right)\overrightarrow j + \left( {t - s} \right)\overrightarrow k \cr & = 0\overrightarrow i - 3t\overrightarrow j + 3t\overrightarrow k \cr & = - 3t\left( {\overrightarrow j - \overrightarrow k } \right) \cr & = - 3.\frac{{ \pm 1}}{{3\sqrt 2 }}\left( {\overrightarrow j - \overrightarrow k } \right) \cr} $$

$${\text{Clearly,}}\,\,\overrightarrow r = \overrightarrow a + \overrightarrow {AP} = \overrightarrow a \pm 6\frac{{\overrightarrow b }}{{\left| {\overrightarrow b } \right|}}.$$