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

$$\eqalign{ & \left| {\overrightarrow p + \overrightarrow q } \right| = \left( {\overrightarrow p + \overrightarrow q } \right).\left( {\overrightarrow p + \overrightarrow q } \right) \cr & = {\left| {\overrightarrow p } \right|^2} + {\left| {\overrightarrow q } \right|^2} + 2\overrightarrow p .\overrightarrow q \cr & = 2 + 2\,\cos \,\alpha , \cr & {\text{where }}\alpha {\text{ is the angle between}}\overrightarrow p {\text{ and }}\overrightarrow q \cr & = 2\left( {1 + \cos \,\alpha } \right) \cr & = 4\,{\cos ^2}\left( {\frac{\alpha }{2}} \right) \cr & {\left| {\overrightarrow p + \overrightarrow q } \right|^2} < 1 \cr & \Rightarrow \left( {4\,{{\cos }^2}\frac{\alpha }{2} - 1} \right) < 0 \cr & \Rightarrow \left( {2\,\cos \frac{\alpha }{2} - 1} \right)\left( {2\,\cos \frac{\alpha }{2} + 1} \right) < 0,\, - \frac{1}{2} < \cos \frac{\alpha }{2} < \frac{1}{2} \cr & \Rightarrow \frac{\pi }{3} < \frac{\alpha }{2} < \frac{{2\pi }}{3} \cr & \Rightarrow \frac{{2\pi }}{3} < \alpha < \frac{{4\pi }}{3} \cr} $$

Since, the moment about $$A$$ is zero, hence $$\overrightarrow F $$ passes through $$A.$$ Taking $$A$$ as origin.
Let the line of action of force $$\overrightarrow F $$ be $$y = mx.$$   (see figure)
Moment about $$B = \frac{{3m}}{{\sqrt {1 + {m^2}} }}\left| {\overrightarrow F } \right| = 9......\left( 1 \right)$$
Moment about $$C = \frac{4}{{\sqrt {1 + {m^2}} }}\left| {\overrightarrow F } \right| = 16......\left( 2 \right)$$
Dividing $$\left( 1 \right)$$ by $$\left( 2 \right),$$ we get :
$$m = \frac{3}{4} \Rightarrow \left| {\overrightarrow F } \right| = 5N.$$

Vector $$a\vec i + a\vec j + c\vec k,\,\vec i + \vec k$$     and $$c\vec i + c\vec j + b\vec k$$   are coplanar
\[\left| \begin{array}{l} a\,\,\,\,\,\,\,a\,\,\,\,\,c\\ 1\,\,\,\,\,\,0\,\,\,\,\,\,1\\ c\,\,\,\,\,\,\,c\,\,\,\,\,\,b \end{array} \right| = 0\,\,\, \Rightarrow {c^2} = ab\,\,\, \Rightarrow c = \sqrt {ab} \]
$$\therefore \,\,c$$  is G.M. of $$a$$ and $$b.$$

$$\eqalign{ & {\text{Let }}\overrightarrow a \times \overrightarrow b = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k \cr & \therefore \overrightarrow a \times \overrightarrow b .\overrightarrow i = x\,\,\overrightarrow a \times \overrightarrow b .\overrightarrow j = y\,\,\overrightarrow a \times \overrightarrow b .\overrightarrow k = z \cr & \therefore \overrightarrow a \times \overrightarrow b = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow i } \right]\overrightarrow i + \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow j } \right]\overrightarrow j + \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow k } \right]\overrightarrow k \cr} $$

$$\eqalign{ & {\text{Here, }}\lambda p = \mu q = \nu r{\text{ and }}\lambda p' = \mu q' = \nu r' \cr & {\text{So, }}\frac{{p'}}{p} = \frac{{q'}}{q} = \frac{{r'}}{r} = \frac{{\sqrt {p{'^2} + q{'^2} + r{'^2}} }}{{\sqrt {{p^2} + {q^2} + {r^2}} }} = \frac{{\sqrt {{b^2}} }}{{\sqrt {{a^2}} }} = \frac{b}{a} \cr & \therefore \,pp' + qq' + rr' = \frac{b}{a}\left( {{p^2} + {q^2} + {r^2}} \right) = \frac{b}{a}.{a^2} = ab \cr} $$

$$\eqalign{ & \overrightarrow F + \overrightarrow {{F_1}} + \overrightarrow {{F_2}} = 7i + 2j - 4k \cr & \overrightarrow d = P.V.\,{\text{ of }}\overrightarrow B - P.V.{\text{ of }}\overrightarrow A = 4i + 2j - 2k \cr & W = \overrightarrow F .\overrightarrow d = 28 + 4 + 8 = 40{\text{ units}} \cr} $$

$$\eqalign{ & {\text{Let }}\overrightarrow \alpha = \overrightarrow a \hat i + \overrightarrow b \hat j + \overrightarrow c \hat k \cr & {\text{Now, }}\overrightarrow \alpha .\hat i = \left( {\overrightarrow a \hat i + \overrightarrow b \hat j + \overrightarrow c \hat k} \right).\hat i = \overrightarrow a \cr & \overrightarrow \alpha .\hat j = \left( {\overrightarrow a \hat i + \overrightarrow b \hat j + \overrightarrow c \hat k} \right).\hat j = \overrightarrow b \cr & \overrightarrow \alpha .\hat k = \left( {\overrightarrow a \hat i + \overrightarrow b \hat j + \overrightarrow c \hat k} \right).\hat k = \overrightarrow c \cr & {\text{Now, }}\overrightarrow a \hat i + \overrightarrow b \hat j + \overrightarrow c \hat k = \overrightarrow \alpha \cr & {\text{Thus, required expression}} = \overrightarrow \alpha \cr} $$

$$\therefore \left( {\overrightarrow a - \overrightarrow d } \right).\left( {\overrightarrow b - \overrightarrow c } \right) = 0$$
$$ \Rightarrow \overrightarrow {DA} $$   and $$\overrightarrow {CB} $$  are perpendicular
$$\left( {\overrightarrow b - \overrightarrow d } \right).\left( {\overrightarrow c - \overrightarrow a } \right) = 0$$
$$ \Rightarrow \overrightarrow {DB} $$   and $$\overrightarrow {AC} $$  are perpendicular
$$\therefore \,D$$  is orthocenter of $$\Delta ABC.$$

$$\eqalign{ & \overrightarrow {PA} + \overrightarrow {AP} = 0{\text{ and }}\overrightarrow {PC} + \overrightarrow {CP} = 0 \cr & \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0{\text{ and }}\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {CP} = 0 \cr} $$
Adding, we get $$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {AC} + \overrightarrow {BC} + 2\overrightarrow {CP} = 0$$
Since $$\overrightarrow {AC} = - \overrightarrow {BC} \,\,\,{\text{& }}\,\,\overrightarrow {CP} = - \overrightarrow {PC} $$
$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0$$

Rewriting the equations
$$\eqalign{ & - x + y - z + 1 = 0 \cr & - x - y - z + 2 = 0 \cr & {\text{Here, }}{d_1} = 1 > 0,\,{d_2} = 2 > 0{\text{ and}} \cr & {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = \left( { - 1} \right)\left( { - 1} \right) + 1\left( { - 1} \right) + \left( { - 1} \right)\left( { - 1} \right) = 1 > 0 \cr} $$
$$\therefore $$  the bisector of the angle containing the origin is not the bisector of the acute angle.
$$\therefore $$  the required bisector has the equation $$\frac{{ - x + y - z + 1}}{{\sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 1} \right)}^2}} }} = - \frac{{ - x - y - z + 2}}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} }}.$$