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

$$\eqalign{ & \left( {\overrightarrow a - \overrightarrow d } \right) \times \left( {\overrightarrow b - \overrightarrow c } \right) \cr & = \overrightarrow a \times \overrightarrow b - \overrightarrow d \times \overrightarrow b - \overrightarrow a \times \overrightarrow c + \overrightarrow d \times \overrightarrow c \cr & = \overrightarrow c \times \overrightarrow d - \overrightarrow d \times \overrightarrow b - \overrightarrow b \times \overrightarrow d - \overrightarrow c \times \overrightarrow d \cr & = - \overrightarrow d \times \overrightarrow b + \overrightarrow d \times \overrightarrow b \cr & = 0 \cr & {\text{Again }}\left( {\overrightarrow a \times \overrightarrow b } \right) = \left( {\overrightarrow c \times \overrightarrow d } \right){\text{ given}} \cr & \Rightarrow \left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) = \left( {\overrightarrow c \times \overrightarrow d } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) = 0\,\,\,\,\,\left( {{\text{as }}\overrightarrow a \times \overrightarrow a = 0} \right) \cr} $$
So both $$\left( 1 \right)$$ and $$\left( 2 \right)$$ are correct.

$$\eqalign{ & P = \left( {0,\,2\cos \,{{45}^ \circ },\,2\sin \,{{45}^ \circ }} \right) = \left( {0,\,\frac{1}{{\sqrt 2 }},\,\frac{1}{{\sqrt 2 }}} \right). \cr & {\text{If }}M = \left( {\alpha ,\,\beta ,\,\gamma } \right){\text{ then}} \cr & 3\alpha + \beta - \sqrt 2 \gamma \, = 2\sqrt 2 {\text{ and }}\frac{{\alpha - 0}}{3} = \frac{{\beta - \frac{1}{{\sqrt 2 }}}}{1} = \frac{{\gamma - \frac{1}{{\sqrt 2 }}}}{{ - \sqrt 2 }} \cr & \therefore \frac{\alpha }{3} = \frac{{\beta - \frac{1}{{\sqrt 2 }}}}{1} = \frac{{\sqrt 2 \gamma - 1}}{{ - \sqrt 2 }} \cr & = \frac{{3\alpha + \beta - \frac{1}{{\sqrt 2 }} - \left( {\sqrt 2 \gamma - 1} \right)}}{{9 + 1 - \left( { - 2} \right)}} \cr & = \frac{{2\sqrt 2 - \frac{1}{{\sqrt 2 }} + 1}}{{12}} \cr & = \frac{{3 + \sqrt 2 }}{{12\sqrt 2 }} \cr & \therefore \,\,\alpha = \frac{{3 + \sqrt 2 }}{{4\sqrt 2 }}, \cr & \beta = \frac{1}{{\sqrt 2 }} + \frac{{3 + \sqrt 2 }}{{12\sqrt 2 }} = \frac{{15 + \sqrt 2 }}{{12\sqrt 2 }}, \cr & \gamma = \frac{1}{{\sqrt 2 }}\left\{ {1 - \frac{{6 + 2\sqrt 2 }}{{12\sqrt 2 }}} \right\} = \frac{{10\sqrt 2 - 6}}{{24}} \cr} $$

Direction ratios of the line $$AB$$  are $$2 - 1,\,6 - 2,\,2 - \left( { - 1} \right),$$     i.e., $$1,\,4,\,3.$$   Direction ratios of the line $$AC$$ are $$\lambda - 1,\, - 2 - 2,\, - 4 - \left( { - 1} \right),$$      i.e., $$\lambda - 1,\, - 4,\, - 3.$$    For collinearity, $$AB||AC\,\,\,\,\, \Rightarrow \frac{{\lambda - 1}}{1} = \frac{{ - 4}}{4} = \frac{{ - 3}}{3}\,\,\,\,\, \Rightarrow \lambda = 0.$$

$$\eqalign{ & \left( {\vec a \times \vec b} \right) \times \vec c = \vec a \times \left( {\vec b \times \vec c} \right),\,\,\,\,\,\vec a.\vec b \ne 0,\,\,\,\vec b.\vec c \ne 0 \cr & \Rightarrow \left( {\vec a.\vec c} \right).\vec b - \left( {\vec b.\vec c} \right).\vec a = \left( {\vec a.\vec c} \right).\vec b - \left( {\vec a.\vec b} \right).\vec c \cr & \Rightarrow \left( {\vec a.\vec b} \right).\vec c = \left( {\vec b.\vec c} \right).\vec a \cr & \Rightarrow \vec a\,||\,\vec c \cr} $$

$$a,\,b,\,c$$   are distinct non negative numbers and the vectors $$a\hat i + a\hat j + c\hat k,\,\hat i + \hat k$$     and $$c\hat i + c\hat j + b\hat k$$   are coplanar.
\[\therefore \left| \begin{array}{l} a\,\,\,\,a\,\,\,\,c\\ 1\,\,\,\,0\,\,\,\,1\\ c\,\,\,\,c\,\,\,\,b \end{array} \right| = 0 \Rightarrow \left| \begin{array}{l} a\,\,\,\,a\,\,\,\,c - a\\ 1\,\,\,\,0\,\,\,\,\,\,\,0\\ c\,\,\,\,c\,\,\,\,b - c \end{array} \right|\]
Operating $${C_3} \to {C_3} - {C_1}$$
Expanding along $${R_2},$$  we get
\[ - \left| \begin{array}{l} a\,\,\,\,c - a\\ c\,\,\,\,b - c \end{array} \right| = c\left( {c - a} \right) - a\left( {b - c} \right) = 0\]
$$\eqalign{ & \Rightarrow {c^2} - ac - ab + ac = 0 \cr & \Rightarrow {c^2} = ab \Rightarrow a,\,c,\,b\,\,{\text{are in G}}{\text{.P}}{\text{.}} \cr} $$
$$\therefore \,\,c$$   is the Geometric Mean of $$a$$ and $$b.$$

$$\eqalign{ & {\text{Here }}\left| {\overrightarrow a \times \overrightarrow b } \right| = q{\text{ and}} \cr & \frac{1}{2}\left| {\overrightarrow b \times \left( {10\overrightarrow a + 2\overrightarrow b } \right)} \right| + {\text{ }}\frac{1}{2}\left| {\overrightarrow a \times \left( {10\overrightarrow a + 2\overrightarrow b } \right)} \right| = p \cr & \therefore \,5\left| {\overrightarrow b \times \overrightarrow a } \right| + \left| {\overrightarrow a \times \overrightarrow b } \right| = p \cr & {\text{or }}6q = p\,\,\,\,\,\,\,\,\,\therefore \,\,\frac{p}{q} = 6. \cr} $$

Given $$\left| {2\hat u \times 3\hat v} \right| = 1$$   and $$\theta $$ is acute angle between $${\hat u}$$ and $$\hat v,\,\,\left| {\hat u} \right| = 1,\,\,\left| {\hat v} \right| = 1$$
$$\eqalign{ & \Rightarrow 6\,\left| {\hat u} \right|\,\left| {\hat v} \right|\left| {\sin \,\theta } \right| = 1 \cr & \Rightarrow 6\left| {\sin \,\theta } \right| = 1 \cr & \Rightarrow \sin \,\theta = \frac{1}{6} \cr} $$
Hence, there is exactly one value of $$\theta $$ for which $$2\hat u \times 3\hat v$$   is a unit vector.

$$\overrightarrow u .\overrightarrow v = \frac{{q - r}}{a} + \frac{{r - p}}{b} + \frac{{p - q}}{c}.....(1)$$
From the question, $$\frac{1}{a} = x + \left( {p - 1} \right)y,\,\frac{1}{b} = x + \left( {q - 1} \right)y,\,\frac{1}{c} = x + \left( {r - 1} \right)y$$
Putting these in $$(1)$$ and simplifying, $$\overrightarrow u .\overrightarrow v = 0$$
Clearly, $$\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right).\overrightarrow v = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ne 0.$$        So, $$\overrightarrow u \times \overrightarrow v \ne \overrightarrow i + \overrightarrow j + \overrightarrow k $$ 

Since, $$\overrightarrow a $$ is $$ \bot $$ to $$\overrightarrow d ,$$ so $$x - y + 2z = 0......\left( 1 \right)$$
Moreover, $$\left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right|,$$   so $$\overrightarrow a .\overrightarrow b = \overrightarrow a .\overrightarrow c $$
as $$\overrightarrow a $$ makes equal angles with $$\overrightarrow b $$ and $$\overrightarrow c .$$ Thus
$$\eqalign{ & xy - 2yz + 3xz = 2xz + 3xy - yz \cr & \Rightarrow xz - 2xy - yz = 0......\left( 2 \right) \cr & {\text{Also, }}{x^2} + {y^2} + {z^2} = 12......\left( 3 \right) \cr & {\text{and }}y < 0 \cr} $$
Put the value of $$y$$ from equation $$\left( 1 \right)$$ in equation $$\left( 2 \right),$$
we get, $${x^2} + 2xz + {z^2} = 0\,;$$
So, $$x = - z{\text{ and }}y = z$$
Again put these values in equation $$\left( 3 \right),$$
we get $${z^2} = 4 \Rightarrow z = \pm 2$$
But $$y < 0$$  and $$y = z$$
Hence, $$z = - 2 = y{\text{ and }}x = 2$$

$$\overrightarrow r = 3\overrightarrow i - \overrightarrow j + t\left( {\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k } \right).$$        So, a vector parallel to the line is $$\overrightarrow b = \overrightarrow i + 2\overrightarrow j + 3\overrightarrow k $$
$$\therefore $$  unit vector along the line is $$\frac{{\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k }}{{\sqrt {{1^2} + {2^2} + {3^2}} }},\,{\text{i}}{\text{.e}}{\text{., }}\frac{{\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k }}{{\sqrt {14} }}$$
$$\therefore $$   the projection $$ = \left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right).\frac{{\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k }}{{\sqrt {14} }} = \frac{{1 + 2 + 3}}{{\sqrt {14} }} = \frac{6}{{\sqrt {14} }}.$$