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

We have $$\overrightarrow {PQ} = 6\hat i + \hat j,\,\overrightarrow {QR} = - \hat i + 3\hat j,\,\overrightarrow {SR} = 6\hat i + \hat j,\,\overrightarrow {PS} = - \hat i + 3\hat j$$
$$\eqalign{ & \Rightarrow \overrightarrow {PQ} = \overrightarrow {SR} \,;\,\overrightarrow {QR} = \overrightarrow {PS} \,\,{\text{and }}\overrightarrow {PQ} .\overrightarrow {PS} \ne 0 \cr & {\text{Also }}\left| {\overrightarrow {PQ} } \right| \ne \left| {\overrightarrow {QR} } \right| \cr} $$
$$ \Rightarrow PQRS$$   is a parallelogram but neither a rhombus nor a rectangle.

Let $$P\left( {{x_1},\,{y_1},\,{z_1}} \right)$$   and $$Q\left( {{x_2},\,{y_2},\,{z_2}} \right)$$   be the initial and final points of the vector whose projections on the three coordinate axes are $$6,\, - 3,\,2$$
then
$${x_2} - {x_1} = 6\,;\,\,\,{y_2} - {y_1} = - 3\,;\,\,\,{z_2} - \,{z_1} = 2$$
So that direction ratios of $$\overrightarrow {PQ} $$  are $$6,\, - 3,\,2$$
$$\therefore $$ Direction cosines of $$\overrightarrow {PQ} $$  are
$$\eqalign{ & \frac{6}{{\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},\,\frac{{ - 3}}{{\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},\,\frac{2}{{\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }} \cr & = \frac{6}{7},\,\frac{{ - 3}}{7},\,\frac{2}{7} \cr} $$

Here, $$\left| {\overrightarrow a } \right| = 2 = \left| {\overrightarrow b } \right|$$    and $$\overrightarrow a .\overrightarrow b = 2.2\cos \,{60^ \circ } = 2$$
If the angle between $$\overrightarrow a $$ and $$\left( {\overrightarrow a + \overrightarrow b } \right)$$  be $$\theta $$ then
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\overrightarrow a .\left( {\overrightarrow a + \overrightarrow b } \right) = \left| {\overrightarrow a } \right|\,\,\left| {\overrightarrow a + \overrightarrow b } \right|\cos \,\theta \cr & {\text{or }}{\left| {\overrightarrow a } \right|^2}\, + \overrightarrow a .\overrightarrow b = 2\,\left| {\overrightarrow a + \overrightarrow b } \right|\cos \,\theta \cr & {\text{or }}4 + 2 = 2\left| {\overrightarrow a + \overrightarrow b } \right|\cos \,\theta \cr & \therefore \,\cos \,\theta = \frac{3}{{\left| {\overrightarrow a + \overrightarrow b } \right|}} \cr & {\text{Now, }}{\left| {\overrightarrow a + \overrightarrow b } \right|^2} = {\left( {\overrightarrow a + \overrightarrow b } \right)^2} \cr & = {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b \cr & = 4 + 4 + 2.2 \cr & = 12 \cr & \therefore \,\,\,\left| {\overrightarrow a + \overrightarrow b } \right| = 2\sqrt 3 \cr & \therefore \,\,\cos \,\theta = \frac{3}{{2\sqrt 3 }} = \frac{{\sqrt 3 }}{2}\,\,\,\,\, \Rightarrow \theta = {30^ \circ } \cr} $$

Let $$\overrightarrow p = \lambda \overrightarrow b + \mu \overrightarrow c $$
The projection of $$\overrightarrow p $$ on $$\overrightarrow a = \frac{{\overrightarrow p .\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \sqrt {\frac{2}{3}} ......\left( 1 \right)$$
$$\eqalign{ & \because \,\overrightarrow p = \lambda \left( {\overrightarrow i + 2\overrightarrow j - \overrightarrow k } \right) + \mu \left( {\overrightarrow i + \overrightarrow j - 2\overrightarrow k } \right) \cr & \therefore \,\overrightarrow p .\overrightarrow a = 2\left( {\lambda + \mu } \right) - 1\left( {2\lambda + \mu } \right) + 1\left( { - \lambda - 2\mu } \right) = - \lambda - \mu \cr & \therefore \,\left( 1 \right){\text{gives}}\frac{{ - \lambda - \mu }}{{\sqrt {{2^2} + {1^2} + {1^2}} }} = \sqrt {\frac{2}{3}} \,\,\,{\text{or }}\lambda + \mu = - 2 \cr & \therefore \overrightarrow p = \lambda \left( {\overrightarrow i + 2\overrightarrow j - \overrightarrow k } \right) + \mu \left( {\overrightarrow i + \overrightarrow j - 2\overrightarrow k } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\lambda + \mu } \right)\overrightarrow i + \left( {2\lambda + \mu } \right)\overrightarrow j - \left( {\lambda + 2\mu } \right)\overrightarrow k \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 2\overrightarrow i + \left( {\lambda - 2} \right)\overrightarrow j + \left( {\lambda + 4} \right)\overrightarrow k , \cr} $$
where $$\lambda $$ is a scalar parameter.
When $$\lambda = 1,\,\overrightarrow p = - 2\overrightarrow i - \overrightarrow j + 5\overrightarrow k .$$      Other options hold for no real $$\lambda .$$

Equation of planes be $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$    & $$\frac{x}{{a'}} + \frac{y}{{b'}} + \frac{z}{{c'}} = 1$$
( $$ \bot \,\,r$$  distance on plane from origin is same. )
$$\eqalign{ & \left| {\frac{{ - 1}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}} \right| = \left| {\frac{{ - 1}}{{\sqrt {\frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}} }}} \right| \cr & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} - \frac{1}{{a{'^2}}} - \frac{1}{{b{'^2}}} - \frac{1}{{c{'^2}}} = 0 \cr} $$

\[\begin{array}{l} \left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow c } \right) = \left( {\overrightarrow a \times \overrightarrow b .\overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a \times \overrightarrow b .\overrightarrow a } \right)\overrightarrow c \\ = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow a - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow c \\ = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\overrightarrow a \\ {\rm{Similarly,}}\,\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow b \times \overrightarrow a } \right) = \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right]\overrightarrow b \\ \left( {\overrightarrow c \times \overrightarrow a } \right) \times \left( {\overrightarrow c \times \overrightarrow b } \right) = \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right]\overrightarrow c \\ \therefore {\rm{ the\, expression}} = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \end{array}\]

We have $$\overrightarrow {OR} = \frac{{3\vec p + 2\vec q}}{{3 + 2}} = \frac{1}{2}\left( {3\vec p + 2\vec q} \right)\,\,\,\,\left[ {{\text{Internal division}}} \right]$$
and $$\overrightarrow {OS} = \frac{{3\vec p - 2\vec q}}{{3 - 2}} = 3\vec p - 2\vec q\,\,\,\,\left[ {{\text{External division}}} \right]$$
$$\eqalign{ & {\text{Given }}\overrightarrow {OR} \, \bot \,\overrightarrow {OS} \Rightarrow \overrightarrow {OR} \,.\,\overrightarrow {OS} = 0 \cr & \Rightarrow \frac{1}{5}\left[ {3\vec p + 2\vec q} \right].\left( {3\vec p - 2\vec q} \right) = 0 \cr & \Rightarrow 9{\left| {\vec p} \right|^2} = 4{\left| {\vec q} \right|^2} \cr & \Rightarrow 9{p^2} = 4{q^2} \cr} $$

$$\vec \beta = {{\vec \beta }_1} - {{\vec \beta }_2}.....(1)$$
Since, $${{\vec \beta }_2}$$ is perpendicular to $${\vec \alpha }.$$
$$\therefore {{\vec \beta }_2}.\vec \alpha = 0$$
Since, $${{\vec \beta }_1}$$ is parallel to $${\vec a}.$$
then $${{\vec \beta }_1} = \lambda \vec \alpha \,\,\left( {{\text{say}}} \right)$$
$$\eqalign{ & \vec a.\vec \beta = \vec a.{{\vec \beta }_1} - \vec \alpha .{{\vec \beta }_2} \cr & \Rightarrow 5 = \lambda {\alpha ^2} \cr & \Rightarrow 5 = 1 \times 10 \cr & \Rightarrow \lambda = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \left| {\vec \alpha } \right| = \sqrt {10} } \right) \cr & \therefore {{\vec \beta }_1} = \frac{{\vec \alpha }}{2} \cr} $$
Cross product with $${{\vec \beta }_1}$$ in equation (1)
\[\begin{array}{l} \Rightarrow \vec \beta \times {{\vec \beta }_1} = - {{\vec \beta }_2} \times {{\vec \beta }_1}\\ \Rightarrow \vec \beta \times {{\vec \beta }_1} = {{\vec \beta }_1} \times {{\vec \beta }_2}\\ \Rightarrow {{\vec \beta }_1} \times {{\vec \beta }_2} = \frac{{ - {{\vec \beta }_1} \times \vec \alpha }}{2}\\ \Rightarrow {{\vec \beta }_1} \times {{\vec \beta }_2} = \frac{1}{2}\left| \begin{array}{l} \hat i\,\,\,\,\,j\,\,\,\,\,k\\ 2\,\,\, - 1\,\,\,\,3\\ 3\,\,\,\,\,1\,\,\,\,\,0 \end{array} \right|\\ \Rightarrow {{\vec \beta }_1} \times {{\vec \beta }_2} = \frac{1}{2}\left[ { - 3\hat i - \hat j\left( { - 9} \right) + \hat k\left( 5 \right)} \right] \end{array}\]

Let $$l,\,m,\,n$$   be the required direction cosines. Then
$$\eqalign{ & l.\frac{1}{3} + m.\frac{2}{3} + n.\frac{2}{3} = l.\frac{2}{7} + m.\frac{3}{7} + n.\frac{6}{7} = l.\frac{3}{{13}} + m.\frac{4}{{13}} + n.\frac{{12}}{{13}} \cr & \Rightarrow \left( {\frac{1}{3} - \frac{2}{7}} \right)l + \left( {\frac{2}{3} - \frac{3}{7}} \right)m + \left( {\frac{2}{3} - \frac{6}{7}} \right)n = 0 \cr & \Rightarrow l + 5m - 4n = 0 \cr & \left( {\frac{1}{3} - \frac{3}{{13}}} \right)l + \left( {\frac{2}{3} - \frac{4}{{13}}} \right)m + \left( {\frac{2}{3} - \frac{{12}}{{13}}} \right)n = 0 \cr & \Rightarrow 4l + 14m - 10n = 0 \cr & \, \Rightarrow 2l + 7m - 5n = 0 \cr & \therefore \frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{{ - 3}}{\text{ or }}\frac{l}{1} = \frac{m}{{ - 1}} = \frac{n}{{ - 1}} \cr & \therefore l = \frac{1}{{\sqrt 3 }},\,m = \frac{{ - 1}}{{\sqrt 3 }},\,n = \frac{{ - 1}}{{\sqrt 3 }}. \cr} $$

$$\eqalign{ & \overrightarrow a \bot \overrightarrow b ,\,\overrightarrow a \bot \overrightarrow c ,\,{\text{i}}{\text{.e}}{\text{., }}\overrightarrow a ||\left( {\overrightarrow b \times \overrightarrow c } \right){\text{ and }}\left| {\overrightarrow b \times \overrightarrow c } \right| = \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2} \cr & \left| {\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]} \right| = \left| {\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)} \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b \times \overrightarrow c } \right|\cos \,{0^ \circ } = 1.\frac{{\sqrt 3 }}{2}.1 = \frac{{\sqrt 3 }}{2} \cr} $$