3D Geometry and Vectors MCQ Questions & Answers in Geometry | Maths
Learn 3D Geometry and Vectors MCQ questions & answers in Geometry are available for students perparing for IIT-JEE and engineering Enternace exam.
251.
The shortest distance from the plane $$12x + 4y + 3z = 327$$ to the sphere $${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is :
A
$$39$$
B
$$26$$
C
$$11\frac{4}{{13}}$$
D
$$13$$
Answer :
$$13$$
Shortest distance $$=$$ perpendicular distance between the plane and sphere $$=$$ distance of plane from centre of sphere $$-$$ radius
$$\eqalign{
& = \left| {\frac{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327}}{{\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} \cr
& = 26 - 13 \cr
& = 13 \cr} $$
252.
A variable plane at a distance of $$1$$ unit from the origin cuts the coordinate axes at $$A,\,B$$ and $$C.$$ If the centroid $$D\left( {x,\,y,\,z} \right)$$ satisfies the relation $${x^{ - 2}} + {y^{ - 2}} + {z^{ - 2}} = k$$ then the value of $$k$$ is :
A
$$3$$
B
$$1$$
C
$$\frac{1}{3}$$
D
$$9$$
Answer :
$$9$$
The plane is $$lx + my + nz = 1$$ where $${l^2} + {m^2} + {n^2} = 1.$$ It cuts axes at $$\left( {\frac{1}{l},\,0,\,0} \right),\,\left( {0,\,\frac{1}{m},\,0} \right),\,\left( {0,\,0,\,\frac{1}{n}} \right).$$
$$\therefore $$ the centroid $$ = \left( {\frac{1}{{3l}},\,\frac{1}{{3m}},\,\frac{1}{{3n}}} \right).$$ It satisfies $$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = k\,\,\, \Rightarrow 9\left( {{l^2} + {m^2} + {n^2}} \right) = k\,\,\, \Rightarrow 9 = k.$$
253.
The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors $$\hat a,\,\hat b,\,\hat c$$ such that $$\hat a.\hat b = \,\hat b.\hat c = \hat c.\hat a = \frac{1}{2}.$$ Then, the volume of the parallelepiped is :
A
$$\frac{1}{{\sqrt 2 }}$$
B
$$\frac{1}{{2\sqrt 2 }}$$
C
$$\frac{{\sqrt 3 }}{2}$$
D
$$\frac{1}{{\sqrt 3 }}$$
Answer :
$$\frac{1}{{\sqrt 2 }}$$
We know that the volume of a parallelepiped with coterminous edges as the vectors $$\hat a,\,\hat b,\,\hat c$$ is given by
\[\begin{array}{l}
V = {\left[ {\vec a\,\vec b\,\vec c} \right]^2} = \left[ \begin{array}{l}
\vec a.\vec a\,\,\,\,\,\vec a.\vec b\,\,\,\,\,\vec a.\vec c\\
\vec b.\vec a\,\,\,\,\,\vec b.\vec b\,\,\,\,\,\vec b.\vec c\\
\vec c.\vec a\,\,\,\,\,\vec c.\vec b\,\,\,\,\,\vec c.\vec c
\end{array} \right]\\
{V^2} = \left| \begin{array}{l}
1\,\,\,\,\,\frac{1}{2}\,\,\,\,\,\frac{1}{2}\\
\frac{1}{2}\,\,\,\,\,1\,\,\,\,\,\frac{1}{2}\\
\frac{1}{2}\,\,\,\,\,\frac{1}{2}\,\,\,\,\,1
\end{array} \right| = \frac{1}{2}\\
\Rightarrow V = \frac{1}{{\sqrt 2 }}
\end{array}\]
254.
The three concurrent edges of a parallelepiped represent the vectors $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ such that $$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = \lambda .$$ Then the volume of the parallelepiped whose three concurrent edges are the three concurrent diagonals of three faces of the given parallelepiped is :
A
$$2\lambda $$
B
$$3\lambda $$
C
$$\lambda $$
D
none of these
Answer :
$$2\lambda $$
The new edges of the new parallelepiped will be $$\left( {\overrightarrow a + \overrightarrow b } \right),\,\left( {\overrightarrow b + \overrightarrow c } \right),\,\left( {\overrightarrow c + \overrightarrow a } \right)$$
Hence,
The parallelepiped thus formed will have a volume
$$\eqalign{
& \left[ {\left( {\overrightarrow a + \overrightarrow b } \right)\,\,\,\left( {\overrightarrow b + \overrightarrow c } \right)\,\,\,\left( {\overrightarrow c + \overrightarrow a } \right)} \right] \cr
& = 2\left[ {\overrightarrow a \,\,\,\overrightarrow b \,\,\,\overrightarrow c } \right] \cr
& = 2\lambda \cr} $$
Hence,
The volume of the new parallelepiped is equal to $$2\lambda $$.
255.
If $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are three vectors of equal magnitude and the angle between each pair of vectors is $$\frac{\pi }{3}$$ such that $$\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 6 $$ then $$\left| {\overrightarrow a } \right|$$ is equal to :
A
$$2$$
B
$$ - 1$$
C
$$1$$
D
$$\frac{1}{3}\sqrt 6 $$
Answer :
$$1$$
$$\eqalign{
& {\text{Let }}\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k.{\text{ Then}} \cr
& {\text{cos}}\frac{\pi }{3} = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}} = \frac{{\overrightarrow a .\overrightarrow b }}{{{k^2}}}{\text{ or }}\overrightarrow a .\overrightarrow b = \frac{{{k^2}}}{2} \cr
& {\text{Similarly, }}\overrightarrow b .\overrightarrow c = \overrightarrow c .\overrightarrow a = \frac{{{k^2}}}{2}.{\text{ So, }}\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a = \frac{{3{k^2}}}{2} \cr
& {\text{Now, }}6 = {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) \cr
& \therefore \,6 = 3{k^2} + 2.\frac{{3{k^2}}}{2} \cr
& \therefore \,k = \pm 1.\,{\text{But }}\left| {\overrightarrow a } \right|\,{\text{is positive}}{\text{. So, }}\left| {\overrightarrow a } \right| = 1. \cr} $$
256.
A unit vector perpendicular to the plane passing through the points
whose position vectors are $$\overrightarrow i - \overrightarrow j + 2\overrightarrow k ,\,2\overrightarrow i - \overrightarrow k $$ and $$2\overrightarrow j + \overrightarrow k $$ is :
A
$$2\overrightarrow i + \overrightarrow j + \overrightarrow k $$
B
$$\frac{1}{{\sqrt 6 }}\left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right)$$
C
$$\frac{1}{{\sqrt 6 }}\left( {\overrightarrow i + 2\overrightarrow j + \overrightarrow k } \right)$$
D
none of these
Answer :
$$\frac{1}{{\sqrt 6 }}\left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right)$$
Let the points be $$A,\,B,\,C.$$ Then $$\overrightarrow {OA} = \overrightarrow i - \overrightarrow j + 2\overrightarrow k ,\,\overrightarrow {OB} = 2\overrightarrow i - \overrightarrow k ,\,\overrightarrow {OC} = 2\overrightarrow j + \overrightarrow k .$$
The required vector $$ = \frac{{\overrightarrow {AB} \times \overrightarrow {AC} }}{{\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|}}.$$ Calculate its value.
257.
Let $$\vec a = \vec i - \vec k,\,\vec b = x\vec i + \vec j + \left( {1 - x} \right)\vec k$$ and $$\vec c = y\vec i + x\vec j + \left( {1 + x - y} \right)\vec k.$$ Then $$\left[ {\vec a\,\vec b\,\vec c} \right]$$ depends on :
A
only $$x$$
B
only $$y$$
C
neither $$x$$ nor $$y$$
D
both $$x$$ and $$y$$
Answer :
neither $$x$$ nor $$y$$
\[\begin{array}{l}
\vec a = \hat i - \hat k,\\
\vec b = x\hat i + \hat j + \left( {1 - x} \right)\hat k,\\
\vec c = y\hat i + x\hat j + \left( {1 + x - y} \right)\hat k\\
\left[ {\vec a\,\vec b\,\vec c} \right] = \left| \begin{array}{l}
1\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - 1\\
x\,\,\,\,\,\,1\,\,\,\,\,\,\,1 - x\\
y\,\,\,\,\,\,x\,\,\,\,\,1 + x - y
\end{array} \right|\\
= 1\left( {1 + x - y - x + {x^2}} \right) - 1\left( {{x^2} - y} \right)\\
= 1
\end{array}\]
\[\therefore \] Depends neither on $$x$$ nor on $$y.$$
258.
Let $$\vec u$$ be a vector coplanar with the vectors $$\vec a = 2\hat i + 3\hat j - \hat k$$ and $$\vec b = \hat j + \hat k.$$ If $${\vec u}$$ is perpendicular to $${\vec a}$$ and $$\vec u.\vec b - 24,$$ then $${\left| {\vec u} \right|^2}$$ is equal to :
A
$$315$$
B
$$256$$
C
$$84$$
D
$$336$$
Answer :
$$336$$
$$\eqalign{
& \because \,\vec u,\,\vec a\,\& \,\vec b{\text{ are coplanar}} \cr
& \therefore \,\vec u = \lambda \left( {\vec a \times \vec b} \right) \times \,\vec a = \lambda \left\{ {{{\vec a}^2}.\vec b - \left( {\vec a.\vec b} \right)\vec a} \right\} \cr
& = \lambda \left\{ { - 4\hat i + 8\hat j + 16\hat k} \right\} = \lambda '\left\{ { - \hat i + 2\hat j + 4\hat k} \right\} \cr
& {\text{Also, }}\vec u.\vec b = 24\,\,\,\, \Rightarrow \lambda ' = 4 \cr
& \therefore \vec u = - 4\hat i + 8\hat j + 16\hat k \cr
& \Rightarrow {\left| {\vec u} \right|^2} = 336 \cr} $$
259.
Let $$\vec a,\,\vec b$$ and $$\vec c$$ be three non-zero vectors such that no two of them are collinear and $$\left( {\vec a \times \vec b} \right) \times \vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\vec a$$ If $$\theta $$ is the angle between vectors $${\vec b}$$ and $${\vec c},$$ then a value of $$\sin \,\theta $$ is :
A
$$\frac{2}{3}$$
B
$$\frac{{ - 2\sqrt 3 }}{3}$$
C
$$\frac{{ 2\sqrt 2 }}{3}$$
D
$$\frac{{ - \sqrt 2 }}{3}$$
Answer :
$$\frac{{ 2\sqrt 2 }}{3}$$
$$\eqalign{
& \left( {\vec a \times \vec b} \right) \times \vec c = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\vec a \cr
& \Rightarrow - \vec c\left( {\vec a \times \vec b} \right) = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\vec a \cr
& \Rightarrow - \left( {\vec c.\vec b} \right)\vec a + \left( {\vec c.\vec a} \right)\vec b = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\vec a \cr
& \Rightarrow - \left| {\vec b} \right|\left| {\vec c} \right|\,\cos \,\theta \,\vec a + \left( {\vec c.\vec a} \right)\vec b = \frac{1}{3}\left| {\vec b} \right|\left| {\vec c} \right|\vec a \cr} $$
$$\because \,\vec a,\,\vec b,\,\vec c$$ are non collinear, the above equation is possible only when
$$\eqalign{
& - \cos \,\theta = \frac{1}{3}\,\,{\text{and }}\,\vec c.\vec a = 0 \cr
& \Rightarrow \cos \,\theta = - \frac{1}{3}\,\,\, \Rightarrow \sin \,\theta = \frac{{2\sqrt 2 }}{3}\,;\theta \, \in \,{\text{II}}\,{\text{quad}} \cr} $$
260.
Resolved part of vector $$\overrightarrow a $$ along vector $$\overrightarrow b $$ is $${\overrightarrow a _1}$$ and that perpendicular to $$\overrightarrow b $$ is $${\overrightarrow a _2}$$ then $${\overrightarrow a _1} \times {\overrightarrow a _2}$$ is equal to :
A
$$\frac{{\left( {\overrightarrow a \times \overrightarrow b } \right) . \overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}}$$
B
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}$$
C
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{{{\left| {\overrightarrow b } \right|}^2}}}$$
D
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{\left| {\overrightarrow b \times \overrightarrow a } \right|}}$$
Answer :
$$\frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{{{\left| {\overrightarrow b } \right|}^2}}}$$
$$\eqalign{
& {\overrightarrow a _1} = \left( {\overrightarrow a .\hat b} \right)\hat b = \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr
& \Rightarrow {\overrightarrow a _2} = \overrightarrow a - {\overrightarrow a _1} = \overrightarrow a - \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr
& {\text{Thus, }}{\overrightarrow a _1} \times {\overrightarrow a _2} = \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}} \times \left( {\overrightarrow a - \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| {\overrightarrow b } \right|}^2}}}} \right) \cr
& = \frac{{\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)}}{{{{\left| {\overrightarrow b } \right|}^2}}} \cr} $$