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.
111.
Let $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ be non-coplanar vectors and $$\overrightarrow p = \frac{{\overrightarrow b \times \overrightarrow c }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}},\,\overrightarrow q = \frac{{\overrightarrow c \times \overrightarrow a }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}},\,\overrightarrow r = \frac{{\overrightarrow a \times \overrightarrow b }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}.$$
What is the value of $$\left( {\overrightarrow a - \overrightarrow b - \overrightarrow c } \right).\overrightarrow p + \left( {\overrightarrow b - \overrightarrow c - \overrightarrow a } \right).\overrightarrow q + \left( {\overrightarrow c - \overrightarrow a - \overrightarrow b } \right).\overrightarrow r \,\, = ?$$
A
$$0$$
B
$$ - 3$$
C
$$3$$
D
$$ - 9$$
Answer :
$$3$$
$$\eqalign{
& {\text{As given }}\overrightarrow p = \frac{{\overrightarrow b \times \overrightarrow c }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}},\,\overrightarrow q = \frac{{\overrightarrow c \times \overrightarrow a }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{\text{ and }}\overrightarrow r = \frac{{\overrightarrow a \times \overrightarrow b }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr
& \therefore \,\left( {\overrightarrow a - \overrightarrow b - \overrightarrow c } \right).\overrightarrow p + \left( {\overrightarrow b - \overrightarrow c - \overrightarrow a } \right).\overrightarrow q + \left( {\overrightarrow c - \overrightarrow a - \overrightarrow b } \right).\overrightarrow r \cr
& = \frac{{\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} + \frac{{\overrightarrow b .\left( {\overrightarrow c \times \overrightarrow a } \right)}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{\text{ + }}\frac{{\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right)}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr
& \left[ {{\text{Since }}\overrightarrow b .\left( {\overrightarrow b \times \overrightarrow c } \right) = 0,\,\overrightarrow c .\left( {\overrightarrow b \times \overrightarrow c } \right) = 0,\overrightarrow c .\left( {\overrightarrow c \times \overrightarrow a } \right) = 0,\,\overrightarrow a .\left( {\overrightarrow c \times \overrightarrow a } \right) = 0,\,\overrightarrow a .\left( {\overrightarrow a \times \overrightarrow b } \right) = 0{\text{ and }}\overrightarrow b .\left( {\overrightarrow a \times \overrightarrow b } \right)} = 0 \right] \cr
& = \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} + \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} + \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} \cr
& = 3 \cr} $$
112.
If $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$ where $$\overrightarrow a ,\,\overrightarrow b $$ and $$\overrightarrow c $$ are any three vectors such that $$\overrightarrow a .\overrightarrow b \ne 0,\,\overrightarrow b .\overrightarrow c \ne 0$$ then $$\overrightarrow a $$ and $$\overrightarrow c $$ are :
A
inclined at an angle of $$\frac{\pi }{3}$$ between them
B
inclined at an angle of $$\frac{\pi }{6}$$ between them
C
perpendicular
D
parallel
Answer :
parallel
$$\eqalign{
& \left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\,\overrightarrow a .\overrightarrow b \ne 0,\,\overrightarrow b .\overrightarrow c \ne 0 \cr
& \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right).\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right).\overrightarrow a = \left( {\overrightarrow a .\overrightarrow c } \right).\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right) .\overrightarrow c \cr
& \Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = \left( {\overrightarrow b .\overrightarrow c } \right).\overrightarrow a \cr
& \Rightarrow \left. {\overrightarrow a } \right|\left| {\overrightarrow c } \right. \cr} $$
113.
Let $$\overrightarrow a ,\,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-coplanar vectors, and let and $$\overrightarrow p ,\,\overrightarrow q $$ and $$\overrightarrow r $$ be the vectors defined by the relations $$\overrightarrow p = \frac{{\overrightarrow b \times \overrightarrow c }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}},\,\overrightarrow q = \frac{{\overrightarrow c \times \overrightarrow a }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}$$ and $$\overrightarrow r = \frac{{\overrightarrow a \times \overrightarrow b }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}.$$ Then the value of the expression $$\left( {\overrightarrow a + \overrightarrow b } \right).\overrightarrow p + \left( {\overrightarrow b + \overrightarrow c } \right).\overrightarrow q + \left( {\overrightarrow c + \overrightarrow a } \right).\overrightarrow r $$ is equal to :
A
0
B
1
C
2
D
3
Answer :
3
$$\eqalign{
& \overrightarrow a .\overrightarrow p = \frac{{\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} = \frac{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} = 1 = \overrightarrow b .\overrightarrow q = \overrightarrow c .\overrightarrow r \cr
& \overrightarrow b .\overrightarrow p = \frac{{\overrightarrow b .\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} = \frac{0}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}} = 0 = \overrightarrow c .\overrightarrow p = \overrightarrow a .\overrightarrow r \cr} $$
Therefore, the given expression is equal to $$1 + 0 + 1 + 0 + 1 + 0 = 3.$$
114.
If $$\overrightarrow {OA} = \overrightarrow a \,;\,\overrightarrow {OB} = \overrightarrow b \,;\,\overrightarrow {OC} = 2\overrightarrow a + 3\overrightarrow b \,;\overrightarrow {OD} = \overrightarrow a - 2\overrightarrow b ,$$ the length of $$\overrightarrow {OA} $$ is three times the length of $$\overrightarrow {OB} $$ and $$\overrightarrow {OA} $$ is perpendicular to $$\overrightarrow {DB} $$ then $$\left( {\overrightarrow {BD} \times \overrightarrow {AC} } \right).\left( {\overrightarrow {OD} \times \overrightarrow {OC} } \right)$$ is :
A
$$7{\left| {\overrightarrow a \times \overrightarrow b } \right|^2}$$
B
$$42{\left| {\overrightarrow a \times \overrightarrow b } \right|^2}$$
C
$$0$$
D
None of these
Answer :
$$42{\left| {\overrightarrow a \times \overrightarrow b } \right|^2}$$
$$\eqalign{
& \overrightarrow {BD} = \overrightarrow a - 3\overrightarrow b ,\,\,\overrightarrow {AC} = \overrightarrow a + 3\overrightarrow b \cr
& \overrightarrow {BD} \times \overrightarrow {AC} = \left( {\overrightarrow a - 3\overrightarrow b } \right) \times \left( {\overrightarrow a + 3\overrightarrow b } \right) = 6\overrightarrow a \times \overrightarrow b \cr
& \overrightarrow {OD} \times \overrightarrow {OC} = \left( {\overrightarrow a - 2\overrightarrow b } \right) \times \left( {2\overrightarrow a + 3\overrightarrow b } \right) = 7\overrightarrow a \times \overrightarrow b \cr
& \left( {\overrightarrow {BD} \times \overrightarrow {AC} } \right).\left( {\overrightarrow {OD} \times \overrightarrow {OC} } \right) = 42{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} \cr} $$
115.
If $$\left| {\overrightarrow a } \right| = 5,\,\left| {\overrightarrow a - \overrightarrow b } \right| = 8$$ and $$\left| {\overrightarrow a + \overrightarrow b } \right| = 10$$ then $$\left| {\overrightarrow b } \right|$$ is :
A
$$1$$
B
$$\sqrt {57} $$
C
$$3$$
D
none of these
Answer :
$$\sqrt {57} $$
$$\eqalign{
& \because \,\,{\left| {\overrightarrow a + \overrightarrow b } \right|^2} = \,{\left| {\overrightarrow a - \overrightarrow b } \right|^2} = 4\overrightarrow a .\overrightarrow b \,\,\,\therefore \,4\overrightarrow a .\overrightarrow b = {10^2} - {8^2} = 36 \cr
& \therefore \,\overrightarrow a .\overrightarrow b = 9 \cr
& \,{\text{Now, }}{\left| {\overrightarrow a - \overrightarrow b } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 2\overrightarrow a .\overrightarrow b \, = {5^2} + {\left| {\overrightarrow b } \right|^2} - 18 \cr
& \therefore \,{8^2} = 7 + {\left| {\overrightarrow b } \right|^2}\,\,\,\,\,\,\,\,\,\therefore \,{\left| {\overrightarrow b } \right|^2} = \sqrt {57} \cr} $$
116.
A vector of magnitude $$3$$, bisecting the angle between the vectors $$\overrightarrow a = 2\hat i + \hat j - \hat k$$ and $$\overrightarrow b = \hat i - 2\hat j + \hat k$$ and making an obtuse angle with $$\overrightarrow b $$ is :
A
$$\frac{{3\hat i - \hat j}}{{\sqrt 6 }}$$
B
$$\frac{{\hat i + 3\hat j - 2\hat k}}{{\sqrt {14} }}$$
C
$$\frac{{3\left( {\hat i + 3\hat j - 2\hat k} \right)}}{{\sqrt {14} }}$$
A vector bisecting the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$\frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} \pm \frac{{\overrightarrow b }}{{\left| {\overrightarrow b } \right|}}\,;$$ in this case $$\frac{{2\hat i + \hat j - \hat k}}{{\sqrt 6 }} \pm \frac{{\hat i - 2\hat j + \hat k}}{{\sqrt 6 }}{\text{ i}}{\text{.e}}{\text{.,}}$$
$$\frac{{3\hat i - \hat j}}{{\sqrt 6 }}{\text{ or }}\frac{{\hat i + 3\hat j - 2\hat k}}{{\sqrt 6 }}$$
A vector of magnitude $$3$$ along these vectors is $$\frac{{3\left( {3\hat i - \hat j} \right)}}{{\sqrt {10} }}{\text{ or }}\frac{{3\left( {\hat i + 3\hat j - 2\hat k} \right)}}{{\sqrt {14} }}$$
Now, $$\frac{3}{{\sqrt {14} }}\left( {\hat i + 3\hat j - 2\hat k} \right).\left( {\hat i - 2\hat j + \hat k} \right)$$ is negative and hence $$\frac{3}{{\sqrt {14} }}\left( {\hat i + 3\hat j - 2\hat k} \right)$$ makes an obtuse angle with $$\overrightarrow b .$$
117.
If $$\left| {\vec a} \right| = 5,\,\left| {\vec b} \right| = 4,\,\left| {\vec c} \right| = 3$$ thus what will be the value of $$\left| {\vec a.\vec b + \vec b.\vec c + \vec c.\vec a} \right|,$$ given that $$\vec a + \vec b + \vec c = 0$$
A
$$25$$
B
$$50$$
C
$$ - 25$$
D
$$ - 50$$
Answer :
$$25$$
$$\eqalign{
& {\text{We have, }}\vec a + \vec b + \vec c = 0\, \cr
& \, \Rightarrow {\left( {\vec a + \vec b + \vec c} \right)^2} = 0 \cr
& \Rightarrow {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + {\left| {\vec c} \right|^2} + 2\left( {\vec a.\vec b + \vec b.\vec c + \vec c.\vec a} \right) = 0 \cr
& \Rightarrow 25 + 16 + 9 + 2\left( {\vec a.\vec b + \vec b.\vec c + \vec c.\vec a} \right) = 0 \cr
& \Rightarrow \left( {\vec a.\vec b + \vec b.\vec c + \vec c.\vec a} \right) = - 25 \cr
& \therefore \left| {\vec a.\vec b + \vec b.\vec c + \vec c.\vec a} \right| = 25 \cr} $$
118.
Let $$\vec u = \hat i + \hat j,\,\vec v = \hat i - \hat j$$ and $$\vec w = \hat i + 2\hat j + 3\hat k.$$ If $${\hat n}$$ is a unit vector such that $$\vec u.\hat n = 0$$ and $$\vec v.\hat n = 0,$$ then $$\left| {\vec w.\hat n} \right|$$ is equal to :
119.
Let two non-collinear unit vectors $$\hat a$$ and $$\hat b$$ form an acute angle. A point $$P$$ moves so that at any time $$t$$ the position vector $$\overrightarrow {OP} $$ (where $$O$$ is the origin) is given by $$\hat a\,\cos \,t + \hat b\,\sin \,t.$$ When $$P$$ is farthest from origin $$O,$$ let $$M$$ be the length of $$\overrightarrow {OP} $$ and $${\hat u}$$ be the unit vector along $$\overrightarrow {OP} .$$ Then,
A
$$\hat u = \frac{{\hat a + \hat b}}{{\left| {\hat a + \hat b} \right|}}{\text{ and }}M = {\left( {1 + \hat a.\hat b} \right)^{\frac{1}{2}}}$$
B
$$\hat u = \frac{{\hat a - \hat b}}{{\left| {\hat a - \hat b} \right|}}{\text{ and }}M = {\left( {1 + \hat a.\hat b} \right)^{\frac{1}{2}}}$$
C
$$\hat u = \frac{{\hat a + \hat b}}{{\left| {\hat a + \hat b} \right|}}{\text{ and }}M = {\left( {1 + 2\hat a.\hat b} \right)^{\frac{1}{2}}}$$
D
$$\hat u = \frac{{\hat a - \hat b}}{{\left| {\hat a - \hat b} \right|}}{\text{ and }}M = {\left( {1 + 2\hat a.\hat b} \right)^{\frac{1}{2}}}$$
Answer :
$$\hat u = \frac{{\hat a + \hat b}}{{\left| {\hat a + \hat b} \right|}}{\text{ and }}M = {\left( {1 + \hat a.\hat b} \right)^{\frac{1}{2}}}$$
120.
If $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are three noncoplanar vectors represented by concurrent edges of a parallelepiped of volume $$4$$ then $$\left( {\overrightarrow a + \overrightarrow b } \right).\left( {\overrightarrow b \times \overrightarrow c } \right) + \left( {\overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow c \times \overrightarrow a } \right) + \left( {\overrightarrow c + \overrightarrow a } \right)\left( {\overrightarrow a \times \overrightarrow b } \right)$$ is equal to :
A
$$12$$
B
$$4$$
C
$$ \pm 12$$
D
$$0$$
Answer :
$$ \pm 12$$
Value $$ = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] + \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right] + \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right] = 3\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 3\left( { \pm 4} \right) = \pm 12.$$