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.
21.
If $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are three noncoplanar vectors then $$\left[ {\overrightarrow a + \overrightarrow b + \overrightarrow c \,\,\overrightarrow a - \overrightarrow c \,\,\overrightarrow a - \overrightarrow b } \right]$$ is equal to :
A
$$0$$
B
$$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]$$
C
$$ - 3\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]$$
D
$$2\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]$$
Answer :
$$ - 3\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]$$
$$\eqalign{
& \,\,\,\,\,\,\,\left\{ {\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \times \left( {\overrightarrow a - \overrightarrow c } \right)} \right\}.\left( {\overrightarrow a - \overrightarrow b } \right) \cr
& = \left\{ { - \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right\}.\left( {\overrightarrow a - \overrightarrow b } \right) \cr
& = \left[ {\overrightarrow a \,\,\overrightarrow c \,\,\overrightarrow b } \right] - \left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right] - \left[ {\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b } \right] \cr
& = - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] - \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr
& = - 3\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] \cr} $$
22.
Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \vec q,\,\overrightarrow {AD} = \vec p$$ and $$\angle BAD$$ bean acute angle. If $${\vec r}$$ is the vector that coincide with the altitude directed from the vertex $$B$$ to the side $$AD,$$ then $${\vec r}$$ is given by :
A
$$\vec r = 3\vec q - \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
B
$$\vec r = - \vec q + \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
C
$$\vec r = \vec q - \frac{{\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
D
$$\vec r = - 3\vec q - \frac{{3\left( {\vec p.\vec q} \right)}}{{\left( {\vec p.\vec p} \right)}}\vec p$$
Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \vec q,\,\overrightarrow {AD} = \vec p$$ and $$\angle BAD$$ be an acute angle.
We have
$$\eqalign{
& \overrightarrow {AX} = \left( {\frac{{\vec p.\vec q}}{{\left| {\vec p} \right|}}} \right)\left( {\frac{{\vec p}}{{\left| {\vec p} \right|}}} \right) = \frac{{\vec p.\vec q}}{{{{\left| {\vec p} \right|}^2}}}\vec p \cr
& {\text{Let }}\vec r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX} = - \vec q + \frac{{\vec p.\vec q}}{{{{\left| {\vec p} \right|}^2}}}\vec p \cr} $$
23.
If $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ be three vectors such that $$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 4$$ then $$\left[ {\overrightarrow a \times \overrightarrow b \,\,\overrightarrow b \times \overrightarrow c \,\,\overrightarrow c \times \overrightarrow a } \right]$$ is equal to :
A
8
B
16
C
64
D
none of these
Answer :
16
$$\eqalign{
& \left[ {\overrightarrow a \times \overrightarrow b \,\,\overrightarrow b \times \overrightarrow c \,\,\overrightarrow c \times \overrightarrow a } \right] = \left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow b \times \overrightarrow c } \right).\overrightarrow c \times \overrightarrow a \cr
& = \left\{ {\left( {\overrightarrow a \times \overrightarrow b .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a \times \overrightarrow b .\overrightarrow b } \right)\overrightarrow c } \right\}.\overrightarrow c \times \overrightarrow a \cr
& = \left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]\,\left[ {\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a } \right] \cr
& = {\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]^2} \cr
& = {4^2} \cr
& = 16 \cr} $$
24.
Consider points $$A,\,B,\,C$$ and $$D$$ with position vectors $$7\hat i - 4\hat j + 7\hat k,\,\hat i - 6\hat j + 10\hat k,\, - \hat i - 3\hat j + 4\hat k$$ and $$5\hat i - \hat j + 5\hat k$$ respectively. Then $$ABCD$$ is a :
25.
A force $${10\overrightarrow i - 5\overrightarrow j + 7\overrightarrow k }$$ displaces a particle from the point $$A$$ to the point $$B.$$ The position vectors of $$A$$ and $$B$$ are $$3\overrightarrow i - \overrightarrow j + 2\overrightarrow k $$ and $$\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k $$ respectively. Then the work done is :
A
40
B
20
C
60
D
none of these
Answer :
40
Work done $$ = \left| {\overrightarrow F .\overrightarrow {AB} } \right| = \left| {\left( {10\overrightarrow i - 5\overrightarrow j + 7\overrightarrow k } \right).\left( { - 2\overrightarrow i + 4\overrightarrow j } \right)} \right| = \left| { - 20 - 20} \right| = 40$$
26.
Let $$\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow a $$ and $$\overrightarrow r .\overrightarrow c = 0,$$ where $$\overrightarrow a .\overrightarrow b \ne 0.$$ Then $$\overrightarrow r $$ is equal to :
A
$$\overrightarrow b + t\overrightarrow a $$ where $$t$$ is a scalar
B
$$\overrightarrow b - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }}\overrightarrow a $$
C
$$\overrightarrow a - \overrightarrow c $$
D
none of these
Answer :
$$\overrightarrow b - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }}\overrightarrow a $$
$$\eqalign{
& {\text{Here }}\left( {\overrightarrow r - \overrightarrow b } \right) \times \overrightarrow a = 0.\,{\text{So, }}\left( {\overrightarrow r - \overrightarrow b } \right)||\overrightarrow a \cr
& \therefore \,\overrightarrow r - \overrightarrow b = t\overrightarrow a {\text{ or }}\overrightarrow r = \overrightarrow b + t\overrightarrow a \cr
& {\text{But }}\overrightarrow r .\overrightarrow c = 0 \cr
& \therefore \,0 = \overrightarrow b .\overrightarrow c + t\overrightarrow a .\overrightarrow c \cr
& \therefore \,t = - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }} \cr
& \therefore \,\overrightarrow r = \overrightarrow b - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }}\overrightarrow a \cr} $$
27.
$$\overrightarrow a ,\,\overrightarrow b $$ and $$\overrightarrow c $$ are three vectors with magnitude $$\left| {\overrightarrow a } \right| = 4,\,\left| {\overrightarrow b } \right| = 4,\,\left| {\overrightarrow c } \right| = 2$$ and such that $$\overrightarrow a $$ is perpendicular to $$\left( {\overrightarrow b + \overrightarrow c } \right),\,\overrightarrow b $$ is perpendicular to $$\left( {\overrightarrow c + \overrightarrow a } \right)$$ and $$\overrightarrow c $$ is perpendicular to $$\left( {\overrightarrow a + \overrightarrow b } \right).$$ It follows that $$\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|$$ is equal to :
A
9
B
6
C
5
D
4
Answer :
6
Since, $$\overrightarrow a ,\,\overrightarrow b $$ and $$\overrightarrow c $$ are three vectors with magnitude $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 4{\text{ and }}\left| {\overrightarrow c } \right| = 2$$
As $$\overrightarrow a $$ is perpendicular to $$\left( {\overrightarrow b + \overrightarrow c } \right)$$
$$ \Rightarrow \overrightarrow a .\left( {\overrightarrow b + \overrightarrow c } \right) = 0{\text{ or }}\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c = 0......\left( {\text{i}} \right)$$
$$\overrightarrow b $$ is perpendicular to $$\left( {\overrightarrow c + \overrightarrow a } \right)$$
$$ \Rightarrow \overrightarrow b .\left( {\overrightarrow c + \overrightarrow a } \right) = 0{\text{ or }}\overrightarrow b .\overrightarrow c + \overrightarrow b .\overrightarrow a = 0......\left( {{\text{ii}}} \right)$$
$$\overrightarrow c $$ is perpendicular to $$\left( {\overrightarrow a + \overrightarrow b } \right)$$
$$ \Rightarrow \overrightarrow c .\left( {\overrightarrow a + \overrightarrow b } \right) = 0{\text{ or }}\overrightarrow c .\overrightarrow a + \overrightarrow c .\overrightarrow b = 0......\left( {{\text{iii}}} \right)$$
From equations $$\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right)$$ and $$\left( {{\text{iii}}} \right),$$ we get
$$ \Rightarrow 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) = 0$$
Further we know that
$$\eqalign{
& {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + \overrightarrow {2a} .\overrightarrow b + \overrightarrow {2b} .\overrightarrow c + \overrightarrow {2c} .\overrightarrow a \cr
& \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {4^2} + {4^2} + {2^2} + 0 \cr
& \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 36 \cr
& {\text{or }}\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = 6 \cr} $$
28.
Three forces $$\overrightarrow P ,\,\overrightarrow Q $$ and $$\overrightarrow R ,$$ each of 15 units, act along $$AB,\,BC$$ and $$CA$$ respectively. The position vectors of $$A,\,B$$ and $$C$$ are $$\overrightarrow {OA} = 2\overrightarrow i - \overrightarrow j + 3\overrightarrow k ,\,\overrightarrow {OB} = 5\overrightarrow i + 3\overrightarrow j - 2\overrightarrow k $$ and $$\overrightarrow {OC} = - 2\overrightarrow i + 2\overrightarrow j + 3\overrightarrow k $$ respectively. The resultant force vector is :
29.
Let $$\alpha ,\,\beta ,\,\gamma $$ be distinct real numbers. The points with position vectors $$\alpha \hat i + \beta \hat j + \gamma \hat k,\,\beta \hat i + \gamma \hat j + \alpha \hat k$$ and $$\gamma \hat i + \alpha \hat j + \beta \hat k$$
A
are collinear
B
form an equilateral triangle
C
form a scalene triangle
D
form a right-angled triangle
Answer :
are collinear
$$\alpha ,\,\beta $$ and $$\gamma $$ be distinct real numbers
$$\alpha \hat i + \beta \hat j + \gamma \hat k\,;\,\beta \hat i + \gamma \hat j + \alpha \hat k\,;\,\gamma \hat i + \alpha \hat j + \beta \hat k$$
$$\overrightarrow a ,\,\overrightarrow b $$ and $$\overrightarrow c $$ are collinear
If $$a = \alpha ,\,b = \beta ,\,c = \gamma \,\,\,\left( {\because \,\alpha = \hat i + \hat j + \hat k} \right)$$
30.
If $$\vec u,\,\vec v,\,\vec \omega $$ are non-coplanar vectors and $$p,\,q$$ are real numbers, then the equality $$\left[ {3\vec u\,p\vec v\,p\vec \omega } \right] - \left[ {p\vec v\,\vec \omega \,q\vec u} \right] - \left[ {2\vec \omega \,q\vec v\,q\vec u} \right] = 0$$ holds for :
A
exactly two values of $$\left( {p,\,q} \right)$$
B
more than two but not all values of $$\left( {p,\,q} \right)$$
C
all values of $$\left( {p,\,q} \right)$$
D
exactly one value of $$\left( {p,\,q} \right)$$
Answer :
exactly one value of $$\left( {p,\,q} \right)$$
