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.
101.
If $$A = \left( {1,\,1,\,1} \right),\,B = \left( {2,\, - 1,\,3} \right),\,C = \left( {0,\,4,\, - 2} \right),\,D = \left( {1,\,2,\,\lambda } \right)$$ and $$AB,\,AC$$ and $$AD$$ are coplanar then $$\lambda $$ is :
A
$$1$$
B
$$0$$
C
$$ - 1$$
D
$$3$$
Answer :
$$0$$
Direction ratios of $$AB,\,AC$$ and $$AD$$ are respectively $$1,\, - 2,\,2;\, - 1,\,3,\, - 3;\,0,\,1,\,\lambda - 1.$$ They are coplanar $$ \Rightarrow $$ there is a line with direction ratios $$l,\,m,\,n$$ which is perpendicular to $$AB,\,AC$$ and $$AD.$$
$$\therefore \,l - 2m + 2n = 0,\, - l + 3m - 3n = 0,\,0.l + m + \left( {\lambda - 1} \right)n = 0$$
Eliminating $$l,\,m,\,n$$ we get \[\left| \begin{array}{l}
\,\,\,\,\,\,1\,\,\, - 2\,\,\,\,\,\,\,\,\,2\\
- 1\,\,\,\,\,\,\,\,3\,\,\,\,\, - 3\\
\,\,\,\,\,\,0\,\,\,\,\,\,\,\,1\,\,\,\,\,\lambda - 1
\end{array} \right| = 0 \Rightarrow \lambda = 0.\]
102.
If $$\overrightarrow p $$ and $$\overrightarrow q $$ are non-collinear unit vectors and $$\left| {\overrightarrow p + \overrightarrow q } \right| = \sqrt 3 ,$$ then $$\left( {2\overrightarrow p - 3\overrightarrow q } \right).\left( {3\overrightarrow p + \overrightarrow q } \right)$$ is equal to :
A
$$0$$
B
$$\frac{1}{3}$$
C
$$ - \frac{1}{3}$$
D
$$ - \frac{1}{2}$$
Answer :
$$ - \frac{1}{2}$$
$$\left| {\overrightarrow p + \overrightarrow q } \right| = \sqrt 3 \Rightarrow \overrightarrow {{p^2}} + \overrightarrow {{q^2}} + 2\overrightarrow p \overrightarrow q = 3$$
Since $$\overrightarrow p $$ and $$\overrightarrow q $$ are unit vectors
$$\eqalign{
& {\text{So, }}1 + 1 + 2pq = 3 \cr
& \Rightarrow 2pq = 1 \cr
& \Rightarrow pq = \frac{1}{2} \cr
& \left( {2\overrightarrow p - 3\overrightarrow q } \right)\left( {3\overrightarrow p + \overrightarrow q } \right) = 6\overrightarrow {{p^2}} + 2\overrightarrow p \overrightarrow q - 9\overrightarrow q \overrightarrow p - 3\overrightarrow {{q^2}} = \frac{{ - 1}}{2} \cr} $$
103.
If $${a^2} + {b^2} + {c^2} = 1$$ where $$a,\,b,\,c\, \in \,R,$$ then the maximum value of $${\left( {4a - 3b} \right)^2} + {\left( {5b - 4c} \right)^2} + {\left( {3c - 5a} \right)^2}{\text{ is :}}$$
104.
A particles is acted upon by constant forces $$4\hat i + \hat j - 3\hat k$$ and $$3\hat i + \hat j - \hat k$$ which displace it from a point $$\hat i + 2\hat j + 3\hat k$$ to the point $$5\hat i + 4\hat j + \hat k.$$ The work done in standard units by the forces is given by :
A
$$15$$
B
$$30$$
C
$$25$$
D
$$40$$
Answer :
$$40$$
Resultant of forces $$\vec F = 7\hat i + 2\hat j - 4\hat k$$
Displacement $$\vec d = 4\hat i + 2\hat j - 2\hat k$$
$$\therefore $$ Work done $$ = \vec F.\vec d = 28 + 4 + 8 = 40$$
105.
If the vectors $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ and $$\overrightarrow d $$ are coplanar then $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right)$$ is equal to :
A
$$\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d $$
B
$$\overrightarrow 0 $$
C
$$\overrightarrow a + \overrightarrow b = \overrightarrow c + \overrightarrow d $$
D
none of these
Answer :
$$\overrightarrow 0 $$
Let their plane be $$\pi .\,\overrightarrow a \times \overrightarrow b $$ is perpendicular to $$\pi $$ and $${\overrightarrow c \times \overrightarrow d }$$ is also perpendicular to $$\pi .$$ So, $$\overrightarrow a \times \overrightarrow b ||\overrightarrow c \times \overrightarrow d .$$ Hence, $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) = \overrightarrow 0 .$$
106.
Let $${P_r}\left( {{x_r},\,{y_r},\,{z_r}} \right);\,r = 1,\,2,\,3;$$ be three points where $${x_1},\,{x_2},\,{x_3};\,{y_1},\,{y_2},\,{y_3}$$ and $${z_1},\,{z_2},\,{z_3}$$ are each in GP with the same common ratio. Then $${P_1},\,{P_2},\,{P_3}$$ are :
A
coplanar points
B
collinear points
C
vertices of an equilateral triangle
D
none of these
Answer :
collinear points
$${P_1} = \left( {{x_1},\,r{x_1},\,{r^2}{x_1}} \right),\,{P_2} = \left( {{y_1},\,r{y_1},\,{r^2}{y_1}} \right),\,{P_3} = \left( {{z_1},\,r{z_1},\,{r^2}{z_1}} \right)$$
Direction ratios of $${P_1}{P_2}$$ are $${y_1} - {x_1},\,r\left( {{y_1} - {x_1}} \right),\,{r^2}\left( {{y_1} - {x_1}} \right).$$
Direction ratios of $${P_2}{P_3}$$ are $${z_1} - {y_1},\,r\left( {{z_1} - {y_1}} \right),\,{r^2}\left( {{z_1} - {y_1}} \right).$$
$${\text{As }}\frac{{{y_1} - {x_1}}}{{{z_1} - {y_1}}} = \frac{{r\left( {{y_1} - {x_1}} \right)}}{{r\left( {{z_1} - {y_1}} \right)}} = \frac{{{r^2}\left( {{y_1} - {x_1}} \right)}}{{{r^2}\left( {{z_1} - {y_1}} \right)}},\,{P_1}{P_2}||{P_2}{P_3}$$
So $${P_1},\,{P_2},\,{P_3}$$ are collinear.
107.
If $$\overrightarrow a .\overrightarrow i = \overrightarrow a .\left( {\overrightarrow i + \overrightarrow j } \right) = \overrightarrow a .\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) = 1$$ then $$\overrightarrow a $$ is :
A
$$\overrightarrow i - \overrightarrow j $$
B
$$\overrightarrow i $$
C
$$\overrightarrow j $$
D
$$\overrightarrow k $$
Answer :
$$\overrightarrow i $$
$$\overrightarrow a .\overrightarrow i = 1,\,\overrightarrow a .\overrightarrow j = 0,\,\overrightarrow a .\overrightarrow k = 0.$$ These imply $$\overrightarrow a = \overrightarrow i $$
108.
$$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right)$$ and $$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)$$ are :
A
linearly dependent
B
equal vectors
C
parallel vectors
D
none of these
Answer :
linearly dependent
$$\eqalign{
& \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) + \,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right) + \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) \cr
& = \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c + \left( {\overrightarrow b .\overrightarrow a } \right)\overrightarrow c - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a + \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = 0 \cr} $$
$$\therefore $$ the given vectors are linearly dependent.
109.
Let $$a,\,b,\,c$$ be three distinct positive real numbers. If $$\overrightarrow p ,\,\overrightarrow q ,\,\overrightarrow r $$ lie in a plane, where $$\overrightarrow p = a\overrightarrow i - a\overrightarrow j + b\overrightarrow k ,\,\overrightarrow q = \overrightarrow i + \overrightarrow k $$ and $$\overrightarrow r = c\overrightarrow i + c\overrightarrow j + b\overrightarrow k ,$$ then $$b$$ is :
110.
If $$\overrightarrow a = \overrightarrow i + \overrightarrow j ,\,\overrightarrow b = 2\overrightarrow j - \overrightarrow k $$
and $$\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow a ,\,\overrightarrow r \times \overrightarrow b = \overrightarrow a \times \overrightarrow b $$ then $$\frac{{\overrightarrow r }}{{\left| {\overrightarrow r } \right|}}$$ is equal to :
A
$$\frac{1}{{\sqrt {11} }}\left( {\overrightarrow i + 3\overrightarrow j - \overrightarrow k } \right)$$
B
$$\frac{1}{{\sqrt {11} }}\left( {\overrightarrow i - 3\overrightarrow j + \overrightarrow k } \right)$$
C
$$\frac{1}{{\sqrt 3 }}\left( {\overrightarrow i - \overrightarrow j + \overrightarrow k } \right)$$
D
none of these
Answer :
$$\frac{1}{{\sqrt {11} }}\left( {\overrightarrow i + 3\overrightarrow j - \overrightarrow k } \right)$$
$$\eqalign{
& {\text{Here, }}\overrightarrow r \times \overrightarrow a + \overrightarrow r \times \overrightarrow b = 0\,\,\,{\text{or }}\overrightarrow r \times \left( {\overrightarrow a + \overrightarrow b } \right) = 0\,\, \cr
& \therefore \overrightarrow r ||\left( {\overrightarrow a + \overrightarrow b } \right) \cr
& \therefore \,\overrightarrow r = t\left( {\overrightarrow a + \overrightarrow b } \right) = t\left( {\overrightarrow i + 3\overrightarrow j - \overrightarrow k } \right) \cr
& \therefore \,\left| {\overrightarrow r } \right| = t.\sqrt {{1^2} + {3^2} + {1^2}} = t.\sqrt {11} \cr
& \therefore \,\frac{{\overrightarrow r }}{{\left| {\overrightarrow r } \right|}} = \frac{1}{{\sqrt {11} }}\left( {\overrightarrow i + 3\overrightarrow j - \overrightarrow k } \right) \cr} $$