$$A, B, C$$ are three points on the axes of $$x, y$$ and $$z$$ respectively at distance $$a, b, c$$ from the origin $$O$$; then the co - ordinates of the point which is equidistant from $$A, B, C$$ and $$O$$ is

$$\displaystyle \left ( \frac{a}{2},\frac{b}{2},\frac{c}{2} \right )$$

$$\displaystyle \left ( a,b,c \right )$$

$$\displaystyle \left ( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right )$$

Find the ratio in which the yz-plane divides the join of the points $$\displaystyle \left ( -2,4,7 \right )$$ and $$\displaystyle \left ( 3,-5,8 \right )$$ and also find the co-ordinates of the point of intersection of this line with the $$yz$$ - plane.

$$\displaystyle \lambda =\frac{2}{3}$$ and $$\displaystyle \left ( 0,\frac{2}{5},\frac{37}{5} \right )$$

$$\displaystyle \lambda =\frac{1}{3}$$ and $$\displaystyle \left ( \frac{-3}{4},\frac{7}{4},\frac{29}{4} \right )$$

$$\displaystyle \lambda =\frac{2}{3}$$ and $$\displaystyle \left ( \frac{-3}{4},\frac{7}{4},\frac{29}{4} \right )$$

$$\displaystyle \lambda =\frac{1}{3}$$ and $$\displaystyle \left ( 0,\frac{2}{5},\frac{37}{5} \right )$$

MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 6

Let the distance between vectors are given as follows :

(i) 4 i + 3 j - 6 k, - 2 i + j - k

$(i)4i +3j-6k, -2i+j-k$ be

\sqrt{k}

$\displaystyle \sqrt{k}$

(i i) - 2 i + 3 j + 5 k, 7 i - k

$(ii) -2i+3j+5k, 7i-k$ be

m \sqrt{n}

$\displaystyle m\sqrt{n}$
Find

k - (m * n)

$k-(m*n)$ ?

0%
20
0%
21
0%
22
0%
23

Explanation

(i) Distance between

4 i + 3 j - 6 k

$4i + 3j - 6k$ and

- 2 i + j - k

$-2i + j - k$ is given by

\sqrt{(4 + 2)^{2} + (3 - 1)^{2} + (- 6 + 1)^{2}} = \sqrt{36 + 4 + 25} = \sqrt{65}

$\sqrt{(4 + 2)^2 + (3 - 1)^2 + (-6 + 1)^2} = \sqrt{36 + 4 + 25} = \sqrt{65}$

\Rightarrow k = 65

$\Rightarrow k = 65$

(ii) Distance between

- 2 i + 3 j + 5 k

$-2i + 3j + 5k$ and

7 i - k

$7i - k$ would be

\sqrt{(- 2 - 7)^{2} + (3)^{2} + (5 + 1)^{2}} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3 \sqrt{14}

$\sqrt{(-2 - 7)^2 + (3)^2 + (5 + 1)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$

\Rightarrow m = 3, n = 14

$\Rightarrow m = 3, n = 14$

∴ k - (m * n) = 65 - (3 \times 14) = 65 - 42 = 23

$\therefore k - (m*n) = 65 - (3 \times 14) = 65 - 42 = 23$

A, B, C

$A, B, C$ are three points on the axes of

x, y

$x, y$ and

z

$z$ respectively at distance

a, b, c

$a, b, c$ from the origin

O

$O$ ; then the co - ordinates of the point which is equidistant from

A, B, C

$A, B, C$ and

O

$O$ is

0%
$\displaystyle \left ( a,b,c \right )$
0%
$\displaystyle \left ( \frac{a}{2},\frac{b}{2},\frac{c}{2} \right )$
0%
$\displaystyle \left ( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right )$
0%
None of these

Explanation

Let

P

$P$ be the required point

(x, y, z)

$\displaystyle \left ( x,y,z \right )$ and the point

A, B, C

$A, B, C$ and

O

$O$ are

(a, 0, 0), (0, b, 0), (0, 0, c)

$\displaystyle \left ( a,0,0 \right ),\left ( 0,b,0 \right ),\left ( 0,0,c \right )$ and

(0, 0, 0)

$\displaystyle \left ( 0,0,0 \right )$ ;

We are given that

P O = P A = P B = P C .

$\displaystyle PO=PA=PB=PC.$
Taking

P O = P A

$\displaystyle PO=PA$ or

P O^{2} = P A^{2},

$\displaystyle PO^{2}=PA^{2},$ we get

x^{2} + y^{2} + z^{2} = {(x - a)}^{2} + y^{2} + z^{2}

$\displaystyle x^{2}+y^{2}+z^{2}=\left ( x-a \right )^{2}+y^{2}+z^{2}$

0 = a^{2} - 2 a x

$\displaystyle 0=a^{2}-2ax$ i.e.

x = \frac{a}{2}

$\displaystyle x=\dfrac {a}{2}$
Similarly taking

P O^{2} = P B^{2}

$\displaystyle PO^{2}=PB^{2}$ and

P O^{2} = P C^{2},

$\displaystyle PO^{2}=PC^{2},$ we get

y = \frac{b}{2}

$\displaystyle y=\dfrac {b}{2}$ and

z = \frac{c}{2}

$\displaystyle z=\dfrac {c}{2}$

If the centroid of the tetrahedron

O A B C

$OABC$ , where

A, B, C

$A, B ,C$ are given by

(α, 5, 6), (1, β, 4), (3, 2, γ)

$\displaystyle (\alpha, 5, 6), (1, \beta, 4), (3, 2, \gamma)$ respectively be

1, - 1, 2

$1, -1, 2$ , then value of

α^{2} + β^{2} + γ^{2}

$\displaystyle \alpha^2 + \beta^2 + \gamma^2$ equals

0%
$\displaystyle \alpha^2 + \beta^2$
0%
$\displaystyle \gamma^2 + \beta^2$
0%
$\displaystyle \alpha^2 + \gamma^2$
0%
None of these

Explanation

The vertices of tetrahedron are

(0, 0, 0), (α, 5, 6), (1, β, 4), (3, 2, γ)

$(0,0,0) , (\alpha , 5,6) , (1,\beta , 4) , (3,2, \gamma)$

The centroid of tetrahedron is

(\frac{α + 4}{4}, \frac{β + 7}{4}, \frac{γ + 10}{4})

$(\frac{\alpha+4}{4},\frac{\beta+7}{4},\frac{\gamma+10}{4})$

Given that centroid is

1, - 1, 2

$1,-1,2$

By equating , we get

α = 0, β = - 11, γ = - 2

$\alpha=0 , \beta=-11 , \gamma=-2$

Since

α = 0

$\alpha =0$ ,

α^{2} + β^{2} + γ^{2} = β^{2} + γ^{2}

${ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }={ \beta }^{ 2 }+{ \gamma }^{ 2 }$

Therefore the correct option is

B

$B$

Perimeter of triangle whose vertices are

(0, 4, 0), (3, 4, 0)

$(0,4,0), (3,4,0)$ and

(0, 4, 4)

$(0,4,4)$ , is

0%
$10$
0%
$12$
0%
$25$
0%
$15$

Explanation

Vertices of triangle are

(0, 4, 0), (3, 4, 0)

$(0,4,0),(3,4,0)$ and

(0, 4, 4)

$(0,4,4)$

then perimeter=?

here AB=

\sqrt{(3 - 0)^{2} + (4 - 4)^{2} + (0 - 0)^{2}}

$\sqrt{(3-0)^2+(4-4)^2+(0-0)^2}$

= 3

$=3$

= \sqrt{(3 - 0)^{2} + (4 - 4)^{2} + (0 - 4)^{2}}

$=\sqrt{(3-0)^2+(4-4)^2+(0-4)^2}$

= \sqrt{9 + 16}

$=\sqrt{9+16}$

= 5

$=5$

= \sqrt{(0 - 0)^{2} + (4 - 4)^{2} + (0 - 4)^{2}} = 4

$=\sqrt{(0-0)^2+(4-4)^2+(0-4)^2}=4$

used distance formula b/w two points

(x_{1}, y_{1}, z_{1}) a n d (x_{2}, y_{2}, z_{2})

$(x_1,y_1,z_1) and (x_2,y_2,z_2)$

= \sqrt{(x_{2} - x_{2})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}

$=\sqrt{(x_2-x_2)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

p e r i m e t e r = A B C + B C + C A

$perimeter =ABC+BC+CA$

= 3 + 5 + 4

$=3+5+4$

= 12 u n i t s

$=12\ units$

P (x, y, z)

$P(x,y,z)$ is a point on the line segment joining

A (2, 2, 4)

$A (2,2,4)$ and

B (3, 5, 6)

$B(3,5,6)$ such that projection of

\vec{O P}

$\vec{OP}$ on axes are

\frac{13}{5}, \frac{19}{5}, \frac{26}{5}

$\displaystyle \frac{13}{5},\frac{19}{5},\frac{26}{5}$ respectively, then P divide AB in the ratio

0%
$3:2$
0%
$2:3$
0%
$1:2$
0%
$1:3$

Explanation

\vec{O P} = \frac{13}{5} \hat{i} + \frac{19}{5} \hat{j} + \frac{26}{5} \hat{k} = \vec{P} - \vec{O}

$\vec{OP} = \dfrac{13}{5}\hat{i} + \dfrac{19}{5}\hat{j} + \dfrac{26}{5}\hat{k} = \vec{P} - \vec{O}$

where

\vec{P} =

$\vec{P} =$ origin vector

= 0 \hat{i} + 0 \hat{j} + 0 \hat{k}

$= 0\hat{i} + 0\hat{j} + 0\hat{k}$

O P = (\frac{13}{5}, \frac{19}{5} . \frac{26}{5})

$OP = (\dfrac{13}{5}, \dfrac{19}{5}. \dfrac{26}{5})$ so,

\vec{P} = \frac{13}{5} \hat{i} + \frac{19}{5} \hat{j} + \frac{26}{5} \hat{k}

$\vec{P} = \dfrac{13}{5}\hat{i} + \dfrac{19}{5}\hat{j} + \dfrac{26}{5}\hat{k}$

(x, y, z)

$(x, y, z)$ divides

(x_{1}, y_{1}, z_{1})

$(x_1, y_1, z_1)$ and

(x_{2}, y_{2}, z_{2})

$(x_2, y_2, z_2)$ in the ratio

α : 1

$\alpha : 1$

∴ x = \frac{α x_{2} + x_{1}}{α + 1}, y = \frac{α y_{2} + y_{1}}{α + 1}, z = \frac{α z_{2} + z_{1}}{α + 1}

$\therefore x = \dfrac{\alpha x_2 + x_1}{\alpha + 1}, y = \dfrac{\alpha y_2 + y_1}{\alpha + 1}, z = \dfrac{\alpha z_2 + z_1}{\alpha + 1}$

Then,

x = \frac{3 α + 2}{α + 1}, y = \frac{5 α + 2}{α + 1}, z = \frac{6 α + 4}{α + 1}

$x = \dfrac{3\alpha + 2}{\alpha + 1}, y = \dfrac{5\alpha + 2}{\alpha + 1}, z = \dfrac{6\alpha + 4}{\alpha + 1}$

(x, y, z)

$(x, y, z)$ given

= (\frac{13}{5}, \frac{19}{5}, \frac{26}{5})

$= (\dfrac{13}{5}, \dfrac{19}{5}, \dfrac{26}{5})$

∴ \frac{3 α + 2}{α + 1} = \frac{13}{5} \Rightarrow 13 α + 13 = 15 α + 10

$\therefore \dfrac{3\alpha + 2}{\alpha + 1} = \dfrac{13}{5} \Rightarrow 13\alpha + 13 = 15\alpha + 10$

\Rightarrow 3 = 2 α \Rightarrow α = \frac{3}{2}

$\Rightarrow 3 = 2\alpha \Rightarrow \alpha = \dfrac{3}{2}$

For

z

$z$ co-ordinates,

\frac{19}{5} = \frac{5 α + 2}{α + 1} \Rightarrow 19 α + 19 = 25 α + 10

$\dfrac{19}{5} = \dfrac{5\alpha + 2}{\alpha + 1} \Rightarrow 19\alpha + 19 = 25\alpha + 10$

\Rightarrow 6 α = 9 \Rightarrow α = \frac{3}{2}

$\Rightarrow 6\alpha = 9 \Rightarrow \alpha = \dfrac{3}{2}$

α : 1 = 3 : 2

$\alpha : 1 = 3 : 2$

Find the ratio in which the yz-plane divides the join of the points

(- 2, 4, 7)

$\displaystyle \left ( -2,4,7 \right )$ and

(3, - 5, 8)

$\displaystyle \left ( 3,-5,8 \right )$ and also find the co-ordinates of the point of intersection of this line with the

y z

$yz$ - plane.

0%
$\displaystyle \lambda =\frac{2}{3}$ and $\displaystyle \left ( 0,\frac{2}{5},\frac{37}{5} \right )$
0%
$\displaystyle \lambda =\frac{1}{3}$ and $\displaystyle \left ( \frac{-3}{4},\frac{7}{4},\frac{29}{4} \right )$
0%
$\displaystyle \lambda =\frac{2}{3}$ and $\displaystyle \left ( \frac{-3}{4},\frac{7}{4},\frac{29}{4} \right )$
0%
$\displaystyle \lambda =\frac{1}{3}$ and $\displaystyle \left ( 0,\frac{2}{5},\frac{37}{5} \right )$

Explanation

Let

λ

$\lambda$ be the ratio in which yz-plane divides the line joining the points

(- 2, 4, 7)

$(-2,4,7)$ and

(3, - 5, 8)

$(3,-5,8)$

The co-ordinates of any point on the line joining the two points are

(\frac{3 λ - 2}{λ + 1}, \frac{- 5 λ + 4}{λ + 1}, \frac{8 λ + 7}{λ + 1})

$\displaystyle \left ( \frac{3\lambda -2}{\lambda +1},\frac{-5\lambda +4}{\lambda +1},\frac{8\lambda +7}{\lambda +1} \right )$
If the point is in the

y z

$yz$ - plane, then its

x

$x$ -coordinate should be zero.

∴ \frac{3 λ - 2}{λ + 1} = 0

$\therefore \displaystyle \frac{3\lambda -2}{\lambda +1}=0$ or

3 λ - 2 = 0

$\displaystyle 3\lambda -2=0$

∴ λ = \frac{2}{3} .

$\therefore \lambda =\dfrac{2}{3}.$
Hence, the required ratio is

2 : 3

$2:3$ and putting

λ = \frac{2}{3}

$\displaystyle \lambda =\frac{2}{3}$ , the required point is

(0, \frac{2}{5}, \frac{37}{5}) .

$\displaystyle \left ( 0,\frac{2}{5},\frac{37}{5} \right ).$

Find the distance between the points whose position vectors are given as follows

$-2\hat i+3\hat j+5\hat k, 7\hat i-\hat k$

0%
$\displaystyle 3\sqrt{14}$
0%
$\displaystyle \sqrt{54}$
0%
$\displaystyle 3\sqrt{19}$
0%
$\displaystyle \sqrt{57}$

Explanation

Distance Between two position vectors

a \hat{i} + b \hat{j} + c \hat{k}

$a\hat i+b\hat j+c\hat k$ and

x \hat{i} + y \hat{j} + z \hat{k}

$x\hat i+y\hat j+z\hat k$ is given by

\sqrt{{(a - x)}^{2} + {(b - y)}^{2} + {(c - z)}^{2}}

$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } }$
Given

- 2 \hat{i} + 3 \hat{j} + 5 \hat{k}

$-2\hat i+3\hat j+5\hat k$ ,

7 \hat{i} - \hat{k}

$7\hat i-\hat k$
Distance

= \sqrt{{(- 2 - 7)}^{2} + {(3 - 0)}^{2} + {(5 + 1)}^{2}}

$=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 3-0 \right) ^{ 2 }+\left( 5+1 \right) ^{ 2 } }$

= \sqrt{126}

$=\sqrt { 126 }$

= 3 \sqrt{14}

$=3\sqrt { 14 }$

The points

A (5, 1, 1), B (7, 4, 7), C (1, 6, 10)

$A(5, 1, 1), B(7, 4, 7), C(1, 6, 10)$ and

D (- 1, 3, 4)

$D(-1, 3, 4)$ are the vertices of a

0%
Parallelogram
0%
Rectangle
0%
Rhombus
0%
Square

Explanation

Given points are

A (5, 1, 1), B (7, 4, 7), C (1, 6, 10)

$A(5, 1, 1), B(7, 4, 7), C(1, 6, 10)$ and

D (- 1, 3, 4)

$D(-1, 3, 4)$

A B = \sqrt{4 + 9 + 36} = 7

$AB=\sqrt{4+9+36}=7$

B C = \sqrt{36 + 4 + 9} = 7

$BC=\sqrt{36+4+9}=7$

C D = \sqrt{4 + 9 + 36} = 7

$CD=\sqrt{4+9+36}=7$

A D = \sqrt{36 + 4 + 9} = 7

$AD=\sqrt{36+4+9}=7$

A C = \sqrt{16 + 25 + 81} = \sqrt{122}

$AC= \sqrt{16+25+81}=\sqrt{122}$

B D = \sqrt{64 + 1 + 9} = \sqrt{74}

$BD=\sqrt{64+1+9}=\sqrt{74}$

All sides are equal and diagonals are not equal.

Hence,

A B C D

$ABCD$ is a Rhombus.

Hence, option C.

P (x, y, z)

$P\left( x,y,z \right)$ is a point on the line segment joining

Q (2, 2, 4)

$Q\left( 2,2,4 \right)$ and

R (3, 5, 6)

$R\left( 3,5,6 \right)$ such that the projection of

\bar{O P}

$\overline { OP }$ on the axes are

\frac{13}{5}, \frac{19}{5}, \frac{26}{5},

$\displaystyle\frac { 13 }{ 5 } ,\frac { 19 }{ 5 } ,\frac { 26 }{ 5 } ,$ respectively, then

P

$P$ divides

Q R

$QR$ in the ratio

0%
$1:2$
0%
$3:2$
0%
$2:3$
0%
$1:3$

Explanation

Since

O P

$OP$ has projections

\frac{13}{5}, \frac{19}{5}

$\displaystyle\frac { 13 }{ 5 } ,\frac { 19 }{ 5 }$ and

\frac{26}{5}

$\displaystyle\frac { 26 }{ 5 }$ on the coordinate axes,

therefore

O P = \frac{13}{5} \hat{i} + \frac{19}{5} \hat{j} + \frac{26}{5} \hat{k}

$\displaystyle OP=\frac { 13 }{ 5 } \hat i+\frac { 19 }{ 5 } \hat j+\frac { 26 }{ 5 } \hat k$

Suppose

P

$P$ divides the join of

Q (2, 2, 4)

$Q\left( 2,2,4 \right)$ and

R (3, 5, 6)

$R\left( 3,5,6 \right)$ in the ratio

λ : 1

$\lambda:1$

Then, the position vector of

P

$P$ is

(\frac{3 λ + 2}{λ + 1}) \hat{i} + (\frac{5 λ + 2}{λ + 1}) \hat{j} + (\frac{6 λ + 4}{λ + 1}) \hat{k}

$\displaystyle \left( \frac { 3\lambda +2 }{ \lambda +1 } \right)\hat i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \hat j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \hat k$

∴ \frac{13}{5} \hat{i} + \frac{19}{5} \hat{j} + \frac{26}{5} \hat{k} = (\frac{3 λ + 2}{λ + 1}) \hat{i} + (\frac{5 λ + 2}{λ + 1}) \hat{j} + (\frac{6 λ + 4}{λ + 1}) \hat{k}

$\displaystyle\therefore \frac { 13 }{ 5 }\hat i+\frac { 19 }{ 5 }\hat j+\frac { 26 }{ 5 } \hat k=\left( \frac { 3\lambda +2 }{ \lambda +1 } \right) \hat i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \hat j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \hat k$

\Rightarrow \frac{3 λ + 2}{λ + 1} = \frac{13}{5}, \frac{5 λ + 2}{λ + 1} = \frac{19}{5}, \frac{6 λ + 4}{λ + 1} = \frac{26}{5}

$\displaystyle\Rightarrow \frac { 3\lambda +2 }{ \lambda +1 } =\frac { 13 }{ 5 } ,\frac { 5\lambda +2 }{ \lambda +1 } =\frac { 19 }{ 5 } ,\frac { 6\lambda +4 }{ \lambda +1 } =\frac { 26 }{ 5 }$

\Rightarrow 2 λ = 3

$\displaystyle\Rightarrow 2\lambda =3$

\Rightarrow λ = \frac{3}{2}

$\Rightarrow \lambda =\dfrac { 3 }{ 2 }$

Find the distance between the points whose position vectors are given as follows

(- 1, 1, 3), (0, 5, 6)

$(-1,1,3), (0,5,6)$

0%
$\displaystyle \sqrt{118}$
0%
$8$
0%
$\displaystyle \sqrt{26}$
0%
none of these

Explanation

Distance Between two points

(a, b, c)

$(a,b,c)$ and

(x, y, z)

$(x,y,z)$ is given by

\sqrt{{(a - x)}^{2} + {(b - y)}^{2} + {(c - z)}^{2}}

$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } }$
Given

(- 1, 1, 3)

$(-1,1,3)$ ,

(0, 5, 6)

$(0,5,6)$
Distance

= \sqrt{{(- 1 - 0)}^{2} + {(1 - 6)}^{2} + {(3 - 6)}^{2}}

$=\sqrt { \left( -1-0 \right) ^{ 2 }+\left( 1-6 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } }$

= \sqrt{26}

$=\sqrt { 26 }$

The plane

X O Z

$XOZ$ divides the join of

(1, - 1, 5)

$(1, - 1, 5)$ and

(2, 3, 4)

$(2, 3, 4)$ in the ratio

λ : 1

$\lambda : 1$ , then

λ

$\lambda$ is -

0%
$- 3$
0%
$-1 / 3$
0%
$3$
0%
$1 / 3$

Explanation

The plane XOZ divides the join of

(1, - 1, 5)

$(1, - 1, 5)$ and

(2, 3, 4)

$(2, 3, 4)$ in the ratio

λ : 1

$λ : 1$

i . e ., y = 0

$i.e.,y = 0$ divide the join of

(1, - 1, 5)

$(1, -1, 5)$ and

(2, 3, 4)

$(2, 3, 4)$ in the ratio

λ : 1.

$λ : 1.$

∴ \frac{3 λ - 1}{λ + 1} = 0 \Rightarrow λ = \frac{1}{3}

$\therefore \dfrac {3λ−1}{λ+1} = 0 ⇒λ=\dfrac{1}{3}$ .

Find the distance between the pairs of points whose cartesian coordinates are

(2, 3, - 1), (2, 6, 2) .

$(2,3,-1), (2,6,2).$

0%
$\displaystyle 3\sqrt{2}.$
0%
$\displaystyle 2\sqrt{3}.$
0%
$\displaystyle 5\sqrt{2}.$
0%
$\displaystyle 2\sqrt{5}.$

Explanation

Distance Between two points

(a, b, c)

$(a,b,c)$ and

(x, y, z)

$(x,y,z)$ is given by

\sqrt{{(a - x)}^{2} + {(b - y)}^{2} + {(c - z)}^{2}}

$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } }$
Given

(2, 3, - 1)

$(2,3,-1)$ ,

(2, 6, 2)

$(2,6,2)$
Distance

= \sqrt{{(2 - 2)}^{2} + {(3 - 6)}^{2} + {(- 1 - 2)}^{2}}

$=\sqrt { \left( 2-2 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } }$

= \sqrt{18}

$=\sqrt { 18 }$

= 3 \sqrt{2}

$=3\sqrt { 2 }$

Find the distance between the points whose position vectors are given as follows

4 \hat{i} + 3 \hat{j} - 6 \hat{k}, - 2 \hat{i} + \hat{j} - \hat{k}

$4\hat i+3\hat j-6\hat k, -2\hat i+\hat j-\hat k$

0%
$\displaystyle \sqrt{65}$
0%
$\displaystyle \sqrt{69}$
0%
$13$
0%
none of these

Explanation

Distance Between two position vectors

a \hat{i} + b \hat{j} + c \hat{k}

$a\hat i+b\hat j+c\hat k$ and

x \hat{i} + y \hat{j} + z \hat{k}

$x\hat i+y\hat j+z\hat k$ is given by

\sqrt{{(a - x)}^{2} + {(b - y)}^{2} + {(c - z)}^{2}}

$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } }$
Given

4 \hat{i} + 3 \hat{j} - 6 \hat{k}

$4\hat i+3\hat j-6\hat k$ ,

- 2 \hat{i} + \hat{j} - \hat{k}

$-2\hat i+\hat j-\hat k$
Distance

= \sqrt{{(4 + 2)}^{2} + {(3 - 1)}^{2} + {(- 6 + 1)}^{2}}

$=\sqrt { \left( 4+2 \right) ^{ 2 }+\left( 3-1 \right) ^{ 2 }+\left( -6+1 \right) ^{ 2 } }$

= \sqrt{65}

$=\sqrt { 65 }$

Find the point which divides the lines joining

A (2, 3, 5)

$A(2,3,5)$ and

B (- 6, 5, 8)

$B(-6,5,8)$ in the ratio

2 : 3

$2:3$ externally

0%
$(18,-1,-1)$
0%
$(12,-1,-1)$
0%
$(3,4,4)$
0%
$(5,8,11)$

Explanation

The given points are

A (x_{1}, y_{1}, z_{1}) = A (2, 3, 5)

$A(x_1,y_1,z_1)=A(2,3,5)$

B (x_{2}, y_{2}, z_{2}) = B (- 6, 5, 8)

$B(x_2,y_2,z_2)=B(-6,5,8)$

Let

P (x, y, z)

$P(x,y,z)$ divides AB in the ratio

m : n = 2 : 3

$m:n=2:3$ externally

Then by using section Formula,

P (x, y, z) = (\frac{m x_{2} - n x_{1}}{m - n}, \frac{m y_{2} - n y_{1}}{m - n}, \frac{m z_{2} - n z_{1}}{m - n})

$P(x,y,z)=\left(\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n},\dfrac{mz_2-nz_1}{m-n}\right)$

P (x, y, z) = (\frac{2 \times (- 6) - 3 \times 2}{2 - 3}, \frac{2 \times 5 - 3 \times 3}{2 - 3}, \frac{2 \times 8 - 3 \times 5}{2 - 3})

$P(x,y,z)=\left(\dfrac{2\times (-6)-3\times 2}{2-3},\dfrac{2\times 5-3\times 3}{2-3},\dfrac{2\times 8-3\times 5}{2-3}\right)$

P (x, y, z) = (\frac{- 12 - 6}{- 1}, \frac{10 - 9}{- 1}, \frac{16 - 15}{- 1})

$P(x,y,z)=\left(\dfrac{-12-6}{-1},\dfrac{10-9}{-1},\dfrac{16-15}{-1}\right)$

P (x, y, z) = (\frac{- 18}{- 1}, \frac{1}{- 1}, \frac{1}{- 1})

$P(x,y,z)=\left(\dfrac{-18}{-1},\dfrac{1}{-1},\dfrac{1}{-1}\right)$

P (x, y, z) = (18, - 1, - 1)

$P(x,y,z)=(18,-1,-1)$

The ordinate of the point which divides the lines joining the origin and the point

(1, 2)

$(1,2)$ externally in the ratio of

3 : 2

$3:2$ is

0%
$-2$
0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{2}{5}$
0%
$6$

Explanation

Co-ordinates of the required point will be

y = \frac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} - m_{2}} = \frac{3 \times 2 - 2 \times 0}{3 - 2} = 6

$\displaystyle y=\frac{m_{1}y_{2}-m_{2}y_{1}}{m_{1}-m_{2}}=\frac{3\times 2-2\times 0}{3-2}=6$

Find the ratio in which (the plane)

2 x + 3 y + 5 z = 1

$2x+3y+5z=1$ divides the line joining the points

(1, 0, - 3)

$(1,0,-3)$ and

(1, - 5, 7)

$(1,-5,7)$ .

0%
$1:2$
0%
$2:3$
0%
$3:1$
0%
None of these

Explanation

Here,

2 x + 3 y + 5 z = 1

$2x+3y+5z=1$ divides

(1, 0, - 3)

$(1,0,-3)$ and

(1, - 5, 7)

$(1,-5,7)$ in the ratio of

k : 1

$k:1$ at point

P .

$P.$

Then,

P = (\frac{k + 1}{k + 1}, \frac{- 5 k}{k + 1}, \frac{7 k - 3}{k + 1})

$\displaystyle P=\left( \frac { k+1 }{ k+1 } ,\frac { -5k }{ k+1 } ,\frac { 7k-3 }{ k+1 } \right)$ which must satisfy

2 x + 3 y + 5 x = 1

$2x+3y+5x=1$

\Rightarrow 2 (\frac{k + 1}{k + 1}) + 3 (\frac{- 5 k}{k + 1}) + 5 (\frac{7 k - 3}{k + 1}) = 1

$\displaystyle \Rightarrow 2\left( \frac { k+1 }{ k+1 } \right) +3\left( \frac { -5k }{ k+1 } \right) +5\left( \frac { 7k-3 }{ k+1 } \right) =1$

\Rightarrow 2 k + 2 - 15 k + 35 k - 15 = k + 1 \Rightarrow 21 k = 14

$\Rightarrow 2k+2-15k+35k-15=k+1\Rightarrow 21k=14$

\Rightarrow k = \frac{2}{3}

$\displaystyle \Rightarrow k=\frac { 2 }{ 3 }$

∴ 2 x + 3 y + 5 z = 1

$\therefore 2x+3y+5z=1$

It divides the joint of

(1, 0 - 3)

$(1,0-3)$ and

(1, - 5, 7)

$(1,-5,7)$ in the radio of

2 : 3.

$2:3.$

A hall has dimensions

24 m \times 8 m \times 6 m

$24 m \times 8 m \times 6 m$ . The length of the longest pole which can be accommodated in the hall is

0%
26 m
0%
28 m
0%
30 m
0%
36 m

Explanation

Given that,

Dimensions of the hall x $=24cm\times 8cm\times 6cm$

Now Leght of the longest pole which can be accommodated in the hall x $=\sqrt{{{24}^{2}}+{{8}^{2}}+{{6}^{2}}}=26cm$

The coordinates of a point which divides the line joining the points

P (2, 3, 1)

$P(2,3,1)$ and

Q (5, 0, 4)

$Q(5,0,4)$ in the ratio

1 : 2

$1:2$ are

0%
$\left(\dfrac{7}{3}, 1, \dfrac{5}{3}\right)$
0%
$(4, 1, 3)$
0%
$(3, 2, 2)$
0%
$(1, -1, 1)$

Explanation

Using section formula,

Coordinate of the point which divides

P (2, 3, 1)

$P(2,3,1)$ and

Q (5, 0, 4)

$Q(5,0,4)$ in ratio

1 : 2

$1:2$ is

(\frac{2 \cdot 2 + 5}{2 + 1}, \frac{2 \cdot 3 + 0}{2 + 1}, \frac{2 \cdot 1 + 4}{2 + 1})

$\left(\dfrac{2\cdot 2+5}{2+1}, \dfrac{2\cdot 3+0}{2+1},\dfrac{2\cdot 1+4}{2+1} \right)$

= (\frac{9}{3}, \frac{6}{3}, \frac{6}{3}) = (3, 2, 2)

$=\left(\dfrac{9}{3}, \dfrac{6}{3},\dfrac{6}{3} \right)=(3,2,2)$

△ A B C

$\triangle ABC$ the mid points of the sides

A B, B C

$AB, BC$ and

C A

$CA$ are respectively

(l, 0, 0), (0, m, 0)

$\left( l,0,0 \right) ,\left( 0,m,0 \right)$ and

(0, 0, n)

$\left( 0,0,n \right)$ . Then,

\frac{{A B}^{2} + {B C}^{2} + {C A}^{2}}{l^{2} + m^{2} + n^{2}}

$\dfrac { { AB }^{ 2 }+{ BC }^{ 2 }+{ CA }^{ 2 } }{ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } }$ is equal to

0%
$2$
0%
$4$
0%
$8$
0%
$16$

Explanation

Let

A = (x_{1}, y_{1}, z_{1}), B = (x_{2}, y_{2}, z_{2}), C = (x_{3}, y_{3}, z_{3})

$A=(x_1,y_1,z_1),B=(x_2,y_2,z_2),C=(x_3,y_3,z_3)$

From the figure,

x_{1} + x_{2} = 2 l, y_{1} + y_{2} = 0, z_{1} + z_{2} = 0

${ x }_{ 1 }+{ x }_{ 2 }=2l, { y }_{ 1 }+{ y }_{ 2 }=0, { z }_{ 1 }+{ z }_{ 2 }=0$ , [midpoint formula]

x_{2} + x_{3} = 0, y_{2} + y_{3} = 2 m, z_{2} + z_{3} = 0

${ x }_{ 2 }+{ x }_{ 3 }=0, { y }_{ 2 }+{ y }_{ 3 }=2m, { z }_{ 2 }+{ z }_{ 3 }=0$

and

x_{1} + x_{3} = 0, y_{1} + y_{3} = 0, z_{1} + z_{3} = 2 n

${ x }_{ 1 }+{ x }_{ 3 }=0, { y }_{ 1 }+{ y }_{ 3 }=0, { z }_{ 1 }+{ z }_{ 3 }=2n$

On solving, we get

x_{1} = l, x_{2} = l, x_{3} = - l

${ x }_{ 1 }=l, { x }_{ 2 }=l, { x }_{ 3 }=-l$ ,

y_{1} = - m, y_{2} = m, y_{3} = m

${ y }_{ 1 }=-m, { y }_{ 2 }=m, { y }_{ 3 }=m$

and

z_{1} = n, z_{2} = - n, z_{3} = n

${ z }_{ 1 }=n, { z }_{ 2 }=-n, { z }_{ 3 }=n$

∴

$\therefore$ Coordinates are

A (l, - m, n), B (l, m, - n)

$A\left( l,-m,n \right) ,B\left( l,m,-n \right)$ and

C (- l, m, n)

$C\left( -l,m,n \right)$

∴ \frac{{A B}^{2} + {B C}^{2} + {C A}^{2}}{l^{2} + m^{2} + n^{2}}

$\therefore \dfrac { { AB }^{ 2 }+{ BC }^{ 2 }+{ CA }^{ 2 } }{ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } }$

= \frac{4 m^{2} + 4 n^{2} + 4 l^{2} + 4 n^{2} + (4 l^{2} + 4 m^{2})}{l^{2} + m^{2} + n^{2}}

$=\dfrac { 4{ m }^{ 2 }+4{ n }^{ 2 }+4{ l }^{ 2 }+4{ n }^{ 2 }+\left( 4{ l }^{ 2 }+4{ m }^{ 2 } \right) }{ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } }$

= 8

$=8$

The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes

x - y + 2 z = 3

$x−y+2z=3$ and

2 x - 2 y + z + 12 = 0

$2x−2y+z+12=0$ is :

0%
$2\sqrt{2}$
0%
$4$
0%
$\sqrt2$
0%
$23$

The coordinates of any point, which lies in

y z

$yz$ plane, are

0%
$(x,y,y)$
0%
$(0,y,y)$
0%
$(0,y,x)$
0%
$(x,y,z)$

Explanation

In a

y - z

$y-z$ plane, x co-ordinate is always

0

$0$

(0, y, y)

$(0,y,y)$ and

(0, y, x)

$(0,y,x)$ are point in a

y - z

$y-z$ plane.

The point

(3, 0, - 4)

$(3, 0, -4)$ lies on the

0%
Y-axis
0%
Z-axis
0%
XY-plane
0%
XZ-plane
0%
YZ-plane

What is the length of the segment in the x-y plane with end points at

(- 2, - 2)

$(-2, -2)$ and

(2, 3)

$(2, 3)$ ?

0%
$3$
0%
$\sqrt{31}$
0%
$\sqrt{41}$
0%
$7$
0%
$9$

If the line joining

A (1, 3, 4)

$A(1, 3, 4)$ and

B

$B$ is divided by the point

(- 2, 3, 5)

$(-2, 3, 5)$ in the ratio

1 : 3

$1 : 3$ , then

B

$B$ is

0%
$(-11, 3, 8)$
0%
$(-11, 3, -8)$
0%
$(-8, 12, 20)$
0%
$(13, 6, -13)$

Explanation

Let the co-ordinates of

B

$B$ be

(x, y, z)

$(x, y, z)$ .

Then applying the section formula gives us

- 2 = \frac{3 (1) + x}{3 + 1}

$-2=\dfrac{3(1)+x}{3+1}$

- 8 = x + 3

$-8=x+3$

x = - 11

$x=-11$ ................(i)

Similarly

3 = \frac{3 (3) + y}{4}

$3=\dfrac{3(3)+y}{4}$

12 = 9 + y

$12=9+y$

y = 3

$y=3$ ..........(ii)

5 = \frac{3 (4) + z}{4}

$5=\dfrac{3(4)+z}{4}$

20 = 12 + z

$20=12+z$

z = 8

$z=8$ ...........(iii)

Hence

B = (- 11, 3, 8)

$B=(-11, 3, 8)$

Graph

x^{2} + y^{2} = 4

$x^2+y^2=4$ in 3D looks like

0%
Circle
0%
Cylinder
0%
Hemisphere
0%
Sphere

Explanation

The given curve is

x^{2} + y^{2} = 4

$x^2+y^2=4$

x

$x$ coordinate and y-coordinate are connected by

x^{2} + y^{2} = 4

$x^2+y^2=4$

which is locus of a circle with radius

2

$2$

But z-coordinate can be anything, so in three dimension the circle

x^{2} + y^{2} = 4

$x^2+y^2=4$ will be

stretched which will be a cylinder with radius same as the radius of the circle .

An equation of sphere with centre at origin and radius

r

$r$ can be represented as

0%
$x^2+y^2+z^2=r$
0%
$x^2+y^2+z^2=r^2$
0%
$x^2+y^2+z^2=2r^2$
0%
None of the above

Explanation

Sphere is locus of a point in 3D whose distance from a fixed point(center) is constant (radius)

\Rightarrow \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 0)^{2}} = | r |

$\Rightarrow \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=|r|$

\Rightarrow x^{2} + y^{2} + z^{2} = r^{2}

$\Rightarrow x^2+y^2+z^2=r^2$ , square both sides

Point

D

$D$ has coordinates as

(3, 4, 5)

$(3,4,5)$ . Referring to the given figure, find the coordinates of point

E

$E$ .

0%
$(0,4,3)$
0%
$(0,4,5)$
0%
$(0,5,4)$
0%
$(0,3,4)$

Explanation

Point D has coordinates

(3, 4, 5)

$(3,4,5)$

Point E is in line with D, lying on the y-z plane having x coordinate 0, such that line DE is perpendicular to the y-z plane.

Thus, the y and coordinates of point E remain same as that of D.

The coordinates of point E are therefore

(0, 4, 5)

$(0,4,5)$

The ratio in which the line joining

(2, - 4, 3)

$(2, -4, 3)$ and

(- 4, 5, - 6)

$(-4, 5, -6)$ is divided by the plane

3 x + 2 y + z - 4 = 0

$3x+2y+z-4=0$ is

0%
$2 : 1$
0%
$4 : 3$
0%
$-1 : 4$
0%
$2 : 3$

Explanation

A line passes through the points

(2, - 4, 3)

$(2,-4,3)$ and

(- 4, 5, - 6)

$(-4,5,-6)$

The equation of the line can be written as

t = \frac{x - 2}{- 4 - 2} = \frac{y + 4}{5 + 4} = \frac{z - 3}{- 6 - 3}

$t = \cfrac{x - 2}{-4 - 2} = \cfrac{y + 4}{5 + 4} = \cfrac{z - 3}{-6 - 3}$

∴ x = - 6 t + 2, y = 9 t - 4, z = - 9 t + 3

$\therefore x = -6t + 2, y = 9t - 4, z = -9t + 3$

Substituting this in equation of the plane, we get

- 18 t + 6 + 18 t - 8 - 9 t + 3 - 4 = 0

$-18t + 6 + 18t - 8 - 9t + 3 - 4 = 0$

\Rightarrow t = \frac{- 1}{3}

$\Rightarrow t = \cfrac{-1}{3}$ and the point becomes

(2 + 2, - 3 - 4, 3 + 3)

$(2 + 2, -3 - 4, 3 + 3)$ i.e.

(4, - 7, 6)

$(4,-7,6)$

Let this point divide the line joining

(2, - 4, 3)

$(2,-4,3)$ and

(- 4, 5, - 6)

$(-4,5,-6)$ in the ratio

m : n

$m:n$ ,

∴ \frac{- 4 m + 2 n}{m + n} = 4

$\therefore \cfrac{-4m + 2n}{m + n} = 4$

∴ - 4 m + 2 n = 4 m + 4 n

$\therefore -4m + 2n = 4m + 4n$

∴ - 8 m = 2 n

$\therefore -8m = 2n$

∴ m : n = - 1 : 4

$\therefore m:n = -1:4$

Calculate the distance between the points

(- 3, 6, 7)

$(-3,6,7)$ and

(2, - 1, 4)

$(2,-1,4)$ in

3 D

$3D$ space.

0%
$4.36$
0%
$5.92$
0%
$7.91$
0%
$9.11$
0%
$22.25$

Explanation

We know the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The coordinates are $(-3, 6, 7)$ and $(2, -1, 4)$

$d=\sqrt{(2-(-3))^2+((-1)-6)^2+(4-7)^2}$

$d=\sqrt{(5)^2+((-7)^2+(-3)^2}$

$d=\sqrt{25+49+9}$

$d=\sqrt{83}$

$d = 9.11$

If the distance between the points

(7, 1, - 3)

$(7,1,-3)$ and

(4, 5, λ)

$(4,5,\lambda)$ is

13

$13$ units, then what is one of the values of

λ

$\lambda$ ?

0%
$20$
0%
$10$
0%
$9$
0%
$8$

Explanation

Given two points A(7,1,-3) and B(4,5,

λ

$\lambda$ )

total distance=13units ,

λ

$\lambda$ =?

distance of AB

= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}} 13 = \sqrt{(4 - 7)^{2} + (5 - 1)^{2} + (λ + 3)^{2}} 169 = 9 + 16 + [λ^{2} + 6 λ + 9] λ^{2} + 6 λ + 9 = 144 \Rightarrow λ^{2} + 6 λ - 135 = 0 ∴ λ = 9, - 15 ∴ λ = 9

$=\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 }+({ z }_{ 2 }-{ z }_{ 1 })^{ 2 } } \\ 13=\sqrt { (4-7)^{ 2 }+(5-1)^{ 2 }+(\lambda +3)^{ 2 } } \\ 169=9+16+\left[ \lambda ^{ 2 }+6\lambda +9 \right] \\ { \lambda }^{ 2 }+6\lambda +9=144\\ \Rightarrow { \lambda }^{ 2 }+6\lambda -135=0\\ \therefore \lambda =9,-15\\ \therefore \lambda =9$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 6 - MCQExams.com

We know the distance formula:

0:0:3

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 6 - MCQExams.com

We know the distance formula:

0:0:3

Practice Class 11 Engineering Maths Quiz Questions and Answers

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