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Questions Related to maths

Perimeter of triangle whose vertices are $(0,4,0), (3,4,0)$ and $(0,4,4)$, is

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. $10$

  2. $12$

  3. $25$

  4. $15$

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Correct Option: B
Explanation:

 Vertices of triangle are $(0,4,0),(3,4,0)$ and $(0,4,4)$

then perimeter=?
here AB=$\sqrt{(3-0)^2+(4-4)^2+(0-0)^2}$
$=3$
BC$=\sqrt{(3-0)^2+(4-4)^2+(0-4)^2}$
$=\sqrt{9+16}$
$=5$
CA$=\sqrt{(0-0)^2+(4-4)^2+(0-4)^2}=4$
used distance formula b/w two points
$(x _1,y _1,z _1) and (x _2,y _2,z _2)$
$=\sqrt{(x _2-x _2)^2+(y _2-y _1)^2+(z _2-z _1)^2}$
$ perimeter =ABC+BC+CA$
$=3+5+4$
$=12\ units$

Let the distance between vectors are given as follows :
$(i)4i +3j-6k, -2i+j-k$  be $\displaystyle \sqrt{k}$
$(ii) -2i+3j+5k, 7i-k $  be  $\displaystyle m\sqrt{n}$
Find $k-(m*n)$ ?

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. 20

  2. 21

  3. 22

  4. 23

Show answer
Correct Option: D
Explanation:

(i) Distance between $4i + 3j - 6k$ and $-2i + j - k$ is given by $\sqrt{(4 + 2)^2 + (3 - 1)^2 + (-6 + 1)^2} = \sqrt{36 + 4 + 25} = \sqrt{65}$

$\Rightarrow k = 65$

(ii) Distance between $-2i + 3j + 5k$ and $7i - k$ would be $\sqrt{(-2 - 7)^2 + (3)^2 + (5 + 1)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$
$\Rightarrow m = 3, n = 14$

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

The name of the figure formed by the points $(3, -5, 1), (-1, 0, 8)$ and $(7, -10, -6)$ is

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. a triangle

  2. a straight line

  3. an isosceles triangle

  4. an equilateral triangle

Show answer
Correct Option: B
Explanation:

Let $A=(3,-5,1), B = (-1,0,8)$ and $C=(7,-10,-6)$

Now, 
$AB =\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}=\sqrt {90}=3\sqrt{10}$,
$BC =\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}=\sqrt{360}=6\sqrt{10}$ 
and $CA =\sqrt{(7-3)^2+(-10+5)^2+(-6-1)^2}=\sqrt{90}=3\sqrt{10}$
Clearly $BC = AB+CA$
$\therefore  $ given points lies on straight line. 

If $(1, 1, a)$ is the centroid of the triangle formed by the points $(1, 2, -3)$ , $(\mathrm{b}, 0, 1)$ and $(-1, 1, -4)$ then $a-b$ $=$

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. $-5$

  2. $-7$

  3. $5$

  4. $1$

Show answer
Correct Option: A
Explanation:

The coordinates of the vertices of the triangle are given by $(1,2,-3) , (b,0,1), (-1,1,-4)$

Accordingly the coordinates of the centroid of this triangle will be given by 

($ \dfrac{b}{3}, 1 , -2 $)

Hence, $ \dfrac{b}{3} $ $= 1$ Or, $b= 3$

and $a = -2$

So $a- b = -5$

The circum centre of the triangle formed by the points $(2, 5, 1), (1, 4, -3)$ and $(-2, 7, -3)$ is

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. $(6,0,1)$

  2. $(0,6,-1)$

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

  4. $(6,1,-2)$

Show answer
Correct Option: B
Explanation:

Let the points $A(2,5,1) B(1,4,-3)$ and $C(-2,7,-3)$
Now using distance formula in $3D$, we have
$AB {=}$$\sqrt{18}$, $BC {=}$$\sqrt{18}$, and $AC {=}$$\sqrt{36}$
Since, ${AB}^{2}+{BC}^{2}$${=}$${AC}^{2}$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e $AC$.
Therefore by using section formula (1:1), circumcentre ${=}(0,6,-1)$

Assertion (A): The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus
Reason (R): $AB = BC = CD = DA$ and $AC = BD$

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. Both A and R are individually true and R is the correct explanation of A

  2. Both A and R individually true but R is not the correct explanation of A

  3. A is true but R is false

  4. Both A and R false

Show answer
Correct Option: D
Explanation:
Given: The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus. 
So using distance formula Reason is not true. 
Thus both A and R false. 

$P(0,5,6),Q(1,4,7),R(2,3,7)$ and $S(3,5,16)$ are four points in the space. The point nearest to the origin $O(0,0,0)$ is

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. $P$

  2. $Q$

  3. $R$

  4. $S$

Show answer
Correct Option: A
Explanation:

The $4$ points are as given.
We calculate their individual distance from the origin.
$OP =$ $ {({5}^{2} + {6}^{2})}^{0.5} $ = $ {(61)}^{0.5} $
$OQ =$ $ {({1}^{2} + {4}^{2} + {7}^{2} )}^{0.5} $ = $ {(66)}^{0.5} $
$OR =$ $ {({2}^{2} + {3}^{2} + {7}^{2} )}^{0.5} $ = $ {(62)}^{0.5} $
$OS =$ $ {({3}^{2} + {5}^{2} + {16}^{2} )}^{0.5} $= $ {(290)}^{0.5} $
Hence, $P$ is the nearest to the origin.

The name of the figure formed by the points $(-1, -3, 4), (5, -1,1), (7, -4, 7)$ and $(1, -6, 10)$ is a

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. square

  2. rhombus

  3. parallelogram

  4. rectangle

Show answer
Correct Option: B
Explanation:

Keeping the above points as vertices and using the distance formula, 
$D=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}+(z _{2}-z _{1})^{2}}$
We get that the sides of the parallelogram formed by the above lines are equal and all the sides are of $7$ units.
Hence the parallelogram will be either a rhombus or a square. For the parallelogram to be square, all the adjacent sides of the parallelogram should make an angle of $\dfrac{\pi}{2}$.
Consider the vector equation of $AB$ as $(5-(-1))i'+(-1-(-3))j'+(1-4)k'$
$6i'+2j'-3k'$
Consider the vector equation of $BC$ as $(7-5)i'+(-4-(-1))j'+(7-1)k'$
$2i'-3j'+6k'$
Taking dot product, we get $6(2)+2(-3)-3(6)$
$=12-6-18$
$=-12$
Thus the parallelogram is a rhombus.

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

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. 26 m

  2. 28 m

  3. 30 m

  4. 36 m

Show answer
Correct Option: A
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 distance of the point (1,−2,4)(1,−2,4) from the plane passing through the point (1,2,2)(1,2,2) and perpendicular to the planes x−y+2z=3x−y+2z=3 and 2x−2y+z+12=0,2x−2y+z+12=0, is :

maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

  1. $2\sqrt{2}$

  2. $4$

  3. $\sqrt2$

  4. $23$

Show answer
Correct Option: A