Tag: scalars and vectors

Questions Related to scalars and vectors

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

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

Find the distance between the pairs of points whose cartesian coordinates are $(2,3,-1), (2,6,2).$

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

  1. $\displaystyle 3\sqrt{2}.$

  2. $\displaystyle 2\sqrt{3}.$

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

  4. $\displaystyle 2\sqrt{5}.$

Show answer
Correct Option: A
Explanation:

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(2,3,-1)$, $(2,6,2)$
Distance $=\sqrt { \left( 2-2 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } } $
                $=\sqrt { 18 } $
                $=3\sqrt { 2 }$

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

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

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

  1. $\displaystyle \sqrt{118}$

  2. $8$

  3. $\displaystyle \sqrt{26}$

  4. none of these

Show answer
Correct Option: C
Explanation:

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(-1,1,3)$, $(0,5,6)$
Distance $=\sqrt { \left( -1-0 \right) ^{ 2 }+\left( 1-6 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } $
                $=\sqrt { 26 } $

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

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

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

  1. $\displaystyle 3\sqrt{14}$

  2. $\displaystyle \sqrt{54}$

  3. $\displaystyle 3\sqrt{19}$

  4. $\displaystyle \sqrt{57}$

Show answer
Correct Option: A
Explanation:

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $-2\hat i+3\hat j+5\hat k$, $7\hat i-\hat k$
Distance $=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 3-0 \right) ^{ 2 }+\left( 5+1 \right) ^{ 2 } } $
                $=\sqrt { 126 } $
                $=3\sqrt { 14 } $

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

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

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

  1. $\displaystyle \sqrt{65}$

  2. $\displaystyle \sqrt{69}$

  3. $13$

  4. none of these

Show answer
Correct Option: A
Explanation:

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $4\hat i+3\hat j-6\hat k$, $-2\hat i+\hat j-\hat k$
Distance $=\sqrt { \left( 4+2 \right) ^{ 2 }+\left( 3-1 \right) ^{ 2 }+\left( -6+1 \right) ^{ 2 } } $
                $=\sqrt { 65 } $

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. $1$

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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.