Tag: scalars and vectors

Questions Related to scalars and vectors

Find the distance between $(12,3,4)$ and $(4,5,2)$

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

  1. $\sqrt {72}$

  2. $\sqrt {62}$

  3. $\sqrt {64}$

  4. None of these

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Correct Option: A
Explanation:

Consider the problem,

Let the given points 
$A(12,3,4)$ and $B(4,5,2)$
So, distance between $A$ and $B$ by distance formula.
$AB=\sqrt{(4-12)^2+(5-3)^2+(2-4)^2}=\sqrt{(-8)^2+2^2+(-2)^2}$ 
$=\sqrt{64+4+4}=\sqrt{72}$
So, distance between the points $(12,3,4)$ and $(4,5,2)$ is $\sqrt{72}$ sq. units.

If $O=(0,0,0),OP=5$ and the d.rs of OP are $1,2,2$ then $P _x+P _y+P _z=$

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

  1. $25$

  2. $\dfrac{25}{9}$

  3. $\dfrac{25}{3}$

  4. $\left(\dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3}\right)$

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Correct Option: D

Find the co-ordinates of a point lying on the line $\dfrac{x -2}{3} = \dfrac{y + 3}{4} = \dfrac{z - 1}{7}$ which is at a distance $10$ units from $(2, -3, 1)$.

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

  1. $(32,37,71)$

  2. $(-28,-43,-69)$

  3. $(-32,-37,-71)$

  4. None of these

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Correct Option: D
Explanation:
Given that the required point lies on the line $\dfrac{x-2}{3}=\dfrac{y+3}{4}=\dfrac{z-1}{7}$.

The required point is at a distance of $10$ units from $(2,-3,1)$

By option verification,

(A) Substituting the point $(32,37,71)$ 

$\implies \dfrac{32-2}{3}=\dfrac{37+3}{4}=\dfrac{71-1}{7}$

$\implies 10=10=10$

Therefore, distance is $\sqrt{(32-2)^2+(37+3)^2+(71-1)^2} =86.02$ units

Hence, $(32,37,71)$ is not the required point.

(B) Substituting the point $(-28,-43,-69)$ 

$\implies \dfrac{-28-2}{3}=\dfrac{-43+3}{4}=\dfrac{-69-1}{7}$

$\implies -10=-10=-10$

Therefore, distance is $\sqrt{(-28-2)^2+(-43+3)^2+(-69-1)^2} =86.02$ units

Hence, $(-28,-43,-69)$ is not the required point.


(C) Substituting the point $(-32,-37,-71)$ 

$\implies \dfrac{-32-2}{3}=\dfrac{-37+3}{4}=\dfrac{-71-1}{7}$

$\implies 11.3=11.3=10.28$

Therefore, distance is $\sqrt{(-32-2)^2+(-37+3)^2+(-71-1)^2} =86.57$ units

Hence, $(-32,-37,-71)$ is not the required point.

The distance between the points $(\cos \, \theta , \, \sin \, \theta) $ and $ (\sin \, \theta - \cos \, \theta)$ is 

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

  1. $\sqrt{3}$

  2. $\sqrt{2}$

  3. $2$

  4. $1$

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Correct Option: B
Explanation:
Distance between the point $(\cos\theta, \sin\theta)$ and $(\sin\theta, -\cos\theta)$ is :
$ = \sqrt{(\sin\theta - \cos\theta)^2 + (-\cos\theta - \sin\theta)^2}$
$ = \sqrt{1 - 2 \sin\theta \cos\theta + 1 + 2 \sin\theta \cos\theta}$
$(\cos^2\theta + \sin^2\theta = 1)$
$= \sqrt{2}$
Option B is the correct answer

If the distance between a point P and the point (1, 1, 1) on the line $\frac{{x\, - \,1}}{3}\, = \,\frac{{y - \,1}}{4}\, = \,\frac{{z\, - 1}}{{12}}$ is 13, then the coordinates of P are

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

  1. (3, 4, 12)

  2. $\left( {\frac{3}{{13}},\,\frac{4}{{13}},\,\frac{{12}}{{13}}} \right)$

  3. (4, 5, 12)

  4. (40, 53, 157)

Show answer
Correct Option: D
Explanation:

$\begin{array}{l} According\, to\, question: \ we\, have\, the\, line\, equ=\frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } \, and\, point\, (P)\, is\, 13. \ Now, \ line\, of\, equ: \ \Rightarrow \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } =13 \ Now\, find\, x,y\, & \, z\, \, coordinates: \ \, \Rightarrow \frac { { x-1 } }{ 3 } =13 \ \, \, \, \, \, \, \, \therefore \, \, \, \, x=39+1=40 \ \Rightarrow \frac { { y-1 } }{ 4 } =13 \ \, \, \, \, \, \, \, \therefore \, \, y=\, 52+1=53 \ \Rightarrow \frac { { z-1 } }{ { 12 } } =13 \ \, \, \, \, \, \, \, \, \therefore \, \, z=156+1=157 \ Now,\, we\, get\, \, new\, coordinates\, of\, P:\, \, \, \left( { 40,53,157 } \right) \, \,  \ so\, that\, \, the\, correct\, option\, is\, D. \end{array}$

If the lines $\frac{x - 0}{1} =\frac{y+1}{2}=\frac{z-1}{-1}$ and $\frac{x+1}{k}=\frac{y-3}{-2}=\frac{z-2}{1}$ are at right angles, then the value of k is

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

  1. $5$

  2. $0$

  3. $3$

  4. $-1$

Show answer
Correct Option: A
Explanation:

If $({ l } _{ 1 }{ m } _{ 1 }{ n } _{ 1 })$ and  $({ l } _{ 2 }{ m } _{ 2 }{ n } _{ 2 })$ are directions of two $\bot$ lines then,

${ l } _{ 1 }{ l } _{ 2 }+{ m } _{ 1 }{ m } _{ 2 }+n _{ 1 }{ n } _{ 2 }=0\ k-4-1=0\ k=5$

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b , a + b)$, then  $bx = ay$.

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

  1. True

  2. False

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Correct Option: A
Explanation:

If a point (x,y) is equidistant from A(a,b) and B(c,d) then,

$x = \frac{{a + c}}{2},y = \frac{{b + d}}{2}$
So,
$\begin{array}{l}x = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{2},y = \frac{{\left( {b - a} \right) + \left( {a + b} \right)}}{2}\x = \frac{{2a}}{2},y = \frac{{2b}}{2}\x = a,y = b\end{array}$

putting values in bx=ay
Solving LHS,
b(a)=ab
Solving RHS,
a(b)=ab
As LHS=RHS the equation bx=ay is true.

The area of triangle whose vertices are $(1, 2, 3), (2, 5, -1)$ and $(-1, 1, 2)$ is

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

  1. $150\ sq. units$

  2. $145\ sq. units$

  3. $\sqrt {155}/2\ sq. units$

  4. $155/2\ sq. units$

Show answer
Correct Option: C
Explanation:

Let the vertices of triangle are 


$A(1,2,3),B(2,5,-1)$ and $C(-1,1,2)$ 

Then, 

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =\hat { i } +3\hat { j } -4\hat { k }  \  \ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =-2\hat { i } -\hat { j } -\hat { k }  \end{array}$

Then, 

$\begin{array}{l} \overrightarrow { AB } \times \overrightarrow { AC } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i }  } & { \hat { j }  } & { \hat { k }  } \ 1 & 3 & { -4 } \ { -2 } & { -1 } & { -1 } \end{array} } \right|  \  \ =-7\hat { i } +9\hat { j } +5\hat { k }  \  \ \left| { \overrightarrow { AB } \times \overrightarrow { AC }  } \right| =\sqrt { { { \left( { -7 } \right)  }^{ 2 } }+{ { \left( 9 \right)  }^{ 2 } }+{ { \left( 5 \right)  }^{ 2 } } }  \  \ =\sqrt { 49+8+25 }  \  \ =\sqrt { 155 }  \end{array}$

Area of triangle $ABC=\frac{1}{2}|\vec {AB} \times \vec {AC}|$

$=\frac{1}{2} \times \sqrt {155}$

$=\frac{\sqrt {155}}{2}$ sq. units 

The points $(10,7,0)$, $(6,6-1)$ and $(6,9,-4)$ form a 

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

  1. Right -angled triangle

  2. Isosceles triangle

  3. Both $(1)$ & $(2)$

  4. Equilateral triangle

Show answer
Correct Option: C
Explanation:
$P(10, 7, 0),\ Q(6, 6, -1),\ R(6, 9, -4)$
$PQ=\sqrt {4^2+(-1)^2 +(-1)^2}$
$=\sqrt {16+1+1}=3\sqrt 3$
$QR=\sqrt {0^2+3^2+-3^2}$
$PR=\sqrt {4^2+2^2+(-4)^2}$
$=\sqrt {16+4+16}=6$
$PQ^2+QR^2=(3\sqrt 2)^2+(3\sqrt 2)^2=6^2=PR^2$
$\therefore \ \triangle PQR$ is a right angle $D$ of $Q,\ PQ=QR$
$\triangle PQR$ is a isoscels traingle
$(C)$ Both $(1)$ & $(2)$


The shortest distance of the point $(1,2,3)$ from ${x}^{2}+{y}^{2}=0$ is 

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

  1. $5$

  2. $\sqrt{5}$

  3. $2$

  4. $\sqrt{14}$

Show answer
Correct Option: B
Explanation:

${x}^{2}+{y}^{2}=0$

$\therefore$ $(x,y)\equiv (0,0)$
Let  point be $\equiv (0,0,z)$
$d\equiv \sqrt { { (1-0) }^{ 2 }+{ (2-0) }^{ 2 }+{ (z-3) }^{ 2 } } =\sqrt { 1+4+{ (z-3) }^{ 2 } } \Rightarrow { (z-3) }^{ 2 }\ge 0$
${ d } _{ min }=\sqrt { 5 } $