Tag: scalars and vectors

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Questions Related to scalars and vectors

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

The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is 

  1. $\sqrt{17}$

  2. $\sqrt{29}$

  3. $\sqrt{34}$

  4. $\sqrt{41}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$P(1,2,3),$ Plane :$x-y+z=5$

$F$ is foot of perpendicular form $P$ to plane and $I$ is image,then $PF=FI$
$\therefore$ If $(x,y,z)=(r+1,-r+2,r+3)$ are foot of perpendicular.
$ \Rightarrow (r+1)-(-r+2)+r+3=5\quad \quad \Rightarrow r=1\ \therefore F=(2,1,4)\ \therefore I=(3,0,5)$
$ \therefore$ distance of $I$ from origin  $=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 34 } $

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

The equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

  1. $x-2z=0$

  2. $2x-z=0$

  3. $2x+y=0$

  4. $x-2y=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the given points be A($1,2,3$) and B($3,2,-1$) and let the point equidistant from A and B be P($x,y,z$) 

then  $PA=PB$
$\sqrt {{{(x - 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}}  = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} + {{(z + 1)}^2}} $
Squaring both sides
${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$
${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$
$-2x-6z+10=-6x+2z+10$
$4x-8z=0$
$x-2z=0$

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

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

  1. $(3, 4, 12)$

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

  3. $(4, 5, 13)$

  4. $(40, 53, 157)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given points be $A(1,1,1)$

Consider,
$\dfrac{{x - 1}}{3} = \dfrac{{y - 1}}{4} = \dfrac{{z - 1}}{{12}} = \lambda $

$\begin{array}{l} x=3\lambda +1 \  \ y=4\lambda + 1\  \ z=12\lambda +1 \end{array}$

General point on the line is 
$3\lambda  + 1,\,4\lambda  + 1,\,12\lambda  + 1$

Given that,
$AP=13$ 

$\sqrt {{{\left( {3\lambda  + 1 - 1} \right)}^2} + {{\left( {\,4\lambda  + 1 - 1} \right)}^2} + {{\left( {12\lambda  + 1 - 1} \right)}^2}}  = 13$

$13\lambda =13$

$\lambda =1$

$\begin{array}{l} 3\lambda +1=4 \  \ 4\lambda +1=5 \  \ 12\lambda +1=13 \end{array}$

Therefore, required point $P$ is $(4,5,13)$.
Hence the correct option is $C$.

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

The x-coordinate of a point on the line joining the points $P(2,2,1)$ and $Q(5,1,-2)$ is $4$. Find its z-coordinate.

  1. $-1$

  2. $-2$

  3. $1$

  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P(2,2,1),Q(5,1,-2)$


$\therefore$ Equation of line through $P$ & $Q$,


$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \\ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r$

$\therefore P$ be point of line 

$\Rightarrow P\equiv (-3r+2,r+2,3r+1)$

$ \therefore -3r+2=4$ ($\because $ since x-coordinate is $4$)

$\Rightarrow r=\cfrac { -2 }{ 3 } $

$\therefore$ z-coordinate $=3r+1=-1$

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

The distance between (5,1,3) and the line x=3, y=7+t, z=1+t is

  1. 4

  2. 2

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} x=3\, \, \, ,y=7+t\, \, ,z=1+t \ A\left( { 3,7+t,1+t } \right)  \ 0.\left( { 3-5 } \right) +1\left( { 7+t-1 } \right) +1\left( { 1+t-3 } \right) =0 \ 6+t+t-2=0 \ 2t=-4 \ t=-2 \ A:\left( { 3,5,-1 } \right)  \ dis\tan  ce=\sqrt { 4+16+16 }  \ =6 \ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

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

The distance between the parallel planes given by the equations, $\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})+3=0$ and $\vec{r}.(4\hat{i}-4\hat{j}+2\hat{k})+5=0$ is-

  1. $1/2$

  2. $1/3$

  3. $1/4$

  4. $1/6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Planes are  $2i-2j+k+3=0,4i-4j+2k+5=0,2i-2j+k+5/2=0$

distance between them is $=\cfrac{|c _1-c _2|}{\sqrt{a^2+b^2+c^2}}\=\cfrac{|3-\cfrac{5}{2}|}{\sqrt{2^2+2^2+1^2}}=\cfrac{1}{6}$

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

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$, then area of the square is

  1. $6$

  2. $3$

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

  4. $\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the extremities of the diagonal of a square be $(1,-2,3)$ and $B(2,-3,5)$.
Then $AB$ is given by $ {({1}^{2} + {1}^{2} + {2}^{2})}^{0.5} $ = $ \sqrt{6} $
Hence, length of the side $ = \sqrt {3} $
So, area of square will be $ \sqrt{3} \times \sqrt{3}  = 3$

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

The point equidistant from the points $(0,0,0), (1,0,0), (0,2,0)$ and $(0,0,3)$ is

  1. $(1,2,3)$

  2. $\left (\displaystyle \dfrac{1}{2},1,\dfrac{3}{2}\right)$

  3. $\left (-\displaystyle \dfrac{1}{2}, -1,-\displaystyle \dfrac{3}{2}\right)$

  4. $(1,-2,3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of sphere passes through origin is given by


$x^2+y^2+z^2+ax+by+cz=0 ...(1)$

Given, it also passes through $(1,0,0),(0,2,0),0,0,3)$

$1+a=0\Rightarrow a = -1$

$4+2b=0\Rightarrow b = -2$

and $ 9+3c=0\Rightarrow c = -3$

$\therefore$the equation of the sphere becomes $x^2+y^2+z^2-x-2y-3z=0$

Comparing with $x^2+y^2+z^2+2gx+2fy+2kz+C=0$

Therefore, the point equidistant from all the given four point will be the centre of the sphere $(1)$ passing through all these points which is $\left(\dfrac{1}{2},1, \dfrac{3}{2}\right)i.e. the \  centre$.

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

Distance between the points $(12,4,7)$ and $(10,5,3)$ is

  1. $\sqrt{21}$

  2. $\sqrt{5}$

  3. $\sqrt{17}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the problem,

Let the given points 
$A(12,4,7)$ and $B(10,5,3)$
So, distance between $A$ and $B$ by distance formula.
$AB=\sqrt{(10-12)^2+(5-4)^2+(3-7)^2}=\sqrt{(-2)^2+1^2+(-4)^2}$ 
$=\sqrt{4+1+16}=\sqrt{21}$
So, distance between the points $(12,4,7)$ and $(10,5,3)$ is $\sqrt{21}$ sq. units.