Tag: scalars and vectors

Questions Related to scalars and vectors

$P(0,5,6),Q(1,4,7),R(2,3,7)$ 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

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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. $23$

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Correct Option: A

Calculate the distance between the points $(-3,6,7)$ and $(2,-1,4)$ in $3D$ space.

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

  1. $4.36$

  2. $5.92$

  3. $7.91$

  4. $9.11$

  5. $22.25$

Show answer
Correct Option: D
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)$

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

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

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

$d=\sqrt{83}$

$d = 9.11$

A point on the line $\displaystyle \frac{{x + 2}}{1} = \frac{{y - 3}}{{ - 4}} = \frac{{z - 1}}{{2\sqrt 2 }}$ at a distance 6 from the point (2, 3, 1) is

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

  1. $(4-21, 1+12\sqrt{2})$

  2. $\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$

  3. $\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$

  4. $\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$

Show answer
Correct Option: B
Explanation:

$(\lambda -2, -4\lambda+3, 2\sqrt 2\lambda+1)\leftrightarrow (-2, 3, 1)$
$(\lambda-2+2)^2+(-4\lambda+3-3)^2+2\sqrt 2\lambda+1-1)^2=36$
$\lambda^2+16\lambda^2+8\lambda^2=36$
$\lambda=\pm \frac {6}{5}$
$\therefore point=(\frac {-4}{5}, \frac {-9}{5}, 1)$

The shortest distance between z-axis and the line 
$x+y+2z-3=0=2x+3y+4z-4$, is _____________

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

  1. $1$

  2. $2$

  3. $4$

  4. $3$

Show answer
Correct Option: A
Explanation:

$x+y+2z-3=0=2x+3y+4z-4$ at $z-axis, x=y=0$

$x+y=3-2z$ and $2x+3y=4(1-z)$
solving $x$ and $y$ in function $z$,
$2x+3(3-2z-x)=4(1-z)$
$\implies 2x+9-6z-3x=4-4z$
$\implies x=9-6z-4+4z$
$\implies x=5-2z\quad equation (2)$
$\implies (5-2z)+y=3-2z$
$\implies y=3-2z-5+2z$
$y=-2\quad equation (2)$
So, $\sqrt {x^2+y^2}=\sqrt {(-2)^2+(5-2z)^2}$
at, $z=\cfrac {5}{2}$, this would be minimum.

In a $\triangle {ABC}$, side $AB$ has the equation $2x+3y=29$ and the side $AC$ has the equation $x+2y=16$. If the mid point of $BC$ is $(5,6)$, then the equation of $BC$ is

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

  1. $2x+y=7$

  2. $x+y=1$

  3. $2x-y=17$

  4. None of these

Show answer
Correct Option: B
Explanation:
Let co-ordinates of $B$ be $(x _1, y _1)$ & $C$ be $(x _2, y _2)$

$\therefore$ $(5,6)$ is the mid point,

so, $\dfrac{x _1 + x _2}{2} = 5, \dfrac{y _1 + y _2}{2} = 6$

$\Rightarrow x _1 + x _2 = 10, y _1 + y _2 = 12$

$B(x _1, y _1)$ lies on the line $2x + 3y = 29$

$\therefore 2x _1 + 3y _1 = 29$  ----(1)

$C(x _2, y _2)$ lies on the line $x + 2y = 16,$

$\therefore x _2 + 2y _2 = 16$  ----(2)

$\therefore$ putting $x _1, y _1$ in the form of $x _2, y _2$ in (1)

$2(10 - x _2) + 3(12 - y _2) = 29$  {$x _1 = 10 - x _2, y _1 = 12 - y _2$}

$\Rightarrow 20 - 2x _2 + 36 - 3y _2 = 29$

$\Rightarrow 2x _2 + 3y _2 = 27$  ----(3)

on subtracting $(3)$ and $(2)$ $\times$ $2$

$-y _2 = -5$

$y _2 = 5$

Putting $y _2 \,  in (2)$

$x _2 + 2(5) = 16$

$x _2 = 6$

$x _1 = 10 - x _2$

      $= 4$

$y _1 = 12 - 5 = 7$

Equation :

$\dfrac{x - x _1}{x _2 - x _1} = \dfrac{y - y _1}{y _2 - y _1}$

$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$

$\Rightarrow -x + 4 = y - 7$

$\Rightarrow x + y = 11$

A swimmer can swim $2$ km in $15$ minutes in a lake and in a river he can swim a distance of $4$ km in $20$ minutes along the stream. If a paper boat is put in the river, then the distance covered by it in $\displaystyle $2$ \, \frac{1}{2}$2 hours will be 

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

  1. $18$ km

  2. $12$ km

  3. $8$ km

  4. $10$ km

Show answer
Correct Option: D
Explanation:

Speed of the man in still water $\cfrac{2}{15/60}$ = $8$ km/hr
Speed of the man in downstream = $\cfrac{4}{20/60}$ = $12$ km/hr
Speed of stream = $4$ km/hr
Distance covered by paper boat in $2$ $\frac{1}{2}$ hours = $\cfrac{5}{2} \, \times$  $4$ = $10$ km 

The point equidistant from the point $O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$ and $C(0, 0, c)$ has the coordinates

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

  1. $(a, b, c)$

  2. $(a/2, b/2, c/2)$

  3. $(a/3, b/3, c/3)$

  4. $(a/4, b/4, c/4)$

Show answer
Correct Option: B
Explanation:
$P(x, y, z)$
$PO=\sqrt{x^2+y^2+z^2}$
$PA=\sqrt{(a-x)^2+(-y)^2+(-z)^2}=\sqrt{(a-z)^2+y^2+z^2}$
$PB=\sqrt{x^2+(b-y)^2+z^2}$
$PC=\sqrt{x^2+y^2+(C-z)^2}$
$\Rightarrow PO=PA$
$\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}$
$x^2+y^2+z^2=(a-x)^2+y^2+z^2$
$x^2=(a-x)^2$
$x=a-x$
$\Rightarrow x=\dfrac{a}{2}$
$y=b/2$
$z=c/2$.