Tag: introduction to three dimensional geometry

Quizzes (6) Questions (119)

Questions Related to introduction to three dimensional geometry

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

$P$ and $Q$ are points on the line joining $A(-2,5)$ and $B(3,1)$ such that $AP=PQ=QB$. Then, the distance of the midpoint of $PQ$ from the origin is

  1. $3$

  2. $\frac {\sqrt 37}{4}$

  3. $4$

  4. $3.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Points P and Q divide AB into three equal parts. P = (-2+1/3(3-(-2)), 5+1/3(1-5)) = (-1/3, 11/3). Q = (-2+2/3(5), 5+2/3(-4)) = (4/3, 7/3). Midpoint of PQ = (1/2, 3). Distance from origin = sqrt(1/4 + 9) = sqrt(37)/2 = sqrt(37/4).

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

A line passes through two point $A (2, -3, -1)$ and $B (8, -1, 2)$. The coordinates of a point on this line at a distance of $14$ units from $A$ are

  1. $(14, 1, 5)$

  2. $(-10, -7, 7)$

  3. $(86, 25, 41)$

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

Given points are $A(2,-3,-1)$ and $B(8,-1,2)$
Therefore direction  ratio of $AB$ are $l = \dfrac{6}{7}, m=\dfrac{2}{7},n= \dfrac{3}{7}$ or $l=\dfrac{-6}{7},m= \dfrac{-2}{7},n= \dfrac{-3}{7}$
Hence, coordinates of a point $14$ unit from point $A$ is given as,$(2+14l,-3+14m,-1+14n)$
$\Rightarrow (14,1,5)$ or $(-10,-7,-7)$
Hence, option 'A' is correct.

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

If  $C _1:{x^2+y^2}-20x+64=0$ and $C _2:{x^2+y^2}+30x+144=0$. Then the length of the shortest line segment $PQ$  which touches $C _1$ at $P$ and  to  $C _2$ at $Q$ is

  1. $10$

  2. $15$

  3. $22$

  4. $27$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $C _1 : x^2+y^2-20x+64=0 \Rightarrow (x-10)^2+y^2=36$
and $C _2 : x^2+y^2+30x+144=0 \Rightarrow (x+15)^2+y^2=81$
So centre and radius of $C _1$ and $C _2$ are $(10,0)$, $(-15,0)$ and $6,9$ respectively.
Then, distance between $C _1$ and $C _2$ is $\sqrt{(15+10)^{2}+(0-0)^{2}}=25$. 

$PQ$ touches $C _{1}$ at $P$ and $C _{2}$ at $Q$.
Then, shortest length of $PQ$ is $=25-(9+6)=25-15=10$

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 the point $(2,1,-1)$ from the line $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-3}{-3}$ measured parallel to the plane $x+2y+z=4$ is

  1. $\sqrt{10}$

  2. $\sqrt{20}$

  3. $\sqrt{5}$

  4. $\sqrt{30}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The distance is found by projecting the vector from the point to a point on the line onto the normal of the plane containing the line and the direction parallel to the given plane.

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

If $L _1$ is the line of intersection of the plane $2x-2y+3z-2=0, x-y+z+1=0$ and $L _2$ is the line of intersection of the plane   $x+2y-z-3=0, 3x-y+2z-1=0$, then the distance of origin from from the plane containing the lines $L _1$ + $L _2$ is :

  1. $\dfrac{1}{\sqrt{2}}$

  2. $\dfrac{1}{4\sqrt{2}}$

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

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

The lines L1 and L2 are found by solving the systems of plane equations. The plane containing these lines is determined, and the distance from the origin is calculated using the standard point-to-plane formula.

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 plane which is passing through the point $(1,2,3)$ and which is at maximum distance from the point $(-1,0,2)$ is

  1. $2x+2y+z=9$

  2. $2x+z=5$

  3. $3x+y-z=2$

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

The plane passes through (1,2,3). The distance from (-1,0,2) is maximized when the normal vector is parallel to the vector from the plane point to the external point. Normal = (1-(-1), 2-0, 3-2) = (2,2,1). Equation: 2(x-1)+2(y-2)+1(z-3)=0 => 2x+2y+z=9.