MCQ 511 Mark
The equation of plane passing through $(-1, 0, -1)$ parallel to $xz$ plane is:
- A
$y = -2$
- ✓
$y = 0$
- C
$-x - z = 0$
- D
AnswerCorrect option: B. $y = 0$
Given, that the plane is parallel to $xz$ plane and the plane passes through $(-1, 0, -1)$ since the plane is parallel to $xz$ plane, the $y-$coordinate should be constant
Given, that it passes through point $(-1, 0, -1)$
$\therefore$ The plane lies on $xz$ plane
$\therefore$ The equation of plane is $y = 0$
View full question & answer→MCQ 521 Mark
Arrange the points. $A(1, 2, -3), B(-1, -2, -3), C(-1, -2, -3)$ and $D(1, -2, -3)$ in the increasing order of their octant numbers:
- ✓
$\text{A, B, C, D}$
- B
$\text{B, C, D, A}$
- C
$\text{C, D, A, B}$
- D
AnswerCorrect option: A. $\text{A, B, C, D}$
View full question & answer→MCQ 531 Mark
The image of the point $P(1, 3, 4)$ in the plane $2x - y + z = 0$ is:
- ✓
$(-3, 5, 2)$
- B
$(3, 5, 2)$
- C
$(3, -5, 2)$
- D
$(3, 5, -2)$
AnswerCorrect option: A. $(-3, 5, 2)$
View full question & answer→MCQ 541 Mark
The ratio in which $xy-$plane divides the line joining the points $(1, 0, -3)$ and $(1, -5, 7)$ is given by:
- A
$7 : 3$
- ✓
$3 : 7$
- C
$3 : 4$
- D
AnswerCorrect option: B. $3 : 7$
Let $xy-$planexy divide the line joining the given points in the ratio $k : 1$ and the point of intersection is $x, y, 0$
$=\frac{\text{7k} - 3}{\text{k} + 1}=0$
$=7\text{k}-3=0$
$\text{k}=\frac{3}{7}$
$\therefore$ The ratio is $3 : 7$
View full question & answer→MCQ 551 Mark
The plane $\text{XOZ}$ divides the join of $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio $\lambda:1$ then $\lambda$ is:
- A
$-3$
- B
$\frac{1}{4}$
- C
$3$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
Given points are $(1, -1, 5)$ and $(2, 3, 4)$ Using section formula the desired points is
$=\Big(\frac{2\lambda+1}{\lambda+1},\frac{3\lambda-1}{\lambda+1},\frac{4\lambda+5}{\lambda+1}\Big)$
Since, this point lies in $\text{XOZ}$ plane then its $yy-$co$-$ordinate should be zero.
$\Rightarrow\frac{3\lambda-1}{\lambda+1}=0$
$\Rightarrow\lambda=\frac{1}{3}$
View full question & answer→MCQ 561 Mark
$D(2, 1, 0), E(2, 0, 0), F(0, 1, 0)$ are mid point of the sides $\text{BC, CA, AB}$ of $\Delta\text{ABC}$ respectively, The the centroid of $\Delta\text{ABC}$ is:
- A
$\Big(\frac{1}{3},\frac{1}{3},\frac{1}{3}\Big)$
- ✓
$\Big(\frac{4}{3},\frac{2}{3},0\Big)$
- C
$\Big(-\frac{1}{3},\frac{1}{3},\frac{1}{3}\Big)$
- D
$\Big(\frac{2}{3},\frac{1}{3},\frac{1}{3}\Big)$
AnswerCorrect option: B. $\Big(\frac{4}{3},\frac{2}{3},0\Big)$
Centroid of triangle coincide with the centroid of triangle formed by joing the mid$-$point of sides of triangle.
So, centroid of $\triangle\text{ABC} =$ centroid of.
$\triangle\text{DEF}=\Big(\frac{2+2+0}{3},\frac{1+0+1}{3},\frac{0+0+0}{3}\Big) = \Big(\frac{4}{3},\frac{2}{3},0\Big)$
View full question & answer→MCQ 571 Mark
If the vertices of a triangle are $A(0, 4, 1), B(2, 3, -1)$ and $C(4, 5, 0),$ then orthocentre of a $\triangle\text{ABC}$ is:
- A
$(4, 5, 0)$
- ✓
$(2, 3, -1)$
- C
$(-2, 3, -1)$
- D
AnswerCorrect option: B. $(2, 3, -1)$
View full question & answer→MCQ 581 Mark
If the $zx-$plane divides the line segment joining $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio $p : 1$ then $p + 1 =$
- A
$\frac{1}{3}$
- B
$1$
- C
$\frac{3}{4}$
- ✓
$\frac{4}{3}$
AnswerCorrect option: D. $\frac{4}{3}$
Let the points be given by $(1, -1, 5)$ and $(2, 3, 4)$
A point that divides the line joining these $2$ points in the ratio $p : 1$ is
given by $\Big(\frac{2\text{p}+1}{\text{p}+1},\frac{3\text{p}-1}{\text{p}+1},\frac{4\text{p}+5}{\text{p}+1}\Big)$
Since, this point has to lie on the $zx-$plane.
so, $3p - 1 = 0$
$\Rightarrow\text{p}=\frac{1}{3}$
$\Rightarrow1+\text{p}=\frac{4}{3}$
View full question & answer→MCQ 591 Mark
An ordered triplet corresponds to in three dimensional space:
- A
- ✓
- C
- D
Infinite number of points
AnswerIt is fundamental fact that, The ordered triplet $(x, y, z)$ represents an unique point in three dimensional space.
View full question & answer→MCQ 601 Mark
Two opposite vertices of a rectangle are $(1, 3)$ and $(5, 1).$ If the rest two vertices lie on the line $y - x + l = 0,$ then $l$ is equal to:
View full question & answer→MCQ 611 Mark
Choose the correct answer. The distance of point $P(3, 4, 5)$ from the $yz-$plane is:
- ✓
$3$ units.
- B
$4$ units.
- C
$5$ units.
- D
$550.$
AnswerCorrect option: A. $3$ units.
Given point is $P(3, 4, 5)$
$\therefore$ Distance of from yz-plane
$=\sqrt{(0-3)^2+(4-4)^2+(5-5)^2}$
$=\sqrt{9}$
$=3$ units
Hence, the correct option is $(a).$
View full question & answer→MCQ 621 Mark
Choose the correct answer. Distance of the point $(3, 4, 5)$ from the origin $(0, 0, 0)$ is:
AnswerCorrect option: A. $\sqrt{50}$
Given point $A(3, 4, 5)$ and the given $O(0, 0, 0)$
$\therefore\sqrt{(3-0)^2+(4-0)^2+(5-0)^2}$
$=\sqrt{9+16+25}$
$=\sqrt{50}$
View full question & answer→MCQ 631 Mark
Distance between the points $(12, 4, 7)$ and $(10, 5, 3)$ is:
- A
$\sqrt{27}$
- B
$\sqrt{5}$
- C
$\sqrt{17}$
- ✓
AnswerConsider the problem,
Let the given points
$A(12, 4, 7)$ and $B(10, 5, 3)$
So, distance between A and B by distance formula.
$\text{AB}=\sqrt{(10-12)^2+(5-4)^2+(3-7)^2}$
$=\sqrt{(-2)^2+1^2+(-4)^2}$
$=\sqrt{4+1+6}$
$=\sqrt{21}$
So, distance between the points $(12, 4, 7)$ and $(10, 5, 3)$ is $\sqrt{21}\text{ Sq. units}$
View full question & answer→MCQ 641 Mark
The points $(2, 5)$ and $(5, 1)$ are the two opposite vertices of a rectangle. If the other two vertices are points on the straight line $y = 2x + k,$ then the value of $k$ is:
AnswerPoints $(2, 5)$ and $(5, 1)$ form a diagonal of the rectangle the mid point of these points will lie on the other diagonal. Mid Point $\Big(\frac{2+5}{2},\frac{5+1}{2}\Big)=\Big(\frac{7}{2},3\Big)$
Equation of the other diagonal is $y = 2x + k$
$\therefore3=2\times\frac{7}{2}+\text{k}$
$\Rightarrow\text{k}=-4$
View full question & answer→MCQ 651 Mark
The plane. $\text{ax + by + cz + (-3) = }0$ meet the co$-$ordinate axes in $\text{A, B, C.}$ The centroid of the triangle is:
- A
$\big(3\text{a, 3b, 3c}\big)$
- B
$\Big(\frac{3}{\text{a}},\frac{3}{\text{b}},\frac{3}{\text{c}}\Big)$
- C
$\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)$
- ✓
$\Big(\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\Big)$
AnswerCorrect option: D. $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\Big)$
View full question & answer→MCQ 661 Mark
If the orthocentre, circumcentre of a triangle are $(-3, 5, 2), (6, 2, 5)$ respectively then the centroid of the triangle is:
AnswerCorrect option: A. $(3, 3, 4)$
Since, the centroid divides the line joining the orthocentre and circumcentre in the ratio $2 : 1$
The coordinates of the centroid will be, $\Big(\frac{9}{3},\frac{9}{3},\frac{12}{3}\Big)$
$=(3, 3, 4)$
View full question & answer→MCQ 671 Mark
The ratio in which the line joining the points $(a, b, c)$ and $(-a, -c, -b)$ is divided by the $xy-$plane is
- A
$a : b$
- B
$b : c$
- C
$a : c$
- ✓
$c : d$
AnswerCorrect option: D. $c : d$
Let $A ≡ (a, b, c)$ and $B ≡ (-a, -c, -b)$
Let the line joining $A$ and $B$ be divided by the $xy-$plane at point $P$ in the ratio $\lambda:1.$
Then, we have,
$\text{P}\equiv\Big(\frac{-\text{a}\lambda+\text{a}}{\lambda+1},\ \frac{-\text{c}\lambda+\text{b}}{\lambda+1},\ \frac{-\text{b}\lambda+\text{a}}{\lambda+1}\Big)$
Since $P$ lies on the $xy-$plane, the $z-$coordinate of $P$ will be zero
$\therefore\frac{-\text{b}\lambda+\text{c}}{\lambda+1}=0$
$\Rightarrow-\text{b}\lambda+\text{c}=0$
$\Rightarrow\lambda=\frac{\text{c}}{\text{b}}$
Hence, the $xz-$plane divides $AB$ in the ratio $c : b$
View full question & answer→MCQ 681 Mark
Plane $ax + by + cz = 1$ intersect axes in $\text{A, B, C}$ respectively. If $\text{G}\Big(\frac{1}{6},-\frac{1}{3},{1}\Big)$ is a centroid of $\triangle\text{ABC}$ then $\text{a + b + 3c:}$
- A
$\frac{4}{3}$
- B
$4$
- ✓
$2$
- D
$\frac{5}{6}$
Answer$\text{A}\Big(\frac{1}{\text{a}},0,0\Big)\text{ B}\Big(0,\frac{1}{\text{b}},0\Big)\text{ C}\Big(0,0,\frac{1}{\text{c}}\Big)$
centroid $\Rightarrow\Big(\frac{1}{3\text{a}},\frac{1}{3\text{b}},\frac{1}{3\text{c}}\Big)=\Big(\frac{1}{6},\frac{-1}{3},1\Big)$ On comparing we get,
$3a = 6$
$\Rightarrow a = 2$
$3b = -3$
$\Rightarrow b = -1$
$3c = 1$
$\Rightarrow\text{c}=\frac{1}{3}$
$\therefore a = 2, b = -1,$
$\Rightarrow\text{c}=\frac{1}{3}$
$\therefore a + b + 3c = 2$
View full question & answer→MCQ 691 Mark
The distance from the origin to the centroid of the tetrahedron formed by the points $(0, 0, 0), (3, 0, 0), (0, 4, 0), (0, 0, 5)$ is:
AnswerCorrect option: D. $\frac{\sqrt{3^2+4^2+5^2}}{4}$
$\text{G}=\Big(\frac{\text{a}}{4},\frac{\text{b}}{4},\frac{\text{c}}{4}\Big)\therefore\sqrt{\Big({\frac{\text{a}^2}{16}+\frac{\text{b}^2}{16}+\frac{\text{c}^2}{16}\Big)}}$
here, $a = 3, b = 4, c = 5$ substituting in above equation we get.
$\text{OG}=\sqrt{\frac{3^2+4^2+5^2}{4}}$
View full question & answer→MCQ 701 Mark
A point $C$ with position vector $\frac{\text{3a}+4\text{b}-5\text{c}}{3} ($where $a, b$ and $c$ are non co$-$planar vectors$)$ divides the line joining $A$ and $B$ in the ratio $2 : 1.$ If the position vector of $A$ is $a - 2b + 3c,$ then the position vector of $B$ is:
- A
$2a + 3b - 4c$
- B
$2a - 3b + 4c$
- C
$2a + 3b + 4c$
- ✓
$a + 3b - 4c$
AnswerCorrect option: D. $a + 3b - 4c$
$\frac{\text{3a}+4\text{b}-5\text{c}}{3}$
$\overrightarrow{\text{c}}=\frac{2\overrightarrow{\text{b}}+\overrightarrow{\text{a}}}{3}$
$\overrightarrow{\text{b}}=\frac{3\overrightarrow{\text{c}}-\overrightarrow{\text{a}}}{2}$
$=\frac{\big(3\overrightarrow{a}+4\overrightarrow{b}-5\overrightarrow{c}\big)-\big(\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}\big)}{2}$
$=\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$
View full question & answer→MCQ 711 Mark
The plane $\text{XOZ}$ divides the join of $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio $\lambda:1$ then $\lambda$ is:
- A
$-3$
- B
$\frac{-1}{3}$
- C
$3$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
The plane $\text{XOZ}$ divides the join of $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio $\lambda:1$
i.e. $y = 0$ divide the join of $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio.
$\lambda:1$
$\therefore\frac{3\lambda−1}{\lambda+1}=0$
$\Rightarrow\lambda=\frac{1}{3}$
View full question & answer→MCQ 721 Mark
If the points $A(3, -2, 4), B(1, 1, 1)$ and $C(-1, 4, -2)$ are collinear, then the ratio in which $C$ divides $AB$ is:
- A
$1 : 2$
- ✓
$-2 : 1$
- C
$-1 : 2$
- D
AnswerCorrect option: B. $-2 : 1$
Let $C$ divide $AB$ in the ratio $x : y$
Let us compare the $x-$coordinate of $C$ by using section formula.
$-1=\frac{\text{x}\times1+\text{y}\times3}{\text{x+y}}$
$\Rightarrow x + 3y = -x - y$
$\Rightarrow x = -2y$
point $C$ divides $AB$ in the ratio $-2 : 1.$
As the ratio is negative,
it means $C$ divides $AB$ externally.
View full question & answer→MCQ 731 Mark
If $A = (2, -3, 1), B = (3, -4, 6)$ and $C$ is a point of trisection of $AB,$ then $C_y=$
- A
$\frac{11}{3}$
- B
$-11$
- C
$\frac{10}{3}$
- ✓
AnswerGiven, $C$ is a point of trisection of $AB.$
$C$ either divides $AB$ in the ratio $2 : 1$ or $1 : 2$
Case $1: C$ divides in the ratio $2 : 1$
The coordinates of $C$ will be $\Big(\frac{8}{3},-\frac{11}{3},\frac{13}{3}\Big)$
Case $2: C$ divides in the ratio $1 : 2$
The coordinates of $C$ will be $\Big(\frac{7}{3},-\frac{10}{3},\frac{8}{3}\Big)$
either $C_y=$ $\frac{-11}{3}$ or $-\frac{10}{3}$
View full question & answer→MCQ 741 Mark
In geometry, we take a point, a line and a plane as undefined terms:
AnswerIn Geometry, we define a point as a location and no size. A line is defined as something that extends infinitely in either direction but has no width and is one dimensional while a plane extends infinitely in two dimensions.
View full question & answer→MCQ 751 Mark
The perpendicular distance of the point $P(6, 7, 8)$ from the $XY-$Plane is:
AnswerLet $Q$ be the foot of perpendicular drawn from the point $P (6, 7, 8)$ to the $XY-$plane.
Thus, the distance of this foot $Q$ from $P$ is $z-$coordinate of $P,$
i.e. $8$ units.
View full question & answer→MCQ 761 Mark
Choose the correct answer. The point $(-2, -3, -4)$ lies in the:
AnswerThe point $(-2, -3, -4)$ lies in seventh octant.
Hence the correct option is $(b).$
View full question & answer→MCQ 771 Mark
If $x-$coordinate of a point $P$ of line joining the points $Q(2, 2, 1)$ and $R(5, 2, -2)$ is $4,$ then the $z-$coordinate of $P$ is:
View full question & answer→MCQ 781 Mark
$(-1, -5, -7)$ lies in Octant:
AnswerHere all the three $x, y, z$ coordinate are negative of the given point.
Therefore, it will lie in the seventh Octant.
View full question & answer→MCQ 791 Mark
There are three points with position vectors $-2a + 3b + 5c, a + 2b + 3c$ and $7a - c.$ What is the relation between the three points:
AnswerThe relation between the three points are collinear.
View full question & answer→MCQ 801 Mark
The length of the perpendicular drawn from the point $P(3, 4, 5)$ on $y-$axis is
- A
$10$
- ✓
$\sqrt{34}$
- C
$\sqrt{113}$
- D
$512$
AnswerCorrect option: B. $\sqrt{34}$
The length of the perpendicular drawn from the point $P (3, 4, 5)$ on $y-$axis is given by
$\sqrt{3^2+5^2}$
$=\sqrt{34}$
Hence, the correct answer is option $(b)$
View full question & answer→MCQ 811 Mark
The point $(3, 0, -4)$ lies on the:
- A
$Y-$axis
- B
$Z-$axis
- C
$XY-$plane
- ✓
$XZ-$plane
AnswerCorrect option: D. $XZ-$plane
$(3, 0, -4)$ Given pointClearly, $y = 0$ and $x$ and $z$ have non$-$zero value.
If the point lies on $x - z$ plane, this condition is possible. the answer is $XZ-$plane.
View full question & answer→MCQ 821 Mark
Which of the following is true for a plane:
AnswerCorrect option: A. A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus
Option $A$ and $C$ are correct $A$ locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus. and also Value of $z$ in a $xy$ plane is zero.
View full question & answer→MCQ 831 Mark
Points $\text{A}\big(3,2,4),\text{B}\Big(\frac{33}{5},\frac{28}{5},\frac{38}{5}\Big),\text{C}\big(9,8,10\big)$ are given The ratio in which $B$ divides $\overline{\text{AC}}$ is:
- A
$5 : 3$
- B
$2 : 1$
- C
$1 : 3$
- ✓
Answer$B$ divides $AC$ in the ratio is $x_1 - x_2: x_2- x_3$
$3-\frac{33}{5}:\frac{33}{5}-9$
$3:2$
View full question & answer→MCQ 841 Mark
The ratio in which the plane $2x + 3y - 2z + 7 = 0$ divides the line segment joining the points $(-1, 1, 3), (2, 3, 5)$ is:
- A
$3 : 5$
- B
$7 : 5$
- C
$9 : 11$
- ✓
View full question & answer→MCQ 851 Mark
The point $A(1, -1, 3), B(2, -4, 5)$ and $C(5, -13, 11)$ are:
View full question & answer→MCQ 861 Mark
The ordinate of the point which divides the lines joining the origin and the point $(1, 2)$ externally in the ratio of $3 : 2$ is:
- A
$-2$
- B
$\frac{3}{5}$
- C
$\frac{2}{5}$
- ✓
$6$
AnswerCo$-$ordinates of the required point will be
$\text{y}=\frac{\text{m}_{1}\text{y}_{2}-\text{m}_{2}\text{y}_{1}}{\text{m}_{1}-\text{m}_{2}}$
$=\frac{3\times2-2\times0}{3-2}$
$=6$
View full question & answer→MCQ 871 Mark
A point at which all the three perpendicular coordinate axes meets is known as:
AnswerThe three perpendicular coordinate axes meets at one of the point which divides the plane into eight quadrant.
The $1^{st}$ quadrant has all positive points, $2^{nd}$ has $x-ve$ and remaining $2\ +ve$ points and so on.
Only $(0, 0, 0)$ is not included in any quadrant and is the intersection point.
Thus, the three axes meet at $(0, 0,0)$ from where the eight quadrants originate.
Hence, the point is known as origin.
View full question & answer→MCQ 881 Mark
In the $\Delta \text{ABC} A = (1, 3, -2)$ and $G (-1, 4, 2)$ is the centroid of the triangle. If $D$ is the mid point of $BC$ then $AD =$
- A
$\frac{\sqrt21}{2}$
- ✓
$\frac{3\sqrt21}{2}$
- C
$\sqrt{21}$
- D
$\frac{63}{2}$
AnswerCorrect option: B. $\frac{3\sqrt21}{2}$
First, we calculate the distance $AG,$
It is $(4 + 1 + 16)^{0.5}= 21^{0.5}$
From the property of the centroid that it divides the line joining $AD$ in the ratio $2 : 1.$
The distance $=\text{AD}=\frac{3}{2}\text{AG}$
$\text{AD}=\frac{3}{2}{ 21}^{0.5}$
View full question & answer→MCQ 891 Mark
Area of quadrilateral whose vertices are $(2, 3), (3, 4), (4, 5)$ and $(5, 6),$ is equal to:
View full question & answer→MCQ 901 Mark
The perpendicular distance of the point $P(6, 7, 8)$ from $xy-$plane is
AnswerThe distance of the point $P(6, 7, 8)$ from the $xy-$plane is equal to the $z-$coordinate of the point.
Here, the value of $z-$coordinate is $8.$
Hence, the correct answer is option $(a)$.
View full question & answer→MCQ 911 Mark
if $P(0, 1, 2), Q (4, -2, 1)$ and $O(0, 0, 0)$ are three points, then $\angle\text{POQ}=$
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
AnswerCorrect option: D. $\frac{\pi}{2}$
$\mathrm{PQ}^2=(4-0)^2+(-2-1)^2+(1-2)^2=16+9+1=26$
$\mathrm{OP}^2=(0-0)^2+(1-0)^2+(2-0)^2=0+1+4=5$
$\mathrm{QO}^2=(0-4)^2+(0+2)^2+(0-1)^2=16+1+4=21$
Since, $\mathrm{PQ}^2=\mathrm{OP}^2+\mathrm{QO}^2$
Hence, $\angle\text{POQ}=\frac{π}{2}$
View full question & answer→MCQ 921 Mark
The position vectors of the four angular point of a tetrahedron $\text{OABC}$ are $(0, 0, 0), (0, 0, 2), (0, 4, 0)$ and $(6, 0, 0)$ respectively. Find the coordinates of cenroid:
- A
$\Big(2,\frac{4}{3},\frac{2}{3}\Big)$
- ✓
$\Big(\frac{6}{4},1,\frac{2}{4}\Big)$
- C
$(0, 0, 0)$
- D
AnswerCorrect option: B. $\Big(\frac{6}{4},1,\frac{2}{4}\Big)$
Angular points of tetrahedron $\text{OABC}$ are.
$(0, 0, 0), (0, 0, 2), (0, 4, 0), (6, 0, 0)$ To find the coordinates of the centroid of the tetrahedron whose vertices are
$\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right),\left(x_3, y_3, z_3\right)$ and $\left(x_4, y_4, z_4\right)$ the centroid is
$\Big(\frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}+\text{x}_{4}}{4}\Big),\Big(\frac{\text{y}_{1}+\text{y}_{2}+\text{y}_{3}+\text{y}_{4}}{4}\Big),\Big(\frac{\text{z}_{1}+\text{z}_{2}+\text{z}_{3}+\text{z}_{4}}{4}\Big)$
Now, substituting the values we get
$\Big(\frac{0+0+0+6}{4}\Big),\Big(\frac{0+0+4+0}{4}\Big),\Big(\frac{0+2+0+0}{4}\Big)$
$\therefore$ The coordinates of the centroid are $\Big(\frac{6}{4},1,\frac{2}{4}\Big)$
View full question & answer→MCQ 931 Mark
Under what condition does the equation $x^2+ y^2+ z^2+ 2ux + 2vy + 2wz + d$ represent a real sphere:
- A
$ u^2+v^2+w^2=d^2 $
- ✓
$ u^2+v^2+w^2>d $
- C
$ u^2+v^2+w^2$
- D
$ u^2+v^2+w^2$
AnswerCorrect option: B. $ u^2+v^2+w^2>d $
View full question & answer→MCQ 941 Mark
Choose the correct answer. If the distance between the points $(a, 0, 1)$ and $(0, 1, 2)$ is $27$, then the value of a is:
AnswerCorrect option: B. $\pm5$
Given points are $A(a, 0, 1)$ and $B(0, 1, 2).$
$\therefore\text{AB}=\sqrt{(\text{a}-0)^2+(0-1)^2+(1-2)^2}$
$=\sqrt{27} ($Given$)$
$\Rightarrow27=\text{a}^2+2$
$\Rightarrow\text{a}^2=25$
$\Rightarrow\text{a}=\pm5$
View full question & answer→MCQ 951 Mark
An equation of sphere with centre at origin and radius $r$ can be represented as:
- A
$x^2+y^2+z^2=r$
- ✓
$x^2+y^2+z^2=r^2$
- C
$x^2+y^2+z^2=2 r^2$
- D
AnswerCorrect option: B. $x^2+y^2+z^2=r^2$
Sphere is locus of a point in $3D$ whose distance from a fixed point $($center$)$ is constant $($radius$)$
$\Rightarrow\sqrt{({\text{x}-0})^2+(\text{y}-0)^2+(\text{z}-0)^2}$
$=\mid\text{r}\mid\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2$ square both sides.
View full question & answer→MCQ 961 Mark
A point is on the $x-$axis. Which of the following represent the point:
- A
$(0, x, 0)$
- B
$(0, 0, x)$
- ✓
$(x, 0, 0)$
- D
AnswerCorrect option: C. $(x, 0, 0)$
At $x-$axis, $y$ and $z$ coordinates are zero.
View full question & answer→MCQ 971 Mark
In three dimensions, the coordinate axes of a rectangular cartesian coordinate system are:
- A
Three mutually parallel lines
- ✓
Three mutually perpendicular lines
- C
Two mutually perpendicular lines and any two parallel
- D
AnswerCorrect option: B. Three mutually perpendicular lines
In three dimensions, the coordinate axes, i.e. $x, y$ and $z$ axes of a rectangular cartesian coordinate system are three mutually perpendicular lines. The word rectangular is used to indicate perpendicularity among the axes.
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Assertion $(A):$ If centroid and circumcentre of a triangle are known its orthocentre can be found
Reason $(R):$ Centriod, orthocentre and circumcentre of a triangle are collinear
- ✓
Both $A$ and $R$ are individually true and $R$ is the correct explanation of $A.$
- B
Both $A$ and $R$ individually true but $R$ is not the correct explanation of $A.$
- C
$A$ is true but $R$ is false.
- D
$A$ is false but $R$ is true.
AnswerCorrect option: A. Both $A$ and $R$ are individually true and $R$ is the correct explanation of $A.$
Centroid, orthocentre and circumcentre are collinear and centroid dividesthe line joining orthocentre and circumcentre in the ratio $2 : 1$ so if any two points are given then this one can be found.
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If the extremities of the diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5),$ then the length of the side is
- A
$\sqrt{6}$
- ✓
$\sqrt{3}$
- C
$\sqrt{5}$
- D
$\sqrt{7}$
AnswerCorrect option: B. $\sqrt{3}$
Length of the diagonal $=\sqrt{(2 − 1)^2 + (−3 + 2)^2 + (5 − 3)^2}$
$=\sqrt{1 + 1 + 4}$
$=\sqrt{6}$
$\therefore$ Length of the side $=\frac{\text{Length of diagonal}}{\sqrt{2}}$
$=\frac{\sqrt{6}}{\sqrt{2}}$
$=\sqrt{3}$
View full question & answer→MCQ 1001 Mark
The coordinates of the foot of the perpendicular drawn from the point $P(3, 4, 5)$ on the $yz-$ plane are
- A
$(3, 4, 0)$
- ✓
$(0, 4, 5)$
- C
$(3, 0, 5)$
- D
$(3, 0, 0)$
AnswerCorrect option: B. $(0, 4, 5)$
We know that the x-coordinate on $yz-$plane is $0.$
The coordinates of the foot of the perpendicular drawn from the point $P(3, 4, 5)$ on the $yz-$plane are $(0, 4, 5).$
Hence, the correct answer is option $(b).$
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