MCQ 11 Mark
If the distance between the points $(a, 0, 1)$ and $(0, 1, 2)$ is $\sqrt{27}$ then the value of $a$ is:
- A
$5$
- ✓
$\underline{+}5$
- C
$-5$
- D
AnswerCorrect option: B. $\underline{+}5$
View full question & answer→MCQ 21 Mark
If $A = (1, 2, 3), B = (2, 3, 4)$ and $AB$ is produced upto $C$ such that $\text{2AB = BC}$ then $C =$
- A
$(5, 4, 6)$
- B
$(6, 2, 4)$
- ✓
$(4, 5, 6)$
- D
AnswerCorrect option: C. $(4, 5, 6)$
Let the point $C$ be $(i, j, k)$
Since, $B$ divides $AC$ in the ratio $1 : 2$
Coordinates of $B$ should be $\Big(\frac{2+\text{i}}{3},\frac{4+\text{j}}{3},\frac{\text{k}+6}{3}\Big)$
Comparing the values given already for $B,$ we get, $i = 4, j = 5$ and $k = 6$
View full question & answer→MCQ 31 Mark
There is one and only one sphere through:
- ✓
$4$ points not in the same plane
- B
$4$ points not lie in the same straight line
- C
- D
$3$ points not lie in the same line
AnswerCorrect option: A. $4$ points not in the same plane
View full question & answer→MCQ 41 Mark
The vector equation of a sphere having centre at origin and radius $5$ is:
AnswerCorrect option: A. $\mid{\text{r}}\mid = 5$
View full question & answer→MCQ 51 Mark
The perpendicular distance of the point $P(3, 3, 4)$ from the $x-$axis is
AnswerThe perpendicular distance of the point $P(3, 3, 4)$ from the $x-$axis is given by
$\sqrt{3^2+4^2}$
$=\sqrt{25}$
$=5$
Hence, the correct answer is option $(b)$
View full question & answer→MCQ 61 Mark
The ratio in which $yz-$plane divides the line segment joining $(-3, 4, 2), (2, 1, 3)$ is:
- A
$-4 : 1$
- ✓
$3 : 2$
- C
$-2 : 3$
- D
AnswerCorrect option: B. $3 : 2$
Let the plane divide the line in the ratio $p : 1$
A point that divides the line joining these $2$ points in the ratio $p : 1$
given by $\Big(\frac{2\text{p}-3}{\text{p}+1},\frac{\text{p}+4}{\text{p}+1},\frac{3\text{p}+2}{\text{p}+1}\Big)$
Since, this point has to lie on the $zy-$plane.
so, $2p - 3 = 0$
$\Rightarrow\text{p}=\frac{3}{2}$
View full question & answer→MCQ 71 Mark
Find the distance between $(12, 3, 4)$ and $(4, 5, 2):$
- ✓
$\sqrt{72}$
- B
$\sqrt{62}$
- C
$\sqrt{64}$
- D
AnswerCorrect option: A. $\sqrt{72}$
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.
$\text{AB}=\sqrt{(4-12)^2+(5-3)^2+(2-4)^2}$
$=\sqrt{(-8)^2+2^2+(-2)^2}$
$=\sqrt{64+4+4}$
$=\sqrt{72}$
So, distance between the points $(12,3,4)$ and $(4,5,2)$ is $\sqrt{72}\text{ Sq. units.}$
View full question & answer→MCQ 81 Mark
Graph $x^2+ y^2= 4$ in $3D$ looks like:
AnswerThe given curve is $x^2+ y^2= 4$
So $x$ coordinate and $y-$coordinate are connected by $x^2+ y^2= 4$ which is locus of a circle with radius $2$ But $z-$coordinate can be anything,
so in three dimension the circle $x^2+ y^2= 4$ will be stretched which will be a cylinder with radius same as the radius of the circle.
View full question & answer→MCQ 91 Mark
The coordinates of any point, which lies on $x$ axis are:
- A
$(0, x, 0)$
- ✓
$(x, 0, 0)$
- C
$(x, x, 0)$
- D
$(x, x, x)$
AnswerCorrect option: B. $(x, 0, 0)$
In $3-$dimensional plane, the point which lies on $x-$axis does not have any part in $y$ and $z$ axes.
At that point, the value of $y$ and $z$ will be $0.$
Hence, coordinate of any point which lies on $x$ axis are $(x, 0, 0)$.
View full question & answer→MCQ 101 Mark
If the plane $7x + 11y + 13z = 3003$ meets the axes in $\ce{A, B, C}$ then the centroid of $\Delta\text{ABC}$ is:
- ✓
$(143, 91, 77)$
- B
$(143, 77, 91)$
- C
$(91, 143, 77)$
- D
AnswerCorrect option: A. $(143, 91, 77)$
View full question & answer→MCQ 111 Mark
$G(1, 1, -2)$ is the centroid of the triangle $\text{ABC}$ and $D$ is the mid point of $\text{BC.}$ If $A = (-1, 1, -4) D=$
AnswerLet the coordinates of $D$ be $(p, q, r)$
Since, the centroid divides the line joining $AD$ in the ratio $2 : 1$ the coordinates of centroid should be,
$\Big(\frac{2\text{p}-1}{3},\frac{2\text{q}+1}{3},\frac{2\text{r}-4}{3}\Big)$
Comparing it with the coordinates of the centroid given, $D (2, 1, -1).$
View full question & answer→MCQ 121 Mark
Point $A$ is $a + 2b,$ and $a$ divides $AB$ in the ratio $2 : 3.$ The position vector of $B$ is:
- A
$2a - b$
- B
$b - 2a$
- ✓
$a - 3b$
- D
AnswerCorrect option: C. $a - 3b$
Let us consider $x$ be the position vector of $B,$ then a divides $AB$ in the ratio $2 : 3$
$\text{a}=\frac{2\text{x}+3(\text{a+2b})}{2+3}$
$\Rightarrow\text{x}=\text{a - 3b}$
View full question & answer→MCQ 131 Mark
$\text{XOZ}$ plane divides the join of $(2, 3, 1)$ and $(6, 7, 1)$ in the ratio:
- A
$3 : 7$
- B
$2 : 7$
- ✓
$-3 : 7$
- D
AnswerCorrect option: C. $-3 : 7$
Let the plane divide the line in the ratio $p : 1$
A point that divides the line joining these $2$ points in the ratio $p : 1$ is
given by $\Big(\frac{6\text{p}+2}{\text{p}+1},\frac{7\text{p}+3}{\text{p}+1},\frac{\text{p}+1}{\text{p}+1}\Big)$
Since, this point has to lie on the $zx$ plane,
so, $7p + 3 = 0$
$\Rightarrow\text{p}=\frac{-3}{7}$
View full question & answer→MCQ 141 Mark
The coordinates of a point which divides the line joining the points $P(2, 3, 1)$ and $Q(5, 0, 4)$ in the ratio $1 : 2$ are:
AnswerCorrect option: C. $(3, 2, 2)$
Using section formula, Coordinate of the point which divides $P (2, 3, 1)$ and $Q (5, 0, 4)$ in ratio.
$1 : 2$ is $\Big(\frac{2.2+5}{2+1},\frac{2.3+0}{2+1},\frac{2.1+4}{2+1}\Big)=\Big(\frac{9}{6},\frac{6}{3},\frac{6}{3}\Big)=(3,2,2)$
View full question & answer→MCQ 151 Mark
What is the length of foot of perpendicular drawn from the point $P(3, 4, 5)$ on $y-$axis:
- A
$\sqrt{41}$
- ✓
$\sqrt{34}$
- C
$5$
- D
AnswerCorrect option: B. $\sqrt{34}$
View full question & answer→MCQ 161 Mark
Find the image of $(-2, 3, 4)$ in the $y z$ plane:
- A
$(-2, 3, 4)$
- ✓
$(2, 3, 4)$
- C
$(-2, -3, 4)$
- D
$(-2, -3, -4)$
AnswerCorrect option: B. $(2, 3, 4)$
View full question & answer→MCQ 171 Mark
If $\alpha,\beta,\text{y}$ are the angles made by a half ray of a line respectively with positive directions of $X-$axis, $Y-$axis and, $Z-$axis, then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \text{y} =$
View full question & answer→MCQ 181 Mark
The distance of the point $P(a, b, c)$ from the $x-$axis is:
- A
$\sqrt{(\text{a}^2 + \text{c}^2)}$
- B
$\sqrt{(\text{a}^2 + \text{b}^2)}$
- ✓
$\sqrt{(\text{b}^2 + \text{c}^2)}$
- D
AnswerCorrect option: C. $\sqrt{(\text{b}^2 + \text{c}^2)}$
The coordinate of the foot of the perpendicular from $P$ on $x-$axis are $(a, 0, 0)$.
So, the required distance $= \sqrt{{(\text{a – a})^2 + (\text{b – 0})^2 + (\text{c – 0})^2}}$
$=\sqrt{(\text{b}^2 + \text{c}^2)}$
View full question & answer→MCQ 191 Mark
What is the distance between the points $(2, -1, 3)$ and $(-2, 1, 3):$
- ✓
$2\sqrt{5}\text{ units}$
- B
$25\text{ units}$
- C
$4\sqrt{5}\text{ units}$
- D
$\sqrt{5}\text{ units}$
AnswerCorrect option: A. $2\sqrt{5}\text{ units}$
View full question & answer→MCQ 201 Mark
If $G$ is centroid of $\triangle\text{ABC}$ then:
- A
$\vec{G}=\vec{a}+\vec{b}+\vec{c}$
- B
$\vec{G}=\frac{\vec{a}+\vec{b}+\vec{c}}{2}$
- ✓
$3 \vec{ G }=\vec{a}+\vec{b}+\vec{c}$
- D
$3 \vec{G}=\frac{\vec{a}+\vec{b}+\vec{c}}{2}$
AnswerCorrect option: C. $3 \vec{ G }=\vec{a}+\vec{b}+\vec{c}$
We have,
In a $\triangle\text{ABC}$
$\overrightarrow {\text{A}}=\overrightarrow{\text{a}}$
$\overrightarrow {\text{B}}=\overrightarrow{\text{b}}$
$\overrightarrow {\text{C}}=\overrightarrow{\text{c}}$
then,
we know that,
$\overrightarrow {\text{G}} = \frac {\overrightarrow{\text{a}} + \overrightarrow {\text{b}} +\overrightarrow{\text{c}}}{3}$
$3\overrightarrow {\text{G}}=\overrightarrow {\text{a}}+ \overrightarrow {\text{b}}+\overrightarrow {\text{c}}$
View full question & answer→MCQ 211 Mark
The point $(0, -2, 5)$ lies on the:
- A
$z$ axis
- B
$x$ axis
- C
$xy$ plane
- ✓
$yz$ plane
AnswerCorrect option: D. $yz$ plane
Given, point is $(0, -2, 5)$
The $X-$coordinate in the given point is zero.
so, the point lies on $yz$ plane.
View full question & answer→MCQ 221 Mark
Three vertices of a parallelogram $\text{ABCD}$ are $A(1, 2, 3), B(-1, -2, -1)$ and $C(2, 3, 2).$ Find the fourth vertex $D:$
- A
$(-4, -7, -6)$
- ✓
$(4, 7, 6)$
- C
$(4, 7, -6)$
- D
AnswerCorrect option: B. $(4, 7, 6)$
View full question & answer→MCQ 231 Mark
The mid$-$points of the sides of a triangle are $(5, 7, 11), (0, 8, 5)$ and $(2, 3, -1)$. Then, the vertices are:
- ✓
$(7, 2, 5), (3, 12, 17), (-3, 4, -7)$
- B
$(7, 2, 5), (3, 12, 17), ( 3, 4, 7)$
- C
$(7, 2, 5), (-3, 12, 17), (-3, -4, -7)$
- D
AnswerCorrect option: A. $(7, 2, 5), (3, 12, 17), (-3, 4, -7)$
View full question & answer→MCQ 241 Mark
The points on the $y-$axis which are at a distance of $3$ units from the point $(2, 3, -1)$ is:
- A
Either $(0, -1, 0)$ or $(0, -7, 0)$
- B
Either $(0, 1, 0)$ or $(0, 7, 0)$
- C
Either $(0, 1, 0)$ or $(0, -7, 0)$
- ✓
Either $(0, -1, 0)$ or $(0, 7, 0)$
AnswerCorrect option: D. Either $(0, -1, 0)$ or $(0, 7, 0)$
View full question & answer→MCQ 251 Mark
Choose the correct answer. What is the length of foot of perpendicular drawn from the point $P(3, 4, 5)$ on $y-$axis.
- A
$\sqrt{41}$
- ✓
$\sqrt{34}$
- C
$5$
- D
$\text{None of these.}$
AnswerCorrect option: B. $\sqrt{34}$
We know that, on the $y-$axis $x = 0$ and $z = 0.$
$\therefore$ Point $\text{A}\equiv(0,4,0)$
$\therefore\text{PA}=\sqrt{(0-3)^2+(4-4)^2+(0-5)^2}$
$=\sqrt{9+0+25}$
$=\sqrt{34}$
View full question & answer→MCQ 261 Mark
If $P (x, y, z)$ is a point on the line segment joining $Q (2, 2, 4)$ and $R (3, 5, 6)$ such that the projections of $OP$ on the axes are $\frac{13}{5}, \frac{19}{5}, \frac{26}{5} $ respectively, then $P$ divides $QR$ in the ration:
- A
$1 : 2$
- ✓
$3 : 2$
- C
$2 : 3$
- D
$1 : 3$
AnswerCorrect option: B. $3 : 2$
View full question & answer→MCQ 271 Mark
Find the ratio in which $2x + 3y + 5z = 1$ divides the line joining the points $(1, 0, -3)$ and $(1, -5, 7):$
- A
$1 : 2$
- B
$2 : 1$
- C
$3 : 2$
- ✓
View full question & answer→MCQ 281 Mark
Choose the correct answer. $L$ is the foot of the perpendicular drawn from a point $P(3, 4, 5)$ on the $xy-$plane. The coordinates of point $L$ are:
- A
$(3, 0, 0).$
- B
$(0, 4, 5).$
- C
$(3, 0, 5).$
- ✓
AnswerWe know that on the $xy-$plane, $z = 0.$
Hence, the coordinates of the points $L$ are $(3, 4, 0).$
View full question & answer→MCQ 291 Mark
$L$ is the foot of the perpendicular drawn from a point $P(3, 4, 5)$ on the $xy-$plane. The coordinates of point $L$ are:
- A
$(3, 0, 0)$
- B
$(0, 4, 5)$
- C
$(3, 0, 5)$
- ✓
View full question & answer→MCQ 301 Mark
Choose the correct answer. $x-$axis is the intersection of two planes:
- ✓
$xy$ and $xz.$
- B
$yz$ and $zx.$
- C
$xy$ and $yz.$
- D
AnswerCorrect option: A. $xy$ and $xz.$
We know that on the $xy$ and $xz-$planes, the line of intersection is $x-$axis.
Hence, the correct option is $(a).$
View full question & answer→MCQ 311 Mark
The locus represented by $xy + yz = 0$ is:
- A
A pair of perpendicular lines
- B
- C
A pair of parallel planes
- ✓
A pair of perpendicular planes
AnswerCorrect option: D. A pair of perpendicular planes
View full question & answer→MCQ 321 Mark
The three planes divides the space into:
AnswerThree planes divides the space into eight regions.
View full question & answer→MCQ 331 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
$1:3$
- C
$\frac{3}{4}$
- ✓
$\frac{4}{3}$
AnswerCorrect option: D. $\frac{4}{3}$
Given, points are $(1, -1, 5)$ and $(2, 3, 4)$ since $ZX-$plane divides the line segment in the ratio $p : 1, y-$coordinate will be $0$ the $y-$coordinate of the point dividing the line segment will be.
$=\frac{3\text{p}-1}{\text{p} + 1}=0,\text{ p}=\frac{1}{3}\text{ p}+1=\frac{1}{3}+1=\frac{4}{3}$
View full question & answer→MCQ 341 Mark
Choose the correct answer. The locus of a point for which $x = 0$ is:
- A
$xy-$plane.
- ✓
$yz-$plane.
- C
$zx-$plane.
- D
AnswerCorrect option: B. $yz-$plane.
On the $yz-$plane, $x = 0$
Hence, the locus of the point is $yz-$plane.
So, the correct option is $(b).$
View full question & answer→MCQ 351 Mark
Choose the correct answer.
If a parallelopiped is formed by planes drawn through the points $(5, 8, 10)$ and $(3, 6, 8)$ parallel to the coordinate planes, then the length of diagonal of the parallelopiped is:
- ✓
$2\sqrt{3}$
- B
$3\sqrt{2}$
- C
$\sqrt{2}$
- D
$\sqrt{3}$
AnswerCorrect option: A. $2\sqrt{3}$
Given parallelepiped passes through $A(5, 8, 10)$ and $B(3, 6, 8)$
$\therefore$ Length of the diagonal,
$\text{AB}=\sqrt{(5-3)^2+(8-6)^2+(10-8)^2}$
$=\sqrt{4+4+4}$
$=2\sqrt{3}$
View full question & answer→MCQ 361 Mark
The equation of the set of point $P,$ the sum of whose distance from $A(4, 0, 0)$ and $B(-4, 0, 0)$ is equal to $10$ is:
- A
$9 x^2+25 y^2+25 z^2+225=0$
- ✓
$9 x^2+25 y^2+25 z^2-225=0$
- C
$9 x^2+25 y^2-25 z^2-225=0$
- D
$9 x^2-25 y^2-25 z^2-225=0$
AnswerCorrect option: B. $9 x^2+25 y^2+25 z^2-225=0$
View full question & answer→MCQ 371 Mark
The plane $x = 0$ divides the joinning of $(-2, 3, 4)$ and $(1, -2, 3)$ in the ratio:
- ✓
$2 : 1$
- B
$1 : 2$
- C
$3 : 2$
- D
AnswerCorrect option: A. $2 : 1$
$\text{R.E.F}$ image
Given, place $x = 0$ and two points
$\Rightarrow (-2, 3, 4)$ and $(1, -2, 3)$
let say a point $(x, y, z)$ in $x = 0$ place
$\text{x}=\frac{\text{m+n}(-2)}{\text{m+n}}=\frac{\text{m - 2n}}{\text{m+n}}$
$0=\frac{\text{m-2n}}{\text{m+n}}$
$\Rightarrow{\text{m}=2{\text{n}}}$
so $\frac{\text{m}}{\text{n}}=\frac{2}{1}$
$\Rightarrow2:1$
View full question & answer→MCQ 381 Mark
$\text{XOZ-}$plane divides the join of $(2, 3, 1)$ and $(6, 7, 1)$ in the ratio
- A
$3 : 7$
- B
$2 : 7$
- ✓
$-3 : 7$
- D
$-2 : 7$
AnswerCorrect option: C. $-3 : 7$
Let $A ≡ (2, 3, 1)$ and $B ≡ (6, 7, 1)$
Let the line joining $A$ and $B$ be divided by the $xz-$plane at point $P$ in the ratio $\lambda:1.$
Then, we have,
$\text{P}\equiv\Big(\frac{6\lambda+2}{\lambda+1},\ \frac{7\lambda+3}{\lambda+1},\ \frac{\lambda+1}{\lambda+1}\Big)$
Since $P$ lies on the $xz-$plane, the $y-$coordinate of $P$ will be zero.
$\therefore\frac{7\lambda+3}{\lambda+1}=0$
$\Rightarrow7\lambda+3=0$
$\Rightarrow\lambda=\frac{-3}{7}$
Hence, the $xz-$plane divides $AB$ in the ratio $-3 : 7$
View full question & answer→MCQ 391 Mark
The points $(5, 2, 4), (6, -1, 2)$ and $(8, -7, k)$ are collinear, if $k$ is equal to:
View full question & answer→MCQ 401 Mark
Distance between $A(4, 5 ,6)$ from origin $O$ is:
- A
$25\sqrt3$
- ✓
$\sqrt{77}$
- C
$3\sqrt{5}$
- D
AnswerCorrect option: B. $\sqrt{77}$
Origin is $O(0, 0, 0)$ and given point is $A(4, 5, 6)$
So, distance $=\sqrt{(4-0)^2+(5-0)^2+(6-0)^2}$
$=\sqrt{4^2+5^2+6^2}$
$=\sqrt{77}$
View full question & answer→MCQ 411 Mark
The locus of a first$-$degree equation in $x, y, z$ is $a:$
View full question & answer→MCQ 421 Mark
The coordinates of any point, which lies in $yz$ plane, are:
- A
$(x, y, y)$
- ✓
$(0, y, y)$
- C
$(0, y, x)$
- D
AnswerCorrect option: B. $(0, y, y)$
In a $y-z$ plane, $x$ co$-$ordinate is always $0$
So $(0, y, y)$ and $(0, y, x)$ are point in a $y-z$ plane.
View full question & answer→MCQ 431 Mark
Which octant do the point $(-5, 4, 3)$ lie:
- A
Octant $I$
- ✓
Octant $II$
- C
Octant $III$
- D
Octant $IV$
AnswerCorrect option: B. Octant $II$
Given, $(-5, 4, 3)$ is the point
Here, the x-coordinate is negative but $y$ and $z$ coordinates are positive
$\therefore (-5, 4, 3)$ lie in octant $II.$
View full question & answer→MCQ 441 Mark
The ratio of $yz-$plane divide the line joining the points $A (3, 1, -5), B (1, 4, -6)$ is:
- A
$3 : 1$
- ✓
$-1 : 3$
- C
$1 : 3$
- D
AnswerCorrect option: B. $-1 : 3$
Let $yz-$plane divide the line segment joining the points $A (3, 1, -5), B (1, 4, -6)$ in the ratio $m : n$
Then, $(0, y, z) =\Big(\frac{3\text{m+n}}{\text{m+n}},\frac{\text{m+4n}}{\text{m+n}},\frac{\text{-5m-6n}}{\text{m+n}}\Big)$
$\Rightarrow\frac{3\text{m+n}}{\text{m+n}}=0$
$\Rightarrow\text{m}:\text{n}=-1:3$
View full question & answer→MCQ 451 Mark
A cube of side $5$ has one vertex at the point $(1, 0, -1)$ and the three edges from this vertex are, respectively, parallel to the negative $x$ and $y-$axes and positive $z-$axis. Find the coordinates of the other vertices of the cube:
- A
$(1, 0, 1)$
- ✓
$(0, -1, 0)$
- C
$(0, 0, -1)$
- D
AnswerCorrect option: B. $(0, -1, 0)$
Consider the problem Below, are four complete cube face on $XZ-$plane, $(y = 0)$
Given, point $(1, 0, -1)$
End of the edge parallel to negative $x-$axis $(0, 0 - 1)$ Origin $(0, 0, 0)$
End of the edge parallel to positive $z-$axis $(1, 0, 0)$
And, below Four point complete the opposite face of cube.
consider $P,$ end of edge parallel to negative $y-$axis $(1, -1, -1)$
Edge from $P$ parallel to positive $z-$axis $(0, -1, 0)$
Edge from $P$ parallel to negative $x-$axis $(0, -1, -1)$ And $(0, -1, 0)$
View full question & answer→MCQ 461 Mark
Coordinate planes divide the space into octants:
AnswerThe coordinate planes divide the three dimensional space into eight octants.
View full question & answer→MCQ 471 Mark
The graph of the equation $y^2+ z^2= 0$ in three dimensional space is:
- ✓
$x-$axis
- B
$y-$axis
- C
$z-$axis
- D
AnswerCorrect option: A. $x-$axis
Consider the problem
$y^2 + z^2= 0$
$x = 0$ and $z = 0$
$\therefore$ The graph of the equation
$y^2+ z^2= 0$ is $x-$axis
View full question & answer→MCQ 481 Mark
If $(0, b, 0)$ is the centroid of the triangle formed by the points $(4, 2, -3) (a, -5, 1)$ and $(2, -6, 2)$ If $a, b$ are the roots of the quadratic equation $x^2+ px + q = 0$, then $p, q$ are:
- ✓
$9, 18$
- B
$-9, -18$
- C
$3, -18$
- D
AnswerCorrect option: A. $9, 18$
Since $a, b$ are the roots of the equation.
$x^2+ px + q = 0$
$\Rightarrow a + b = -p$ and $ab = q$ Centroid of triangle is $\Big(\frac{\text{a}+6}{3},-3,0\Big)$
Given, centroid $(0, b, 0)$ Comparing, we get $b = -3$ and
$\frac{\text{a}+6}{3}=0\Rightarrow\text{a}=-6$
$p = 9, q = 18$
View full question & answer→MCQ 491 Mark
Name three undefined terms:
AnswerThe basic undefined term is point. Line is formed from points and plane is formed from many lines. Undefined terms are point, line and plane.
View full question & answer→MCQ 501 Mark
The coordinate of any point, which lies in $xy$ plane, is:
- A
$(x, 0, y)$
- ✓
$(x, x, 0)$
- C
$(x, 0, x)$
- D
AnswerCorrect option: B. $(x, x, 0)$
Given, that the point lies in $xy$ plane In $xy$ plane, the coordinate of $z$ will be zero so $(x, x, 0)$ represents a point which lies in $xy$ plane.
View full question & answer→