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:
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 2AB = BC then C =
AnswerLet 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
- 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:
AnswerLet 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:
Answer
- Cylinder
Solution:
The 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:
AnswerIn 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 A, B, C then the centroid of $\Delta\text{ABC}$ is:
View full question & answer→MCQ 111 Mark
G(1, 1, -2) is the centroid of the triangle ABC and D is the mid point of 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:
AnswerLet 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
XOZ plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio:
AnswerLet 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:
AnswerUsing 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
- 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:
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
$\overrightarrow {\text{G}}=\overrightarrow {\text{a}}+ \overrightarrow {\text{b}}+\overrightarrow{\text{c}}$
- B
$\overrightarrow {\text{G}} = \frac {\overrightarrow{\text{a}} + \overrightarrow {\text{b}} +\overrightarrow{\text{c}}}{2}$
- ✓
$3\overrightarrow {\text{G}}=\overrightarrow {\text{a}}+ \overrightarrow {\text{b}}+\overrightarrow {\text{c}}$
- D
$3\overrightarrow {\text{G}}=\frac{\overrightarrow {\text{a}}+ \overrightarrow {\text{b}}+\overrightarrow {\text{c}}}{2}$
AnswerCorrect option: C. $3\overrightarrow {\text{G}}=\overrightarrow {\text{a}}+ \overrightarrow {\text{b}}+\overrightarrow {\text{c}}$
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:
AnswerGiven, 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 ABCD are A(1, 2, 3), B(-1, -2, -1) and C(2, 3, 2). Find the fourth vertex D:
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)
- (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)
- 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:
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):
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:
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:
MCQ 301 Mark
Choose the correct answer. $x-$axis is the intersection of two planes:
- A
$xy$ and $xz.$
- B
$yz$ and $zx.$
- ✓
$xy$ and $yz.$
- D
AnswerCorrect option: C. $xy$ and $yz.$
$xy$ and $xz.$
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
- 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:
AnswerOn 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. $9 x^2+25 y^2+25 z^2-225=0$
- $9 x^2+25 y^2+25 z^2-225=0$
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:
AnswerR.E.F image
Given, place x = 0 and two points
⇒ (-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
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
AnswerLet 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:
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:
MCQ 421 Mark
The coordinates of any point, which lies in yz plane, are:
AnswerIn 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:
AnswerGiven, (-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:
AnswerLet 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:
AnswerConsider 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:
Answer
- x-axis
Solution:
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 $x^2+p x+q=0$, then $p$, $q$ are:
- ✓
$9, 18$
- B
$-9,-18$
- C
$3,-18$
- D
$-3,18$
AnswerCorrect option: A. $9, 18$
- $9, 18$
Solution:
Since $a, b$ are the roots of the equation.
$x^2+p x+q=0$
$\Rightarrow \mathrm{a}+\mathrm{b}=-\mathrm{p}$ and $\mathrm{ab}=\mathrm{q}$ Centroid of triangle is $\left(\frac{\mathrm{a}+6}{3},-3,0\right)$
Given, centroid $(0, b, 0)$ Comparing, we get $b=-3$ and
$\frac{a+6}{3}=0 \Rightarrow a=-6$
$p=9, q=18$ 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:
AnswerGiven, 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→MCQ 511 Mark
The equation of plane passing through (-1, 0, -1) parallel to xz plane is:
AnswerGiven, 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:
MCQ 531 Mark
The image of the point P(1, 3, 4) in the plane 2x - y + z = 0 is:
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:
AnswerLet 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 XOZ divides the join of (1, -1, 5) and (2, 3, 4) in the ratio $\lambda:1$ then $\lambda$ is:
- A
- B
$\frac{1}{4}$
- C
- ✓
$\frac{1}{3}$
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 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 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:
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:
MCQ 611 Mark
Choose the correct answer. The distance of point P(3, 4, 5) from the yz-plane is:
AnswerGiven point is P(3, 4, 5)
$\therefore$ Distance of from yz-plane
$=\sqrt{(0-3)^2+(4-4)^2+(5-5)^2}$
$=\sqrt{9}=3\text{ units}$
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)
$\therefore\sqrt{(3-0)^2+(4-0)^2+(5-0)^2}$
$=\sqrt{9+16+25}=\sqrt{50}$
Hence, the correct is a.
View full question & answer→MCQ 631 Mark
Distance between the points (12, 4, 7) and (10, 5, 3) is:
- ✓
$\sqrt{27}$
- B
$\sqrt{5}$
- C
$\sqrt{17}$
- D
AnswerCorrect option: A. $\sqrt{27}$
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.
$\text{AB}=\sqrt{(10-12)^2+(5-4)^2+(3-7)^2}$
$=\sqrt{(-2)^2+1^2+(-4)^2}$
$=\sqrt{4+1+6}=\sqrt{21}$
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. ax + by + cz + (-3) = 0 meet the co-ordinate axes in 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. $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\Big)$
- $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\Big)$
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:
- ✓
$(3, 3, 4)$
- B
$\Big(\frac{3}{2},\frac{7}{2},\frac{9}{2}\Big)$
- C
$(9, 9, 12)$
- D
$\Big(\frac{9}{2}\frac{-3}{2},\frac{3}{2}\Big)$
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
AnswerLet 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 A, B, C respectively. If $\text{G}\Big(\frac{1}{6},-\frac{1}{3},{1}\Big)$ is a centroid of $\triangle\text{ABC}$ then 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 ⇒ a = 2
3b = -3 ⇒ 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:
Answera - 2b + 3c
$\frac{\text{3a}+4\text{b}-5\text{c}}{3}$
$\overrightarrow{\text{c}}=\frac{2\overrightarrow{\text{b}}+\overrightarrow{\text{a}}}{3}$
$\overrightarrow{\text{b}}=\frac{3\overrightarrow{\text{c}}-\overrightarrow{\text{a}}}{2}$
$=\frac{\big(3\overrightarrow{a}+4\overrightarrow{b}-5\overrightarrow{c}\big)-\big(\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}\big)}{2}$
$=\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$
View full question & answer→MCQ 711 Mark
The plane XOZ divides the join of (1, -1, 5) and (2, 3, 4) in the ratio $\lambda:1$ then $\lambda$ is:
- A
- B
$\frac{-1}{3}$
- C
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
The plane 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:
AnswerLet C divide AB in the ratio x : y
Let us compare the x-coordinate of C by using section formula.
$-1=\frac{\text{x}\times1+\text{y}\times3}{\text{x+y}}$
⇒ x + 3y = -x - y
⇒ 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 $A B$, then $C_y=$
- A
$\frac{11}{3}$
- B
$-11$
- C
$\frac{10}{3}$
- ✓
$\frac{-11}{3}$
AnswerCorrect option: D. $\frac{-11}{3}$
- $\frac{-11}{3}$
Solution:
Given, 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 $C_y = \frac{-11}{3}\text{ or}-\frac{10}{3}$ View full question & answer→MCQ 741 Mark
In geometry, we take a point, a line and a plane as undefined terms:
Answer
- True
Solution:
In 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:
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:
Answer(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:
- ✓
A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus
- B
Value of y in a zx plane is non-zero
- C
Value of z in a xy plane is zero
- D
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:
Answer
- 3 : 2
Solution:
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:
MCQ 851 Mark
The point A(1, -1, 3), B(2, -4, 5) and C(5, -13, 11) are:
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
- B
$\frac{3}{5}$
- C
$\frac{2}{5}$
- ✓
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 1st quadrant has all positive points, 2nd 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. $\frac{3\sqrt21}{2}$
- $\frac{3\sqrt21}{2}$
Solution:
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:
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. $\frac{\pi}{2}$
- $\frac{\pi}{2}$
Solution:
$\mathrm{PQ}^2=(4-0)^2+(-2-1)^2+(1-2)^2=16+9+1=26$
$\mathrm{OP}^2=(0-0)^2+(1-0)^2+(2-0)^2=0+1+4=5$
$\mathrm{QO}^2=(0-4)^2+(0+2)^2+(0-1)^2=16+1+4=21$
Since, $\mathrm{PQ}^2=\mathrm{OP}^2+\mathrm{QO}^2$
Hence, $\angle \mathrm{POQ}=\frac{\pi}{2}$ View full question & answer→MCQ 921 Mark
The position vectors of the four angular point of a tetrahedron 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
- D
AnswerCorrect option: B. $\Big(\frac{6}{4},1,\frac{2}{4}\Big)$
- $\Big(\frac{6}{4},1,\frac{2}{4}\Big)$
Solution:
Angular points of tetrahedron 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
$(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_{4})$ 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 $x^2+y^2+z^2+2 u x+2 v y+2 w z+d$ represent a real sphere:
- A
$u^2+v^2+w^2=d^2$
- ✓
$u^2+v^2+w^2>d$
- C
$u^2+v^2+w^2<d$
- D
$u^2+v^2+w^2<d^2$
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. $x^2+y^2+z^2=r^2$
- $x^2+y^2+z^2=r^2$
Solution:
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:
AnswerAt 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.
View full question & answer→MCQ 981 Mark
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.
View full question & answer→MCQ 991 Mark
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
AnswerWe 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).
View full question & answer→MCQ 1011 Mark
The points (3, 2, 0), (5, 3, 2) and (-9, 6, -3), are the vertices of a triangle ABC.AD is the internal bisector of $\angle\text{BAC}$ which meets BC at D. Then the co-ordinates of D, are:
- A
$\Big[\frac{17}{16},\frac{57}{16},\frac{19}{8}\Big]$
- ✓
$\Big[\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big]$
- C
$\Big[0,0,\frac{17}{16}\Big]$
- D
$\Big[\frac{17}{16},0,0\Big]$
AnswerCorrect option: B. $\Big[\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big]$
- $\Big[\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big]$
View full question & answer→MCQ 1021 Mark
Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio 2 : 3 externally:
MCQ 1031 Mark
A = (1, -1, 2) and B = (2, 3, 7) are two points. lf P, O divide AB in the ratios 2 : 3, -2 : 3 respectively then $P_x+Q_y=$
- ✓
$\frac{-38}{5}$
- B
$\frac{38}{5}$
- C
$\frac{-2}{5}$
- D
$\frac{-47}{6}$
AnswerCorrect option: A. $\frac{-38}{5}$
- $\frac{-38}{5}$
Solution:
P divides line joining A (1, -1, 2) and B (2, 3, 7) in the ratio 2 : 3
$\therefore\text{P}_{\text{x}}=\frac{2\times2+3\times1}{2+3}=\frac{7}{5}$ Similarly, Q divides line joining A (1, -1, 2) and B (2, 3, 7) in the ratio -2 : 3
$\therefore\text{Q}_{\text{y}}=\frac{-2\times3+3\times-1}{-2+3}=-9$
$\Rightarrow\text{P}_{\text{x}}+\text{Q}_{\text{y}}=-9+\frac{7}{5}=\frac{-38}{5}$ View full question & answer→MCQ 1041 Mark
Choose the correct answer. L is the foot of the perpendicular drawn from a point (3, 4, 5) on x-axis. The coordinates of L are:
AnswerOn the x-axis, y = 0 and z = 0.
Hence, the required coordinates are (3, 0, 0).
View full question & answer→MCQ 1051 Mark
The coordinates of the foot of the perpendicular from a point P(6, 7, 8) on x-axis are
AnswerWe know that the y and z coordinates on x-axis are 0
The coordinates of the foot of the perpendicular from a point P(6, 7, 8) on x-axis are (6, 0, 0)
Hence, the correct answer is option (a).
View full question & answer→MCQ 1061 Mark
If point p lies in first octant, then the sign of x- coordinate will always be:
- ✓
- B
- C
- D
x coordinate can be + or -
AnswerIn the first octant, the values of x, y and z axes are positive.
Any point which lies in first octant has all their coordinate values as positive.
Since point p lies in first octant, so, p will have all its coordinates as positive.
Hence, sign of x-coordinate will always be +.
View full question & answer→MCQ 1071 Mark
The xy-plane divides the line joining the points (-1, 3, 4) and (2, -5, 6):
- A
Internally in the ratio 2 : 3
- ✓
Externally in the ratio 2 : 3
- C
Internally in the ratio 3 : 2
- D
Externally in the ratio 3 : 2
AnswerCorrect option: B. Externally in the ratio 2 : 3
- Externally in the ratio 2 : 3
Solution:
The ratio that $x y$-plane divides the line joining the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)=-z_1: z_2$
IF the result is positive, it divides internally otherwise externally The ratio that $x y$-plane divides the line joining the points $(-1,3,4)$ and $(2,-5,6)=-4: 6=-2: 3$ View full question & answer→MCQ 1081 Mark
Choose the correct answer. A plane is parallel to yz-plane so it is perpendicular to:
AnswerAny plane parallel to yz-plane, so it is perpendicular to x-axis.
Hence, the correct option is (a)
View full question & answer→MCQ 1091 Mark
If the line joining A(1, 3, 4) and B is divided by the point (-2, 3, 5) in the ratio 1 : 3, then B is:
MCQ 1101 Mark
The points A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are vertices of a:
MCQ 1111 Mark
A = (2, 4, 5) and B = (3, 5, -4) are two points. lf the XY-plane, YZ-plane divide AB in the ratio a : b and p : q respectively, then $\frac{\text{a}}{\text{b}}+\frac{\text{p}}{\text{q}}=$
- A
$\frac{23}{12}$
- B
$\frac{-7}{12}$
- ✓
$\frac{7}{12}$
- D
$\frac{-22}{15}$
AnswerCorrect option: C. $\frac{7}{12}$
View full question & answer→MCQ 1121 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 1131 Mark
The planes 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 are:
AnswerPlanes are 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6
Multiply both sides by 2 to the second equation
⇒ 10x - 5y + 20 = 12 Now divide both sides by 2
$\Rightarrow2\text{x - y + 4z}=\frac{12}{5}$
Clearly both planes are parallel.
View full question & answer→MCQ 1141 Mark
The distance of point P(3, 4, 5) from the yz-plane is:
MCQ 1151 Mark
Four vertices of a tetrahedron are (0, 0, 0), (4, 0, 0), (0, -8, 0) and (0, 0, 12) Its centroid has the coordinates:
AnswerThe centroid of the coordinates is
$\Big(\frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}+\text{x}_{4}}{4},\frac{\text{y}_{1}+\text{y}_{2}+\text{y}_{3}+\text{y}_{4}}{4},\frac{\text{z}_{1}+\text{z}_{2}+\text{z}_{3}+\text{z}_{4}}{4}\Big)$
Thus by substituting the vertices we get
$=\Big(\frac{0+4+0+0}{4},\frac{0+0-8+0}{4},\frac{0+0+0+12}{4}\Big)$
$=\Big(\frac{4}{4},\frac{-8}{4},\frac{12}{4}\Big)$
$\therefore$ The centroid of the coordinates is (1, -2, 3)
View full question & answer→MCQ 1161 Mark
Find the distance between the points whose position vectors are given as follows: $4\hat{\text{i}} + 3\hat{\text{j}} - 6\hat{\text{k}},-2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
- ✓
$ \sqrt{65}$
- B
$ \sqrt{69}$
- C
- D
AnswerCorrect option: A. $ \sqrt{65}$
View full question & answer→MCQ 1171 Mark
Let (3, 4, -1) and (-1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to
Answer$\text{d}^2=(−1−3)^2+(2−4)^2+(3+1)^2$
$\Rightarrow\text{d}^2=(−4)^2+(−2)^2+(4)^2$
$\Rightarrow\text{d}^2=16+4+16$
$\Rightarrow\text{d}^2=36$
$\Rightarrow\text{d}^2=6$
Hence, radius of the sphere is 3 units.
View full question & answer→MCQ 1181 Mark
Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear:
MCQ 1191 Mark
A plane intersects the co ordinate axes at A, B, C. If O = (0, 0, 0) and (1, 1, 1) is the centroid of the tetrahedron OABC, then the sum of the reciprocals of the intercepts of the plane:
- A
$12$
- B
$\frac{4}{3}$
- C
$1$
- ✓
$\frac{3}{4}$
AnswerCorrect option: D. $\frac{3}{4}$
Let the point of intersections be,
A (a, 0, 0), B (0, b, 0) and C (0, 0, c)
The coordinates of the centroid are $\Big(\frac{\text{a}}{4},\frac{\text{b}}{4},\frac{\text{c}}{4}\Big)$
Comparing it with the coordinates given, we get
a = 4, b = 4, c = 4
$\Big(\frac{1}{\text{a}}, \frac{1}{\text{b}},\frac{1}{\text{c}}\Big)=\frac{3}{4}$
View full question & answer→MCQ 1201 Mark
In a three dimensional space, the equation 3x - 4y = 0 represents:
- A
A plane containing y axis
- B
- ✓
A plane containing z axis
- D
A plane containing x axis
AnswerCorrect option: C. A plane containing z axis
- A plane containing z axis
View full question & answer→MCQ 1211 Mark
The ratio in which the line joining (3, 4, -7) and (4, 2, 1) is dividing the xy-plane:
AnswerLet the given points be
A (3, 4, -7), B (4, 2, 1)
Let a point on XY-plane be P (x, y, 0) and the line AB in the ratio k : 1
then by section formula
$0=\frac{\text{k}\times1+1\times-7}{\text{k}+1}$
k - 7 = 0
k = 7
$\therefore$ ratio is 7 : 1
View full question & answer→MCQ 1221 Mark
If (1, -1, 0), (-2, 1, 8) and (-1, 2, 7) are three consecutive vertices of a parallelogram then the fourth vertex is:
MCQ 1231 Mark
Locus of a point P which such that PA = PB where A = (0, 3, 2) and B = (2, 4, 1) is:
MCQ 1241 Mark
The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is:
- A
$\sqrt{17}$
- B
$\sqrt{29}$
- ✓
$\sqrt{34}$
- D
$\sqrt{41}$
AnswerCorrect option: C. $\sqrt{34}$
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.
⇒ (r + 1) - (-r + 2) + r + 3 = 5 ⇒ r = 1
$\therefore$ F = (2, 1, 4)
$∴$ I = (3, 0, 5)
$∴$ Distance of I from origin $=\sqrt{3^2+0^2+5^2}=\sqrt{34}$
View full question & answer→MCQ 1251 Mark
If the plane a 2x - 3y + 5Z - 2 = 0 divides the line segment joining (1, 2, 3) and (2, 1, k) in the ratio 9 : 11, then k is:
AnswerCoordinate of the point which divides the line segment joining the points
(1, 2, 3) and (2, 1, k) in the ratio 9 : 11 are $\Big(\frac{29}{20},\frac{31}{20},\frac{9\text{k}+33}{20}\Big)$
Also, this point will lie on the given plane
$\Rightarrow2\times\frac{29}{20}-3\times\frac{31}{20}+5\times\frac{9\text{k}+33}{20}-2=0$
$\Rightarrow\text{k}=-2$
View full question & answer→MCQ 1261 Mark
A point on XOZ-plane divides the join of (5, -3, -2) and (1, 2, -2) at:
- ✓
$\Big(\frac{13}{5},0,-2\Big)$
- B
$\Big(\frac{13}{5},0,2\Big)$
- C
$\Big(5, 0, 2\Big)$
- D
$\Big(5, 0, -2\Big)$
AnswerCorrect option: A. $\Big(\frac{13}{5},0,-2\Big)$
- $\Big(\frac{13}{5},0,-2\Big)$
View full question & answer→MCQ 1271 Mark
If A = (1, 2, 3), B = (2, 3, 4) and C is a point of trisection of AB such that $\text{C}_{\text{x}} + \text{C}_{\text{y}} = \frac{13}{3}$ then $\text{C}_\text{z}=$
- A
$\frac{10}{3}$
- ✓
$\frac{11}{3}$
- C
$\frac{11}{2}$
- D
$11$
AnswerCorrect option: B. $\frac{11}{3}$
View full question & answer→MCQ 1281 Mark
The points (5, -4, 2), (4, -3, 1), (7, -6, 4) and (8, -7, 5) are the vertices of:
MCQ 1291 Mark
In the tetrahedron ABCD, A = (1, 2, -3) and G (-3, 4, 5) is the centroid of the tetrahedron. If P is the centroid of the $\Delta\text{BCD}$ then AP =
- ✓
$\frac{8\sqrt{21}}{3}$
- B
$ \frac{4\sqrt{21}}{3}$
- C
$4\sqrt{21}$
- D
$\frac{\sqrt{21}}{3}$
AnswerCorrect option: A. $\frac{8\sqrt{21}}{3}$
Given, A = (1, 2, -3), G (-3, 4, 5)
$\therefore\text{AG}=\sqrt{(-3-1)^2+(4-2)^2+(5-(-3))^2}$
$\text{and AG}=\sqrt { 84 } =2\sqrt { 21 }$
P is the centroid of $\triangle\text{BCD}$
So, G divides AP in 3 : 1
Let AG = 3x, then, GP = x
$\text{3x}=2\sqrt{21}$
$\text{x}=\frac{2\sqrt2}{3}$
Now AP = AG + GP
⇒ AP = 3x + x
⇒ AP = 4x
$\Rightarrow\text{AP}=4\Big(\frac{2\sqrt2}{3}\Big)=\frac{8\sqrt21}{3}$
View full question & answer→MCQ 1301 Mark
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8):
MCQ 1311 Mark
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6):
MCQ 1321 Mark
The centroid of triangle A(3, 4, 5), B(6, 7, 2), C(0, -5, 2) is:
AnswerA (3, 4, 5), B (6, 7, 2), C (0, -5, 2) Centroid is given as.
$\Big(\frac{3+6+0}{3},\frac{4+7-5}{3},\frac{5+2+2}{3}\Big)$
$=\Big(\frac{9}{3},\frac{6}{3},\frac{9}{3}\Big)=(3, 2, 3)$
View full question & answer→MCQ 1331 Mark
The cartesian equation of the line is 3x + 1 = 6y - 2 = 1 - z then its direction ratio are:
- ✓
$\frac{1}{3},\frac{1}{6},1$
- B
$\frac{-1}{3},\frac{1}{6},1$
- C
$\frac{1}{3},\frac{-1}{6},1$
- D
$\frac{1}{3},\frac{1}{6},-1$
AnswerCorrect option: A. $\frac{1}{3},\frac{1}{6},1$
- $\frac{1}{3},\frac{1}{6},1$
View full question & answer→MCQ 1341 Mark
What is the locus of a point for which y = 0, z = 0?
AnswerWe know that on x-axis both y = 0, z = 0.
Hence, the correct answer is option (a)
View full question & answer→MCQ 1351 Mark
If the vertices of a triangle are (-1, 6, -4), (2, 1, 1) and (5, -1, 0) then the centroid of the triangle is:
AnswerThe centroid of the triangle is
$\Big(\frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}}{3},\frac{\text{y}_{1}+\text{y}_{2}+\text{y}_{3}}{3},\frac{\text{z}_{1}+\text{z}_{2}+\text{z}_{3}}{3}\Big)$
Thus by substituting the vertices we get =
$\Big(\frac{-1+2+5}{6},\frac{6+1-1}{3},\frac{-4+1+0}{3}\Big)=\Big(\frac{6}{3},\frac{6}{3},\frac{-3}{3}\Big)$
$\therefore$ The centroid of the triangle is (2, 2, -1)
View full question & answer→MCQ 1361 Mark
The ratio in which the line joining the points (1, 2, 3) and (-3, 4, -5) is divided by the xy-plane is:
MCQ 1371 Mark
If $x^2+y^2=1$, then the distance from the point $\left(x, y, 1-x^2-y^2\right)$ to the origin is:
View full question & answer→MCQ 1381 Mark
The points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) are the vertices of
AnswerSuppose:
A(5, -4, 2)
B(4, -3, 1)
C(7, 6, 4)
D(8, -7, 5)
$\text{AB}=\sqrt{(4 − 5)^2 + (−3 + 4)^2 + (1 − 2)^2}$
$=\sqrt{(−1)^2 + (1)^2 + (−1)^2}$
$=\sqrt{1 + 1 + 1}=\sqrt{3}$
$\text{BC}=\sqrt{(7 − 4)^2 + (6 + 3)^2 + (4 − 1)^2}$
$=\sqrt{(3)^2 + (9)^2 + (3)^2}$
$=\sqrt{9 + 81 + 9}=\sqrt{99}=3\sqrt{11}$
$\text{CD}=\sqrt{(8 − 7)^2 + (−7 − 6)^2 + (5 − 4)^2}$
$=\sqrt{(1)^2 + (-13)^2 + (1)^2}$
$=\sqrt{1 + 169 + 1}=\sqrt{171}$
$\text{DA}=\sqrt{(8 − 5)^2 + (−7 + 4)^2 + (5 − 2)^2}$
$=\sqrt{(3)^2 + (-3)^2 + (3)^2}$
$=\sqrt{9 + 9 + 9}=\sqrt{27}=3\sqrt{3}$
We see that none of the sides are equal.
View full question & answer→MCQ 1391 Mark
The maximum distance between points $ (3\sin \theta, 0, 0)$ and $(4\cos \theta, 0, 0)$ is:
View full question & answer→MCQ 1401 Mark
Choose the correct answer. Equation of y-axis is considered as:
AnswerOn y-axis, x = 0 and z = 0
Hence, the correct option is (c).
View full question & answer→MCQ 1411 Mark
The locus of a point for which y = 0, z = 0 is:
MCQ 1421 Mark
The image of the point P(1,3,4) in the plane 2x - y + z = 0 is:
MCQ 1431 Mark
The triangle formed by the points (0, 7, 10), (-1, 6, 6), (-4, 9, 6) is:
MCQ 1441 Mark
Find the centroid of a triangle, the mid-point of whose sides are D(1, 2, -3), E(3, 0, 1) and F(-1, 1, -4):
MCQ 1451 Mark
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. $2\sqrt{3}$
View full question & answer→MCQ 1461 Mark
The points A(2a, 4a), B(2a, 6a) and C(2a + 3a, 5a), a > 0 are the vertices of:
MCQ 1471 Mark
A plane is parallel xy-plane, so it is perpendicular to:
MCQ 1481 Mark
The length of the perpendicular drawn from the point P(a, b, c) from z-axis is
- ✓
$\sqrt{\text{a}^2+\text{b}^2}$
- B
$\sqrt{\text{b}^2+\text{c}^2}$
- C
$\sqrt{\text{a}^2+\text{c}^2}$
- D
$\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
AnswerCorrect option: A. $\sqrt{\text{a}^2+\text{b}^2}$
The length of the perpendicular drawn from the point P(x, y, z) from z-axis is given by $\sqrt{\text{y}^2+\text{x}^2}$
Thus, the length of the perpendicular drawn from the point P(a, b, c) from z-axis is $\sqrt{\text{a}^2+\text{b}^2}$
Hence, the correct answer is option (a)
View full question & answer→MCQ 1491 Mark
The plane ax + by + cz + (-3) = 0 meet the co-ordinate axes in A, B, C. Then centroid of the triangle is:
- A
$\big(3\text{a},3\text{b},3\text{c}\big)$
- 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. $\Big(\frac{1}{\text{a}}, \frac{1}{\text{b}},\frac{1}{\text{c}}\Big)$
For finding the coordinates of the point where the plane ax + by + cz - 3 = 0 cuts the x-axis, we equate y and z to zero.
The point becomes $\Big(\frac{3}{\text{a}},0,0\Big)$ Similarly, the point on y-axis becomes $\Big(0,\frac{3}{\text{b}},0\Big)$ And that on z axis becomes $\Big(0,0,\frac{3}{\text{c}}\Big)$ The centroid of the triangle.
formed by these points would be $\Bigg(\frac{\frac{3}{\text{a}}+0+0}{3},\frac{0+\frac{3}{\text{b}}+0}{3},\frac{0+0+\frac{3}{\text{c}}}{3}\Bigg)=\bigg(\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\bigg)$
View full question & answer→MCQ 1501 Mark
A(3, 2, 0), B(5, 3, 2), C(-9, 6, -3) are three points forming a triangle. If AD, the bisector of $\angle\text{BAC}$ meets BC in D then coordinates of D are:
- ✓
$\Big(-\frac {19}{8}, \frac {57}{16}, \frac {17}{16}\Big)$
- B
$\Big( \frac {19}{8}, -\frac {57}{16}, \frac {17}{16}\Big)$
- C
$\Big( \frac {19}{8}, \frac {57}{16}, \frac {17}{16}\Big)$
- D
AnswerCorrect option: A. $\Big(-\frac {19}{8}, \frac {57}{16}, \frac {17}{16}\Big)$
According to question,
$\text{AB}=\sqrt{4+1+4}=3$
$\text{AC}=\sqrt{144+16+9}=13$
$\text{BD}:\text{DC}=\text{AB}:\text{AC}=3:13$
$\text{D}=\Big(\frac{3(-9)+13(15)}{3+13},\frac{3(6)+13(3)}{3+13},\frac{3(-3)+13(2)}{3+13}\Big)$
$=\Big(\frac{-38}{16},\frac{57}{16},\frac{17}{16}\Big)$
$\therefore\Big(-\frac {19}{8}, \frac {57}{16}, \frac {17}{16}\Big)$
View full question & answer→MCQ 1511 Mark
Solve the following differential equation. $\frac{\text{dy}}{\text{dx}}=\text{x}-1$
- A
$y=x^2+x$
- B
$y=x^2$
- ✓
$y=x^2-x$
- D
AnswerCorrect option: C. $y=x^2-x$
- $y=x^2-x$
Solution:
Given, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}-1$
Integrating on both sides
$\int \frac{d y}{d x}=\int x-1 d x$
$y=x^2-x+c$ View full question & answer→MCQ 1521 Mark
If P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear, then R divides PQ in the ratio:
MCQ 1531 Mark
Choose the correct answer. The locus of a point for which y = 0, z = 0 is:
AnswerWe know that one equation of x-axis, y = 0, z = 0
Hence, the locus of the point is equation of x-axis.
So, the correct option is (a).
View full question & answer→MCQ 1541 Mark
In a three dimensional space the equation $x^2-5 x+6=0$ represents
Answer
- Curves.
Solution:
Since, there is only one variable in the given equation.
Also, it is quadratic equation.
Hence, It represents curves in yz plane.
View full question & answer→MCQ 1551 Mark
The points (-5, 12), (-2, -3), (9, -10), (6, 5) taken in order, form:
MCQ 1561 Mark
The ratio in which the line joining (2, 4, 5) and (3, 5, -9) is divided by the yz-plane is
AnswerLet A ≡ (2, 4, 5) and B ≡ (3, 5, 9)
Let the line joining A and B be divided by the yz-plane at point P in the ratio $\lambda:1.$
Then, we have,
$\text{P}\equiv\Big(\frac{3\lambda+2}{\lambda+1},\ \frac{5\lambda+4}{\lambda+1},\ \frac{-9\lambda+5}{\lambda+1}\Big)$
Since P lies on the yz-plane, the x-coordinate of P will be zero
$\therefore\frac{3\lambda+2}{\lambda+1}=0$
$\Rightarrow3\lambda+2=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Hence, the yz-plane divides AB in the ratio -2 : 3
View full question & answer→MCQ 1571 Mark
A = (1, 1, 4) and B = (5, -3, 4) are two points. If the points P, Q are on the line AB such that AP = PQ = QB then PQ =
- A
$2\sqrt{2}$
- B
$4$
- ✓
$\sqrt{\frac{32}{9}}$
- D
$\sqrt{2}$
AnswerCorrect option: C. $\sqrt{\frac{32}{9}}$
$\text{AB}=\sqrt{(1-5)^2+(1+3)^2+(4+4)^2}$
$\text{AB}=\sqrt{(-4)^2+4^2}$
$\text{AB}=\sqrt{32}$
$\text{AB}=3\times\text{PQ},=\frac{\sqrt{132}}{3}=\sqrt{\frac{32}{9}}$
View full question & answer→MCQ 1581 Mark
The point (-2, -3, -4) lies in the:
MCQ 1591 Mark
The ratio in which the line joining the points (1, 2, 3) and (-3, 4, -5) is divided by the xy-plane is: