MCQ 1011 Mark
The points $(3, 2, 0), (5, 3, 2)$ and $(-9, 6, -3),$ are the vertices of a triangle $\text{ABC.AD}$ 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. $\Big[\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big]$
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:
- A
$(-8, -17, 3)$
- ✓
$(-8, 17, 3)$
- C
$(8, -17, 3)$
- D
AnswerCorrect option: B. $(-8, 17, 3)$
View full question & answer→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. $\frac{-38}{5}$
$P$ 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:
- ✓
$(3, 0, 0).$
- B
$(0, 4, 0).$
- C
$(0, 0, 5).$
- D
AnswerCorrect option: A. $(3, 0, 0).$
On 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
- ✓
$(6, 0, 0)$
- B
$(0, 7, 0)$
- C
$(0, 0, 8)$
- D
$(0, 7, 8)$
AnswerCorrect option: A. $(6, 0, 0)$
We 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:
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 $-2 : 3$
The ratio that $xy-$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 $xy-$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:
- ✓
$x-$axis.
- B
$y-$axis.
- C
$z-$axis.
- D
AnswerCorrect option: A. $x-$axis.
Any 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:
- ✓
$(-11, 3, 8)$
- B
$(-11, 3, -8)$
- C
$(-8, 12, 20)$
- D
AnswerCorrect option: A. $(-11, 3, 8)$
View full question & answer→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:
View full question & answer→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
$\Rightarrow 10x - 5y + 20 = 12$ 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:
- ✓
$3$ units
- B
$4$ units
- C
$5$ units
- D
$550$ units
AnswerCorrect option: A. $3$ units
View full question & answer→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:
AnswerCorrect option: C. $(1, -2, 3)$
The 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
$1$
- 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:
View full question & answer→MCQ 1191 Mark
A plane intersects the co ordinate axes at $\text{A, B, C.}$ If $O = (0, 0, 0)$ and $(1, 1, 1)$ is the centroid of the tetrahedron $\text{OABC,}$ 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 $x$ axis
- ✓
A plane containing $z$ axis
- D
AnswerCorrect option: C. 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:
- A
$3 : 4$
- B
$2 : 1$
- ✓
$7 : 1$
- D
AnswerCorrect option: C. $7 : 1$
Let 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:
- ✓
$(2, 0, -1)$
- B
$(1, 0, -1)$
- C
$(1, -2, 0)$
- D
AnswerCorrect option: A. $(2, 0, -1)$
View full question & answer→MCQ 1231 Mark
Locus of a point $P$ which such that $\text{PA = PB}$ where $A = (0, 3, 2)$ and $B = (2, 4, 1)$ is:
- ✓
$2x + y - z = 4$
- B
$x - 2y + z + 1 = 0$
- C
$9x - 2y + 4z - 5 = 0$
- D
AnswerCorrect option: A. $2x + y - z = 4$
View full question & answer→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 $\text{PF = FI}$
$\therefore$ If $(x, y, z) = (r + 1, -r + 2, r + 3)$ are foot of perpendicular.
$\Rightarrow (r + 1) - (-r + 2) + r + 3 = 5$
$\Rightarrow r = 1$
$\therefore F = (2, 1, 4)$
$\therefore I = (3, 0, 5)$
$\therefore$ 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 $\text{XOZ-}$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
AnswerCorrect option: A. $\Big(\frac{13}{5},0,-2\Big)$
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:
View full question & answer→MCQ 1291 Mark
In the tetrahedron $\text{ABCD, A} = (1, 2, -3)$ 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
AnswerCorrect option: A. $\frac{8\sqrt{21}}{3}$
Given, $A = (1, 2, -3), G (-3, 4, 5)$
$\therefore\text{AG}=\sqrt{(-3-1)^2+(4-2)^2+(5-(-3))^2}$
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 $\text{AP = AG + GP}$
$\Rightarrow AP = 3x + x$
$\Rightarrow AP = 4x$
$\Rightarrow\text{AP}=4\Big(\frac{2\sqrt2}{3}\Big)=\frac{8\sqrt21}{3}$
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 $(3, -5, 8):$
- A
Externally $2 : 3$
- ✓
Internally $2 : 3$
- C
Internally $3 : 2$
- D
Externally $3 : 2$
AnswerCorrect option: B. Internally $2 : 3$
View full question & answer→MCQ 1311 Mark
Find the coordinates of the points which trisect the line segment joining the points $P(4, 2, -6)$ and $Q(10, -16, 6):$
- A
$(6, -4, -2), (8, -10, 2)$
- ✓
$(6, 4, -2), (8, -10, 2)$
- C
$(6, -4, -2), (8, 10, 2)$
- D
AnswerCorrect option: B. $(6, 4, -2), (8, -10, 2)$
View full question & answer→MCQ 1321 Mark
The centroid of triangle $A(3, 4, 5), B(6, 7, 2), C(0, -5, 2)$ is:
- ✓
$(3, 2, 3)$
- B
$(5, 2, 1)$
- C
$(2, 5, 1)$
- D
$(3, 4, 1)$
AnswerCorrect option: A. $(3, 2, 3)$
$A (3, 4, 5), B (6, 7, 2), C (0, -5, 2)$ 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. $\frac{1}{3},\frac{1}{6},1$
View full question & answer→MCQ 1341 Mark
What is the locus of a point for which $y = 0, z = 0?$
- ✓
$x-$axis
- B
$y-$axis
- C
$z-$axis
- D
$yz-$plane
AnswerCorrect option: A. $x-$axis
We 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:
AnswerCorrect option: B. $(2, 2, -1)$
The 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:
- A
$2 : 5$
- ✓
$3 : 5$
- C
$5 : 2$
- D
$5 : 3$
AnswerCorrect option: B. $3 : 5$
View full question & answer→MCQ 1371 Mark
If $x^2+ y^2= 1$, then the distance from the point $(x, y, 1 - x^2- y^2)$ 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:
- A
$x = 0, y = 0.$
- B
$y = 0, z = 0.$
- ✓
$z = 0, x = 0.$
- D
AnswerCorrect option: C. $z = 0, x = 0.$
On $y-$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:
- ✓
Equation of $x-$axis
- B
Equation of $y-$axis
- C
Equation of $z-$axis
- D
AnswerCorrect option: A. Equation of $x-$axis
View full question & answer→MCQ 1421 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 1431 Mark
The triangle formed by the points $(0, 7, 10), (-1, 6, 6), (-4, 9, 6)$ is:
View full question & answer→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 $F(-1, 1, -4):$
- A
$(1, 1, 2)$
- ✓
$(1, 1, -2)$
- C
$(-1, -1, -2)$
- D
$(1, -1, -2)$
AnswerCorrect option: B. $(1, 1, -2)$
View full question & answer→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:
View full question & answer→MCQ 1471 Mark
A plane is parallel $xy-$plane, so it is perpendicular to:
- ✓
$z-$axis
- B
$y-$axis
- C
$x-$axis
- D
AnswerCorrect option: A. $z-$axis
View full question & answer→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 $\text{A, B, C.}$ 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,
$\text{AB}=\sqrt{4+1+4}=3$
$\text{AC}=\sqrt{144+16+9}=13$
$\text{BD}:\text{DC}=\text{AB}:\text{AC}=3:13$
$\text{D}=\Big(\frac{3(-9)+13(15)}{3+13},\frac{3(6)+13(3)}{3+13},\frac{3(-3)+13(2)}{3+13}\Big)$
$=\Big(\frac{-38}{16},\frac{57}{16},\frac{17}{16}\Big)$
$\therefore\Big(-\frac {19}{8}, \frac {57}{16}, \frac {17}{16}\Big)$
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