MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 2

Ratio in which the $$xy-$$plane divides the join of $$(1, 2, 3)$$ and $$(4, 2, 1)$$ is

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$$3:1$$ internally
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$$3:1$$ externally
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$$1:2$$ internally
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$$2:1$$ externally

The equation of the set of points which are equidistant from the points $$(1, 2, 3)$$ and $$(3, 2, -1)$$.

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$$x-2z=0$$
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$$2x-z=0$$
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$$2x+y=0$$
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$$x-2y=0$$

$$A(3, 2, 0), B(5, 3, 2), C(-9, 6, -3)$$ are three points forming a triangle. If $$AD$$, the bisector of $$\angle BAC$$ meets $$BC$$ in $$D$$ then coordinates of $$D$$ are

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$$\left (-\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$$
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$$\left (\dfrac {19}{8}, -\dfrac {57}{16}, \dfrac {17}{16}\right )$$
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$$\left (\dfrac {19}{8}, \dfrac {57}{16}, \dfrac {17}{16}\right )$$
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None of these

$$A=\left(2,4,5\right)$$ and $$B=\left(3,5,-4\right)$$ are two points. If the $$xy$$-plane, $$yz$$-plane divide $$AB$$ in the ratios $$a:b,p:q$$ respectively then $$\dfrac{a}{b}+\dfrac{p}{q}$$=

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$$\dfrac{7}{15}$$
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$$\dfrac{-7}{12}$$
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$$\dfrac{7}{12}$$
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$$\dfrac{22}{25}$$

If $$z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$$, then

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$$\left| z \right| = 1,\arg z = \dfrac{\pi }{4}$$
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$$\left| z \right| = 1,\arg z = \dfrac{\pi }{6}$$
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$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = \dfrac{{5\pi }}{{24}}$$
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$$\left| z \right| = \dfrac{{\sqrt 3 }}{2},\arg z = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$$

The x-coordinate of a point on the line joining the points $$P(2,2,1)$$ and $$Q(5,1,-2)$$ is $$4$$. Find its z-coordinate.

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$$-1$$
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$$-2$$
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$$1$$
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$$2$$

Solve the following differential equation.

$$\dfrac{dy}{dx}=x-1$$.

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$$y=x^2+x$$
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$$y=x^2$$
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$$y=x^2-x$$
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None of the above

The distance between (5,1,3) and the line x=3, y=7+t, z=1+t is

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4
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2
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6
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8

The perimeter of triangle with vertices at $$(1,0,0) , (0,1,0) and (0,0,1)$$ is :

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$$3$$
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$$2$$
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$$2\sqrt {2}$$
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$$3\sqrt {2}$$

If the points $$A(9, 8, -10), B(3, 2, -4)$$ and $$C(5, 4, -6)$$ be collinear, then the point $$C$$ divides the line $$AB$$ in the ratio

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$$2:1$$
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$$3:1$$
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$$1:2$$
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$$-1:2$$

Algebraic sum of intercepts made by the plane x+3y-4z+6=0 on the axes is

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7
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0
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$$\frac{13}{2}$$
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$$-\frac{13}{2}$$

If $$OA$$ is equally inclined to $$OX, OY$$ and $$OZ$$ and if $$A$$ is $$\sqrt 3 $$ units from the origin then $$A$$ is

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$$\left( {3,3,3} \right)$$
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$$\left( { - 1,\,1,\, - 1} \right)$$
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$$\left( { - 1,\,1,\,1} \right)$$
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$$\left( {1,\,1,\,1} \right)$$

The distance between the points $$(-1,2,3)$$ and P is $$13$$. Then $$P=$$

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$$(2,6,-9)$$
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$$(-2,6,9)$$
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$$(2,6,9)$$
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$$(2,-6,9)$$

The distance between the parallel planes given by the equations, $$\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})+3=0$$ and $$\vec{r}.(4\hat{i}-4\hat{j}+2\hat{k})+5=0$$ is-

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$$1/2$$
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$$1/3$$
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$$1/4$$
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$$1/6$$

The number of octants in which $$Z$$ coordinate is positive is

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$$2$$
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$$3$$
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$$4$$
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$$1$$

If $$\overline { a } ,\overline { b } $$ are the position vectors of $$A$$ and $$B$$ then one of the following points lie on $$\overline { AB } $$

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$$\cfrac{2(\overline { a } +\overline { b } )}{3}$$
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$$\cfrac{(\overline { a } -\overline { b } )}{3}$$
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$$\cfrac{(\overline { a } +\overline { b } )}{3}$$
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$$\cfrac{2\overline { a } +2\overline { b } }{3}$$
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None of these

The distance between the points(4,3,7) and (1,-1,-5) is

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7
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12
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13
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25

If $$L, M$$ are the feet of the perpendiculars from $$(2, 4, 5)$$ to the $$xy$$-plane, $$yz$$-plane respectively, then the distance $$LM$$ is:

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$$\sqrt{41}$$
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$$\sqrt{20}$$
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$$\sqrt{29}$$
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$$3\sqrt{5}$$

The name of the figure formed by the points $$(0, 0, 0), (1, 0, 1)$$ and $$(0, 1, 1)$$ is

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a straight line
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an isosceles triangle
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an equilateral triangle
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a scalene triangle

The name of the figure formed by the points $$(3, -5, 1), (-1, 0, 8)$$ and $$(7, -10, -6)$$ is

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a triangle
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a straight line
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an isosceles triangle
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an equilateral triangle

If two vertices of an equilateral triangle are $$(2, 1, 5)$$ and $$(3, 2, 3)$$, then its third vertex is:

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$$(1, 2, 4)$$
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$$(4, 0, 4)$$
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$$(0, -4, 4)$$
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$$(4, 4, 1)$$

The point which is equidistant from the points $$(a, 0, 0), (0, b, 0), (0, 0, c)$$ and $$(0, 0, 0)$$ is:

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$$(a,b,c)$$
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$$(\sqrt{{a}},\sqrt{{b}},\sqrt{{c}})$$
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$$(2a,2b,2c)$$
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$$\left (\dfrac {a}{2}, \dfrac {b}{2}, \dfrac {c}{2}\right)$$

Explanation

$${\textbf{Step - 1: Writing equations using distance formula}}$$

$${\text{Let the point which is equidistant from A(a,0,0), B(0,b,0), C(0,0,c) and O(0,0,0) be P(x, y, z)}}$$

$$\therefore {\text{ AP = BP = CP = OP}}$$

$$ \Rightarrow {\text{ }}\sqrt {{{\left( {{\text{x - a}}} \right)}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\left( {{\text{y - b}}} \right)}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\left( {{\text{z - c}}} \right)}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}} $$

$${\text{Squaring all sides,}}$$

$$ \Rightarrow {\text{ }}{\left( {{\text{x - a}}} \right)^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{\left( {{\text{y - b}}} \right)^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{\left( {{\text{z - c}}} \right)^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}$$

$$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ - 2ax + }}{{\text{a}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ - 2by + }}{{\text{b}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}{\text{ - 2zc + }}{{\text{c}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}$$

$$ \Rightarrow {\text{ }}{{\text{a}}^{\text{2}}}{\text{ - 2ax = }}{{\text{b}}^{\text{2}}}{\text{ - 2by = }}{{\text{c}}^{\text{2}}}{\text{ - 2cz = 0}}$$

$${\textbf{Step - 2: Solving for x, y and z}}$$

$$\because {\text{ }}{{\text{a}}^{\text{2}}}{\text{ - 2ax = }}{{\text{b}}^{\text{2}}}{\text{ - 2by = }}{{\text{c}}^{\text{2}}}{\text{ - 2cz = 0}}$$

$$\therefore {\text{ }}{{\text{a}}^{\text{2}}}{\text{ - 2ax = 0}}$$

$$ \Rightarrow {\text{ 2ax = }}{{\text{a}}^{\text{2}}}$$

$$ \Rightarrow {\text{ x = }}\dfrac{{\text{a}}}{{\text{2}}}$$

$${\text{Similarly, }}{{\text{b}}^{\text{2}}}{\text{ - 2by = 0}}$$

$$ \Rightarrow {\text{ 2by = }}{{\text{b}}^{\text{2}}}$$

$$ \Rightarrow {\text{ y = }}\dfrac{{\text{b}}}{{\text{2}}}$$

$${\text{Similarly, }}{{\text{c}}^{\text{2}}}{\text{ - 2cz = 0}}$$

$$ \Rightarrow {\text{ 2cz = }}{{\text{c}}^{\text{2}}}$$

$$ \Rightarrow {\text{ z = }}\dfrac{{\text{c}}}{{\text{2}}}$$

$${\text{So the point is }}$$ $$\left( {\dfrac{{\text{a}}}{{\text{2}}}{\text{, }}\dfrac{{\text{b}}}{{\text{2}}}{\text{, }}\dfrac{{\text{c}}}{{\text{2}}}} \right)$$

$$\textbf{Hence option D is correct}$$

The point which is equidistant from the points $$(-1, 1, 3), (2, 1, 2), (0, 5, 6)$$ and $$(3,2, 2)$$ is

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$$(-1, 3, 4)$$
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$$(3, 1, 4)$$
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$$(1, 3, 4)$$
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$$(4,1, 3)$$

If the extremities of a diagonal of a square are $$(1, -2, 3)$$ and $$(2, -3, 5)$$, then the length of its side is:

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$$\sqrt{6}$$
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$$\sqrt{3}$$
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$$\sqrt{5}$$
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$$\sqrt{7}$$

If $$A, B$$ are the feet of the perpendiculars from $$(2, 4, 5)$$ to the $$x$$-axis, $$y$$-axis respectively, then the distance $$AB$$ is

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$$2\sqrt{5}$$
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$$\sqrt{29}$$
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$$\sqrt{41}$$
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$$3\sqrt{5}$$

If $$(p, q, r)$$ is equidistant from $$(1, 2, -3), (2, -3, 1)$$ and $$(-3, 1, 2)$$, then $$p +q + r$$ $$=$$

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$$-1$$
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$$1$$
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$$0$$
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$$2$$

Explanation

If $$A(0, 4, 1), B(a, b, c), C(4, 5, 0), D(2, 6, 2)$$ are the consecutive vertices of a square, then the distance $$BD$$ is:

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$$\sqrt{34}$$
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$$6$$
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$$18$$
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$$3\sqrt{2}$$

$$A = (1, -1, 2)$$ and $$B =$$ $$(2, 3, 7)$$ are two points. lf $$P,\ O$$ divide $$AB$$ in the ratios $$2:3, -2:3$$ respectively then $$P_x+Q_y=$$

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$$\displaystyle \frac{-38}{5}$$
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$$\displaystyle \frac{38}{5}$$
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$$\displaystyle \frac{-2}{5}$$
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$$\displaystyle \frac{-47}{6}$$

$$\mathrm{If} \mathrm{A}=(1, 2, 3)$$ , $$\mathrm{B}=(2,3, 4)$$ and $$\mathrm{C}$$ is a point of trisection of$$\mathrm{A}\mathrm{B}$$ such that $$\displaystyle \mathrm{C}_{\mathrm{x}}+\mathrm{C}_{\mathrm{y}}=\frac{13}{3}$$ then $$\mathrm{C}_{\mathrm{z}}=$$

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$$\displaystyle \dfrac{10}{3}$$
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$$\displaystyle \dfrac{11}{3}$$
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$$\displaystyle \dfrac{11}{2}$$
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$$11$$

If $$(1, 1, a)$$ is the centroid of the triangle formed by the points $$(1, 2, -3)$$ , $$(\mathrm{b}, 0, 1)$$ and $$(-1, 1, -4)$$ then $$a-b$$ $$=$$

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$$-5$$
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$$-7$$
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$$5$$
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$$1$$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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