MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 8

The point equidistant from the point

$O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$ and

$C(0, 0, c)$ has the coordinates

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$(a, b, c)$
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$(a/2, b/2, c/2)$
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$(a/3, b/3, c/3)$
0%
$(a/4, b/4, c/4)$

Explanation

$P(x, y, z)$

$PO=\sqrt{x^2+y^2+z^2}$

$PA=\sqrt{(a-x)^2+(-y)^2+(-z)^2}=\sqrt{(a-z)^2+y^2+z^2}$

$PB=\sqrt{x^2+(b-y)^2+z^2}$

$PC=\sqrt{x^2+y^2+(C-z)^2}$

$\Rightarrow PO=PA$

$\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}$

$x^2+y^2+z^2=(a-x)^2+y^2+z^2$

$x^2=(a-x)^2$

$x=a-x$

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

$y=b/2$

$z=c/2$ .

If the distance between a point P and the point (1, 1, 1) on the line

$\frac{{x\, - \,1}}{3}\, = \,\frac{{y - \,1}}{4}\, = \,\frac{{z\, - 1}}{{12}}$ is 13, then the coordinates of P are

0%
(3, 4, 12)
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$\left( {\frac{3}{{13}},\,\frac{4}{{13}},\,\frac{{12}}{{13}}} \right)$
0%
(4, 5, 12)
0%
(40, 53, 157)

Explanation

$\begin{array}{l} According\, to\, question: \\ we\, have\, the\, line\, equ=\frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } \, and\, point\, (P)\, is\, 13. \\ Now, \\ line\, of\, equ: \\ \Rightarrow \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } =13 \\ Now\, find\, x,y\, \& \, z\, \, coordinates: \\ \, \Rightarrow \frac { { x-1 } }{ 3 } =13 \\ \, \, \, \, \, \, \, \therefore \, \, \, \, x=39+1=40 \\ \Rightarrow \frac { { y-1 } }{ 4 } =13 \\ \, \, \, \, \, \, \, \therefore \, \, y=\, 52+1=53 \\ \Rightarrow \frac { { z-1 } }{ { 12 } } =13 \\ \, \, \, \, \, \, \, \, \therefore \, \, z=156+1=157 \\ Now,\, we\, get\, \, new\, coordinates\, of\, P:\, \, \, \left( { 40,53,157 } \right) \, \, \\ so\, that\, \, the\, correct\, option\, is\, D. \end{array}$

$A$ point

$C$ with position vector

$\frac{{3a + 4b - 5c}}{3}$ (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

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$2a+3b-4c$
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$2a-3b+4c$
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$2a+3b+4c$
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$a+3b-4c$

Explanation

$\begin{array}{l} a-2b+3c \\ \frac { { 3a+4b-5c } }{ 3 } \\ \vec { c } =\frac { { 2\vec { b } +\vec { a } } }{ 3 } \\ \vec { b } =\frac { { 3\vec { c } -\vec { a } } }{ 2 } \\ =\frac { { \left( { 3\vec { a } +4\vec { b } -5\vec { c } } \right) -\left( { \vec { a } -2\vec { b } +3\vec { c } } \right) } }{ 2 } \\ =\vec { a } +3\vec { b } -4\vec { c } \end{array}$

The xy-plane divides the line joining the points (-1, 3, 4) and (2,-5,6).

0%
internally in the ratio $2 : 3$
0%
externally in the ratio $2 : 3$
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internally in the ratio $3 : 2$
0%
externally in the ratio $3 : 2$

Explanation

The ratio that xy-plane divides the line joining the points

$(x_1, y_1, z_1)$ and

$(x_2, y_2, z_2) = -z_1 : z_2$

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

$(2, -5, 6) = -4 : 6 = -2 : 3$

i.e., It divides externally in the ratio

$2:3$

$a,b$ and

$c$ are non coplanar vectors then

$-\bar {a}+4\bar {b}-3\bar {c},3\bar {a}+2\bar {b}-5\bar {c},-3\bar {a}+8\bar {b}-5\bar {c},-3\bar {a}+2\bar {b}+\bar {c}$ are collinear.

0%
True
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False

Let

$O$ be the origin and

$P$ be the point at a distance

$3$ units from origin. If d.x.s' of OP are 1, - 2, - 2, then coordinates of P is given by

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$1, - 2, - 2$
0%
$3, - 6, - 6$
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$\dfrac{1}{3}, - \dfrac{2}{3}, - \dfrac{2}{3}$
0%
$\dfrac{1}{9}, - \dfrac{2}{9}, - \dfrac{2}{9}$

Explanation

$r=3$ ,directions

$(l,m,n)=(\cfrac { 1 }{ 3 } ,\cfrac { -2 }{ 3 } ,\cfrac { -2 }{ 3 } )$

$\therefore$

$P(x,y,z)$ be coordinates of point .

$therefore (x-0)=\cfrac { 1 }{ 3 } \times 3,\quad y-0=\cfrac { -2 }{ 3 } \times 3,\quad z-0=\cfrac { -2 }{ 3 } \times 3\\ (x,y,z)=(1,-2,-2)$

$A(2, 2, -3), B(5, 6, 9), C(2, 7, 9)$ be the vertices of a triangle. The internal bisector of the angle

$A$ meets

$BC$ at the point

$D$ , then find the coordinates of

$D$ .

0%
$(\dfrac{13}{2},\dfrac{7}{2},9)$
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$(\dfrac{7}{2},\dfrac{13}{2},9)$
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$(\dfrac{9}{2},\dfrac{7}{2},9)$
0%
$(\dfrac{13}{2},\dfrac{9}{2},9)$

Explanation

We have,

Point

$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 2,2,-3 \right)$

$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 5,6,9 \right)$

$C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( 2,7,9 \right)$

Let the coordinates of point $D\left( x,y,z \right).$

So,

According to question,

$AB=\sqrt{{{\left( 5-2 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}+{{\left( 9-3 \right)}^{2}}}$

$AB=\sqrt{9+16+149}=\sqrt{{{13}^{2}}}=13$

$AC=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 7-2 \right)}^{2}}+{{\left( 9+3 \right)}^{2}}}$

$AC=\sqrt{0+25+144}=\sqrt{169}=13$

Thus $ABC$ is isosceles triangle with $AB=AC$

So, angle bisector $AD$ bisects $BC$

$D\equiv \left( \dfrac{5+2}{2},\dfrac{6+7}{2},\dfrac{9+9}{2} \right)\equiv \left( \dfrac{7}{2},\dfrac{13}{2},9 \right)$

Hence, this is the ansewr.

If the lines

$\frac{x - 0}{1} =\frac{y+1}{2}=\frac{z-1}{-1}$ and

$\frac{x+1}{k}=\frac{y-3}{-2}=\frac{z-2}{1}$ are at right angles, then the value of k is

0%
$5$
0%
$0$
0%
$3$
0%
$-1$

Explanation

$({ l }_{ 1 }{ m }_{ 1 }{ n }_{ 1 })$ and

$({ l }_{ 2 }{ m }_{ 2 }{ n }_{ 2 })$ are directions of two

$\bot$ lines then,

${ l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+n_{ 1 }{ n }_{ 2 }=0\\ k-4-1=0\\ k=5$

The vector

$\vec{AF}$ , is given by?

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$-\left|\dfrac{\vec{a}}{\vec{c}}\right|\vec{c}$
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$\left|\dfrac{\vec{a}}{\vec{c}}\right|\vec{c}$
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$\dfrac{2|\vec{a}|}{|\vec{c}|}\vec{c}$
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$\dfrac{1}{3}\left|\dfrac{\vec{a}}{\vec{c}}\right|\vec{c}$

The position vector of the vertices of a triangle

$ABC$ are

$\hat { i } ,\hat { j } ,\hat { k }$ then the position vector of its orthocentre is

0%
$\hat { i } +\hat { j } +\hat { k }$
0%
$2(\hat { i } +\hat { j } +\hat { k } )$
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$\cfrac{1}{3}(\hat { i } +\hat { j } +\hat { k } )$
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$\cfrac{1}{\sqrt{3}}(\hat { i } +\hat { j } +\hat { k } )$

Explanation

We have, the position vectors of the vertices of a triangle

$ABC$ are

$\hat { i } ,\hat { j } ,\hat { k }$ .

Let

$O$ be the fixed point.

$\hat { i }$ = position vector of

$A=\overrightarrow{OA}$

$\hat { j }$ = position vector of

$B=\overrightarrow{OB}$

$\hat { k }$ = position vector of

$C=\overrightarrow{OC}$

Let

$AD$ be the median ofthe triangle

$ABC$ .

Since,

$D$ is the mis point of

$BC$ .

Position vector of

$D=$ vector

$OD=\dfrac{(\overrightarrow{OB}+\overrightarrow{OC})}{2}$

Now, position vector of

$G$ ,

$=\dfrac{2(\overrightarrow{OD})+1(\overrightarrow{OA})}{3}$

$=\dfrac{2(\dfrac{\overrightarrow{OB}+\overrightarrow{OC}}{2})+\overrightarrow{OA}}{3}$

$=\dfrac{{\overrightarrow{OA}+\overrightarrow{OB}}+\overrightarrow{OC}}{3}$

So,

$=\dfrac{{\hat{i}+\hat{j}}+\hat{k}}{3}$

Hence, the position vector of the orthocentre is

$\dfrac{1}{3}\left(\hat i+\hat j + \hat k\right)$ .

979838_1079404_ans_ba4deb652ded4a1b863a3b1965ad90d9.png

The graph of the equation

$y^{2}+z^{2}=0$ in three dimensional space is

0%
x- axis
0%
y- axis
0%
z- axis
0%
yz-plane

Explanation

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

$x-axis$ .

Hence, the correct option is

$x-axis$ .

The area of triangle whose vertices are

$(1, 2, 3), (2, 5, -1)$ and

$(-1, 1, 2)$ is

0%
$150\ sq. units$
0%
$145\ sq. units$
0%
$\sqrt {155}/2\ sq. units$
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$155/2\ sq. units$

Explanation

Let the vertices of triangle are

$A(1,2,3),B(2,5,-1)$ and

$C(-1,1,2)$

Then,

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =\hat { i } +3\hat { j } -4\hat { k } \\ \\ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =-2\hat { i } -\hat { j } -\hat { k } \end{array}$

Then,

$\begin{array}{l} \overrightarrow { AB } \times \overrightarrow { AC } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 1 & 3 & { -4 } \\ { -2 } & { -1 } & { -1 } \end{array} } \right| \\ \\ =-7\hat { i } +9\hat { j } +5\hat { k } \\ \\ \left| { \overrightarrow { AB } \times \overrightarrow { AC } } \right| =\sqrt { { { \left( { -7 } \right) }^{ 2 } }+{ { \left( 9 \right) }^{ 2 } }+{ { \left( 5 \right) }^{ 2 } } } \\ \\ =\sqrt { 49+8+25 } \\ \\ =\sqrt { 155 } \end{array}$

Area of triangle

$ABC=\frac{1}{2}|\vec {AB} \times \vec {AC}|$

$=\frac{1}{2} \times \sqrt {155}$

$=\frac{\sqrt {155}}{2}$ sq. units

If the point

$(x, y)$ is equidistant from the points

$(a + b, b - a)$ and

$(a - b , a + b)$ , then

$bx = ay$ .

0%
True
0%
False

Explanation

If a point (x,y) is equidistant from A(a,b) and B(c,d) then,

$x = \frac{{a + c}}{2},y = \frac{{b + d}}{2}$

So,

$\begin{array}{l}x = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{2},y = \frac{{\left( {b - a} \right) + \left( {a + b} \right)}}{2}\\x = \frac{{2a}}{2},y = \frac{{2b}}{2}\\x = a,y = b\end{array}$

putting values in bx=ay

Solving LHS,

b(a)=ab

Solving RHS,

a(b)=ab

As LHS=RHS the equation bx=ay is true.

A tangent to the curve

$y = f(x)$ at

$p(x, y)$ meets

$x - axis$ at

$A$ and

$y-axis$ at

$B$ . If

$\overline {AP} : \overline {BP} = 1 : 3$ and

$f(1) = 1$ then the curve also passes through the point.

0%
$\left (\dfrac {1}{2}, 4\right )$
0%
$\left (\dfrac {1}{3}, 24\right )$
0%
$\left (2, \dfrac {1}{8}\right )$
0%
$\left (3, \dfrac {1}{28}\right )$

Explanation

$\dfrac{(y - y_1)}{(x - x_1)} = f'(x_1)$

$y - y_1 = f'(x_1)(x - x_1)$

$y = 0, \dfrac{-y}{f'(x_1)} = x - x_1$

$\Rightarrow x = x_1 = \dfrac{y_1}{f'(x_1)}$

$A = \left(x_1 - \dfrac{y_1}{r'(x_1)}, 0\right)$

$x = 0, y - y_1 = f'(x_1). (-x_1)$

$\Rightarrow y = y_1 - x_1 f'(x_1)$

$B = (0, y_1 - x_1 f'(x_1))$

P divides

$AB$ in ration

$1 : 3$

$x_1 = \dfrac{3\left[x_1 - \dfrac{y_1}{f'(x_1)} \right]}{4} , y_1 = \dfrac{[y_1 - x_1 f'(x_1)]}{4}$

$4y_1 = y_1 x_1 f'(x_1)$

$f'(x_1) = \dfrac{-3y_1}{x_1}$

$f'(x) = \left(\dfrac{-3y}{x} \right)$

$\dfrac{dy}{dx} = \dfrac{-3y}{x}$

$\dfrac{dy}{y} = \dfrac{-3dx}{x}$

$\ln y = -3 \ln x + c$

$y = kx^{-3}$

$y(1) = 1$

$\Rightarrow k = 1 \Rightarrow y = \dfrac{1}{x^3}$ put

$x = 2 , y = \dfrac{1}{8} \left(2, \dfrac{1}{8} \right)$

If the distance between a point

$P$ and the point

$(1, 1, 1)$ on the line

$\dfrac {x - 1}{3} = \dfrac {y - 1}{4} = \dfrac {z - 1}{12}$ is

$13$ , then the coordinates of

$P$ are

0%
$(3, 4, 12)$
0%
$\left (\dfrac {3}{13}, \dfrac {4}{13}, \dfrac {12}{13}\right )$
0%
$(4, 5, 13)$
0%
$(40, 53, 157)$

Explanation

Let the given points be

$A(1,1,1)$

Consider,

$\dfrac{{x - 1}}{3} = \dfrac{{y - 1}}{4} = \dfrac{{z - 1}}{{12}} = \lambda$

$\begin{array}{l} x=3\lambda +1 \\ \\ y=4\lambda + 1\\ \\ z=12\lambda +1 \end{array}$

General point on the line is

$3\lambda + 1,\,4\lambda + 1,\,12\lambda + 1$

Given that,

$AP=13$

$\sqrt {{{\left( {3\lambda + 1 - 1} \right)}^2} + {{\left( {\,4\lambda + 1 - 1} \right)}^2} + {{\left( {12\lambda + 1 - 1} \right)}^2}} = 13$

$13\lambda =13$

$\lambda =1$

$\begin{array}{l} 3\lambda +1=4 \\ \\ 4\lambda +1=5 \\ \\ 12\lambda +1=13 \end{array}$

Therefore, required point

$P$ is

$(4,5,13)$ .

Hence the correct option is

$C$ .

The points

$(10,7,0)$ ,

$(6,6-1)$ and

$(6,9,-4)$ form a

0%
Right -angled triangle
0%
Isosceles triangle
0%
Both $(1)$ & $(2)$
0%
Equilateral triangle

Explanation

$P(10, 7, 0),\ Q(6, 6, -1),\ R(6, 9, -4)$

$PQ=\sqrt {4^2+(-1)^2 +(-1)^2}$

$=\sqrt {16+1+1}=3\sqrt 3$

$QR=\sqrt {0^2+3^2+-3^2}$

$PR=\sqrt {4^2+2^2+(-4)^2}$

$=\sqrt {16+4+16}=6$

$PQ^2+QR^2=(3\sqrt 2)^2+(3\sqrt 2)^2=6^2=PR^2$

$\therefore \ \triangle PQR$ is a right angle

$D$ of

$Q,\ PQ=QR$

$\triangle PQR$ is a isoscels traingle

$(C)$ Both

$(1)$ &

$(2)$

The ratio in which the line joining

$(3,4,-7)$ and

$(4,2,1)$ is dividing the xy-plane

0%
$3:4$
0%
$2:1$
0%
$7:1$
0%
$4:3$

Explanation

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

$0=\dfrac{k \times 1+1 \times -7}{k+1}$

$k-7 =0$

$k=7$

Therefore, ratio is

$7:1$

Hence, option

$C$ is correct.

The ratio in which the line joining points

$(2,4,5)$ and

$(3,5,-4)$ divide YZ -plane is

0%
$-2:3$
0%
$2:3$
0%
$-3:2$
0%
$3:2$

Explanation

Let the ratio be

$\lambda:1$

The dividing plane is

$YZ$ plane.

So, the co-ordinate of

$x$ after dividing the line

$= 0$

Now According to internal division rule of line segment

$x=\dfrac{3\times \lambda - 1\times 2}{\lambda + 1}$

$\Rightarrow \dfrac{3\lambda - 2}{\lambda + 1}=0$

$\Rightarrow 3\lambda - 2=0 \ \Rightarrow \lambda=\dfrac{2}{3}$ .

So. Ratio

$=2:3$

1382673_1131309_ans_6120b455e917495ca5d42d90253fc089.png

The vertices of a triangle are

$)2, 3, 5), (-1, 3, 2), (3, 5, -2)$ , then the angles are

0%
$30^o, 30^o, 30^o$
0%
$\cos^{-1}\left(\dfrac{1}{\sqrt{5}}\right), 90^o, \cos^{-1} \left(\dfrac{\sqrt{5}}{\sqrt{3}}\right)$
0%
$30^o, 60^o, 90^o$
0%
$\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right), 90^o, \cos^{-1} \left(\dfrac{\sqrt{2}}{\sqrt{3}}\right)$

Explanation

$\overrightarrow { AB } =-3\hat { i } -3\hat { k } \\ \overrightarrow { BC } =4\hat { i } +2\hat { j } -4\hat { k } \\ \overrightarrow { CA } =-\hat { i } -2\hat { j } +7\hat { k } \\ \cos { A } =\cfrac { \left| \overrightarrow { AB } .\overrightarrow { CA } \right| }{ \left| \overrightarrow { AB } \right| .\left| \overrightarrow { CA } \right| } =\cfrac { 1 }{ \sqrt { 3 } } \\ \Rightarrow A=\cos ^{ -1 }{ (\cfrac { 1 }{ \sqrt { 3 } } ) } \\ \cos { B } =\cfrac { \left| \overrightarrow { AB } .\overrightarrow { BC } \right| }{ \left| \overrightarrow { AB } \right| .\left| \overrightarrow { BC } \right| } =0\\ \Rightarrow B={ 90 }^{ \circ }\\ \cos { C } =\cfrac { \left| \overrightarrow { CA } .\overrightarrow { BC } \right| }{ \left| \overrightarrow { CA } \right| .\left| \overrightarrow { BC } \right| } =\cfrac { \sqrt { 2 } }{ \sqrt { 3 } } \\ \Rightarrow C=\cos ^{ -1 }{ (\cfrac { \sqrt { 2 } }{ \sqrt { 3 } } ) }$
1089431_1144096_ans_b2f309fc9f7d460a8044dd98b99e3222.png

1089431_1144096_ans_b2f309fc9f7d460a8044dd98b99e3222.png

The values of a for which

$(8, -7, a), (5, 2, 4)$ and

$(6, -1, 2)$ are collinear, is given by?

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

Explanation

$\begin{matrix} then\, a=? \\ The\, equation\, lineAB\, is\, \\ \Rightarrow \dfrac { { x-8 } }{ { 5-8 } } =\dfrac { { y+7 } }{ { 2+7 } } =\dfrac { { z-a } }{ { 4-a } } ....\left( 1 \right) \\ po{ int }\, c\, lies\, in\, the\, line\, AB \\ po{ int }\left( { 6,-1,2 } \right) satisfy\, eq\left( 1 \right) \\ \Rightarrow \dfrac { { -2 } }{ { -3 } } =\dfrac { { -1+7 } }{ 9 } =\dfrac { { 2-a } }{ { 4-a } } \\ \Rightarrow \dfrac { 2 }{ 3 } =\dfrac { { 2-a } }{ { 4-a } } \\ \Rightarrow 8-2a=6-3a \\ \Rightarrow 3a-2a=6-8 \\ a=-2 \\ \end{matrix}$

The shortest distance of the point

$(1,2,3)$ from

${x}^{2}+{y}^{2}=0$ is

0%
$5$
0%
$\sqrt{5}$
0%
$2$
0%
$\sqrt{14}$

Explanation

${x}^{2}+{y}^{2}=0$

$\therefore$

$(x,y)\equiv (0,0)$

Let point be

$\equiv (0,0,z)$

$d\equiv \sqrt { { (1-0) }^{ 2 }+{ (2-0) }^{ 2 }+{ (z-3) }^{ 2 } } =\sqrt { 1+4+{ (z-3) }^{ 2 } } \Rightarrow { (z-3) }^{ 2 }\ge 0$

${ d }_{ min }=\sqrt { 5 }$

A triangle

$ABC$ is placed so that the mid-points of the sides are on the

$x,y,z$ axes. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively

$\alpha, \beta, \gamma$ . Coordinates of the centroid of the triangle

$ABC$ are

0%
$(-\alpha/ 3, \beta/ 3,\gamma/3)$
0%
$(\alpha/3,-\beta/3,\gamma/3)$
0%
$(\alpha/ 3, \beta/3, -\gamma/ 3)$
0%
$(\alpha/3, \beta/ 3, \gamma/ 3)$

Explanation

$\triangle ABC$ has

$P$ as mid-point of

$AB$ on x-axis.

$Q$ as midpoint of

$BC$ on y-axis.

$R$ as midpoint of

$AC$ on z-axis.

$\therefore A=({ x }_{ 1 }{ y }_{ 1 }{ z }_{ 1 })$

$\therefore B=({ x }_{ 2 }{ y }_{ 2 }{ z }_{ 2 })$

$\therefore C=({ x }_{ 3 }{ y }_{ 3 }{ z }_{ 3 })$

$\Rightarrow P=(\cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } ,\cfrac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } )\\ Q=(\cfrac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\cfrac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } ,\cfrac { { z }_{ 2 }+{ z }_{ 3 } }{ 2 } )\\ R=(\cfrac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } ,\cfrac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } ,\cfrac { { z }_{ 3 }+{ z }_{ 1 } }{ 2 } )\\ P=(\alpha ,0,0)\\ Q=(0,\beta ,0)\\ R=(0,0,\gamma )\\ \therefore \cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =\alpha \quad ,\quad \cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =0\quad ,\quad \cfrac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } =0\\ \cfrac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =0\quad ,\quad \cfrac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =\beta \quad ,\quad \cfrac { { z }_{ 2 }+{ z }_{ 3 } }{ 2 } =0\\ \cfrac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =0\quad ,\quad \cfrac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } =0\quad ,\quad \cfrac { { z }_{ 3 }+{ z }_{ 1 } }{ 2 } =\gamma \\ \therefore { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=\alpha ,\quad { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }=\beta ,\quad { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=\gamma$

$\therefore$ centroid

$(G)=(\cfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } ,\cfrac { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }{ 3 } )\\ =(\cfrac { \alpha }{ 3 } ,\cfrac { \beta }{ 3 } ,\cfrac { \gamma }{ 3 } )$

1092050_1162938_ans_83e726a53a2c4ef1b4472a4e0618b7b6.png

The triangle with vertices

$A (1, 0, 1), B (2, - 1, 4) \ and \ C (3, - 4, - 1),$ is right-angled.

0%
True
0%
False

Explanation

The Vertices are

$A(1,0,1) \quad B(2,-1,4)$ and

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

$AB=\sqrt{(1-2)^2+(0+1)^2+(1-4)^2}=\sqrt{1+1+9}=\sqrt {11}$

$BC=\sqrt{(3-2)^2+(-4+1)^2+(-1-4)^2}=\sqrt{1+9+25}=\sqrt {35}$

$AC=\sqrt{(1-3)^2+(0+4)^2+(1+1)^2}=\sqrt{4+16+4}=\sqrt {24}$

$\implies AB^2+AC^2=BC^2$

So it is a right-angled triangle.

The distance between two points

$(1,1)$ and

$\left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{\left( {1 - t} \right)}^2}}}{{1 + {t^2}}}} \right)$ is

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4t
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3t
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1
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none of these

Explanation

Two point are

$A(1,1)$ and

$B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

$AB=\sqrt{\left(1-\dfrac{2t^{2}}{1+t^{2}}\right)^{2}+\left(1-\dfrac{(1-t^{2})}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1+t^{2}-2t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{1+t^{2}-1-t^{2}+2t}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1-t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{RT}{1+t^{2}}\right)^{2}}$

$=\sqrt{\dfrac{1+t^{4}-2t^{2}+4t^{2}}{(1+t^{2})^{2}}}$

$=\sqrt{\dfrac{(1+t^{2})^{2}}{(1+t^{2})^{2}}}$

$=1$

The plane

$x = 0$ divides the joinning of

$( - 2, 3, 4)$ and

$(1, - 2, 3)$ in the ratio :

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$2 : 1$
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$1 : 2$
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$3 : 2$
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$- 4 : 3$

Explanation

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

So,

$x = \dfrac{m+n(-2)}{m+n } = \dfrac{m-2n}{m+n}$

$0 = \dfrac{m-2n}{m+n}\Rightarrow m = 2n$

So,

$\dfrac{m}{n} = \dfrac{2}{1}\Rightarrow 2:1$ option -A

The points (-5,12), (-2,-3),(9,-10),(6,5) taken in order, form

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Parallelogram
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rectangle
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rhombus
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square

Explanation

Given points

$A(-5, 12)\quad B(-2, -3), C(9, -10), D(6, 5)$

Distance

$AB=\sqrt{(-5+2)^2+(12+3)^2}=\sqrt{9+225}=\sqrt{234}$

Distance

$BC=\sqrt{(-2-9)^2+(-3+10)^2}=\sqrt{121+49}=\sqrt{170}$

Distance

$CD=\sqrt{(9-6)^2+(-10-5)^2}=\sqrt{9+225}=\sqrt{234}$

Distance

$AD=\sqrt{(-5-6)^2+(12-5)^2}=\sqrt{121+49}=\sqrt{170}$

Distance

$AC=\sqrt{(-5-9)^2+(12+10)^2}=\sqrt{196+484}=\sqrt{680}$

Distance

$BD=\sqrt{(-2-6)^2+(-3-5)^2}=\sqrt{64+64}=\sqrt{128}$

These points forms a parallelogram, opposite pair of sides are equal and adjacent sides do not form right angles.

The vertices of a triangle are

$(2, 3, 5), (-1, 3, 2), (3, 5, -2)$ , then the angles are

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$30^{\circ}, 30^{\circ}, 120^{\circ}$
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$\cos^{-1} \left (\dfrac {1}{\sqrt {5}}\right ), 90^{\circ}, \cos^{-1} \left (\dfrac {\sqrt {5}}{\sqrt {3}}\right )$
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$30^{\circ}, 60^{\circ}, 90^{\circ}$
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$\cos^{-1} \left (\dfrac {1}{\sqrt {3}}\right ), 90^{\circ}, \cos^{-1} \left (\sqrt{\dfrac { {2}}{ {3}}}\right )$

Explanation

Given vertices:

$A(2,3,5);B(-1,3,2);C(3,5,-2)$

$AB=\sqrt { 9+0+9 } =3\sqrt { 2 }$

$BC=\sqrt { 16+4+16 } =6$

$AC=\sqrt { 1+4+49 } =3\sqrt { 6 }$

$\cos { A } =\cfrac { { CA }^{ 2 }+{ BA }^{ 2 }-{ BC }^{ 2 } }{ 2CA\times BA } =\cfrac { 18+54-36 }{ 2\times 3\sqrt { 2 } \times 3\sqrt { 6 } } =\cfrac { 36 }{ 18\times 2\sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } }$

$\cos { B } =\cfrac { { AB }^{ 2 }+{ BC }^{ 2 }-{ AC }^{ 2 } }{ 2AB\times BC } =\cfrac { 18+36-54 }{ 2.3\sqrt { 2 } \times 6 } =0$

$\cos { C } =\cfrac { 54+36-18 }{ 2\times 3\sqrt { 6 } \times 6 } =\cfrac { 72 }{ 36\sqrt { 6 } } =\cfrac { 2 }{ \sqrt { 6 } } =\sqrt { \cfrac { 2 }{ 3 } }$

$\therefore A=\cos ^{ -1 }{ \left( \cfrac { 1 }{ \sqrt { 3 } } \right) } ;B=90°;C=\cos ^{ -1 }{ \left( \sqrt { \cfrac { 2 }{ 3 } } \right) }$

1162792_1170852_ans_0c6141b9400a4f608062512d28eb51bd.png

The plane passing through the point

$\left(-2,-2,2\right)$ and containing the line joining the points

$\left(1,1,1\right)$ and

$\left(1,-1,2\right)$ makes intercepts on the coordinates axes, the sum whose lengths is ?

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$3$
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$4$
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$6$
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$12$

The nearest point from the origin is

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$(2,3,-1)$
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$(-3,2,1)$
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$(2,2,2)$
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$(1,2,-1)$

Explanation

Distance from the origin

$=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$

Distance from

$\left(2,3,-1\right)$ is

$\sqrt{{2}^{2}+{3}^{2}+{\left(-1\right)}^{2}}=\sqrt{4+9+1}=\sqrt{14}$

Distance from

$\left(-3,2,1\right)$ is

$\sqrt{{\left(-3\right)}^{2}+{2}^{2}+{1}^{2}}=\sqrt{9+4+1}=\sqrt{14}$

Distance from

$\left(2,2,2\right)$ is

$\sqrt{{2}^{2}+{2}^{2}+{2}^{2}}=\sqrt{8}$

Distance from

$\left(1,2,-1\right)$ is

$\sqrt{{1}^{2}+{2}^{2}+{\left(-1\right)}^{2}}=\sqrt{1+4+1}=\sqrt{6}$

$\sqrt{6}<\sqrt{8}<\sqrt{14}$

Thus

$\left(1,2,-1\right)$ is nearest point from the origin.

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

$\angle\ BAC$ which meets

$BC$ at

$D$ . Then the co-ordinates of

$D$ , are

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$\left[ {\dfrac{{17}}{{16}},\ \dfrac{{57}}{{16}},\ \dfrac{{19}}{8}} \right]$
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$\left[ {\dfrac{{19}}{{8}},\ \dfrac{{57}}{{16}},\ \dfrac{{17}}{16}} \right]$
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$\left[0,\ 0,\ {\dfrac{{17}}{{16}}}\right]$
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$\left[{\dfrac{{17}}{{16}}},\ 0,\ 0\right]$

Explanation

$AD$ is bisector of

$\angle BAC$

$\Rightarrow$ Ratio at

$D$ is

$c:b$ where

$c = AB = \sqrt{(3-5)^{2}+(2-3)+(0-2)^{2}}$

$= \sqrt{4+1+4} = \sqrt{9} = 3$

$b = AC = \sqrt{(3+9)^{2}+(2-6)^{2}+(0+3)^{2}}$

$= \sqrt{144+16+9} = \sqrt{169} = 13$

for point

$D$

$x = \dfrac{c(-9)+b(6)}{c+b} = \dfrac{3(-9)+13(5)}{3+13} = \dfrac{38}{16} = \dfrac{19}{8}$

$y = \dfrac{c(6)+b(3)}{c+b} = \dfrac{3(6)+13(3)}{3+13} = \dfrac{57}{16}$

$z = \dfrac{c(-3)+b(2)}{c+b} = \dfrac{3(-3)+13(2)}{3+13} = \dfrac{17}{16}$

Hence, point

$D$ is

$[\dfrac{19}{8},\dfrac{57}{16},\dfrac{17}{6}]$

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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