MCQ Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 10

The distance of P(2,-3) from the x-axis is......

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2
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-3
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3
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$\sqrt{13}$

If the distance between the points (2,-3) and (5,b) is 5 , then b=.........

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1
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2
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7
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5

Find the centroid of a triangle, mid-points of whose sides are

$(!,2,-3),(3,0,1)$ and

$(-1,1,-4)$

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

The point in the

$xy -$ plane which is equidistant from

$(2, 0, 3), (0, 3, 2)$ and

$(0,0, 1)$ is

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$(1, 2,3)$
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$(-3, 2, 0)$
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$(3, -2. 0)$
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$(3, 2, 0)$
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$(3, 2, 1)$

Explanation

Let the points are

$A(2,0,3),B(0,3,2)$ and

$D(0,0,1)$

We know that Z-coordinate of every point an xy-plane is zero so let

$p(x,y,0)$ be a point on xy-plane such that

$PA=PB=PC$

Now,

$PA=PB$

$\Rightarrow PA^2=PB^2$

$\Rightarrow (x-2)^2+(y-0)^2+(0-3)^2=(x-0)^2+(y-3)^2+(0-2)$

$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0$ ......(i)

and ,

$PB=PC$

$\Rightarrow PB^2=PC^2$

$\Rightarrow (x-0)^2+(y-3)^2(0-2)^2=(x-0)^2+(y-0)^2+(0-1)^2$

$\Rightarrow -6y+12=0$

$\Rightarrow y=2$ ......(ii)

Putting

$y=2$ in equation (i), we get

$x=3$

Hence, the required point is

$(3,2,0)$

The foot of the perpendicular from the point

$A(7, 14, 5)$ to the plane

$2x+4y-z=2$ is?

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$(3, 1, 8)$
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$(1, 2, 8)$
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$(3, -3, 5)$
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$(5, -3, -4)$

Explanation

Let N be the foot of the perpendicular drawn from the point

$A(7, 14, 5)$ and perpendicular to the plane

$2x+4y-z=2$

Then, the equation of the line PN is

$\dfrac{x-7}{2}=\dfrac{y-14}{4}=\dfrac{z-5}{-1}=\lambda$ (say)

Let the coordinates of N be

$N(2\lambda +7, 4\lambda +14, -\lambda +5)$

Since N lies on the plane

$2x+4y-z=2$ , so

$2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2$

$\Rightarrow 21\lambda =-63$

$\Rightarrow \lambda =-3$

$\therefore$ required foot of the perpendicular is

$N(-6+7, -12+14, 3+5)$ , i.e.,

$N(1, 2, 8)$ .

The coordinates of the point where the line through the points

$A(5,1,6)$ and

$B(3,4,1)$ crosses the yz-plane is

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$(0,17,-13)$
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$\left( 0,\cfrac { -17 }{ 2 } ,\cfrac { 13 }{ 2 } \right)$
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$\left( 0,\cfrac { 17 }{ 2 } ,\cfrac { -13 }{ 2 } \right)$
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none of these

Explanation

Equation of the line passing through the points

$A(5, 1, 6)$ and

$B(3,4, 1)$ is

$\dfrac{x-5}{2}=\dfrac{4-1}{-3} = \dfrac{7-6}{5}$ ....(i)

Let the line (i) cross the yz plane at point

$(o, h, k)$

Then

$(o, h, k)$ point lie on the line (I)

So,

$\dfrac{0-5}{2} = \dfrac{h-1}{-3} = \dfrac{k-6}{5}$

Then,

$\dfrac{k-6}{5} = - \dfrac{5}{2}$

or,

$k-6 = -\dfrac{25}{2}$

or,

$k = 6- \dfrac{-25}{2} =\dfrac{12-25}{2} = -\dfrac{13}{2}$

Again,

$\dfrac{h-1}{-3} = \dfrac{-5}{2}$

or,

$h-1 = \dfrac{15}{2}$

or,

$h=\dfrac{15}{2} + 1 = \dfrac{17}{2}$

Therefore the crossing points is

$\left(0, \dfrac{17}{2}, -\dfrac{13}{2}\right)$

The point on the line

$\dfrac {x - 2}{1} = \dfrac {y + 3}{-2} = \dfrac {z + 5}{-2}$ at a distance of 6 from the point

$\left ( 2,-3,-5 \right )$ is

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$\left ( 3,-5,-3 \right )$
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$\left ( 4,-7,-9 \right )$
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$\left ( 0,2,-1 \right )$
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$\left ( -3,5,3 \right )$

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are

$d_1,d_2,d_3$ . Then

$d_1^2 +d_2^2 + d_3^2 = kd^2$ where K is

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$1$
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$5$
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$3$
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$2$

The shortest distance between the lines

$\dfrac {x - 3}{2} = \dfrac {y + 15}{-7} = \dfrac {z - 9}{5}$ and

$\dfrac {x + 1}{2} = \dfrac {y - 1}{1} = \dfrac {z - 9}{-3}$ is

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$2 \sqrt 3$
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$4 \sqrt 3$
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$3 \sqrt 6$
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$5 \sqrt 6$

The ratio in which the plane

$\overrightarrow{r} \cdot \left (\overrightarrow{i} - 2 \overrightarrow{j} + 3 \overrightarrow{k} \right ) = 17$ divides the line joining the points

$-2 \overrightarrow{i} + 4 \overrightarrow{j} + 7 \overrightarrow{k}$ and

$3 \overrightarrow{i} - 5 \overrightarrow{j} + 8 \overrightarrow{k}$

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1 : 5
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1 : 10
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3 : 5
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3 : 10

Explanation

Let the plane

$\overrightarrow{r} \cdot \left ( \overrightarrow{i} - 2 \overrightarrow{j} + 3 \overrightarrow{k} \right ) = 17$ divided the line joining the points

$-2 \overrightarrow{i} + 4 \overrightarrow{j} + 7 \overrightarrow{k}$ and

$3 \overrightarrow{i} - 5 \overrightarrow{j} + 8 \overrightarrow{k}$ in the ratio

$t : 1$ at point P.

Therefore, point P is

$\dfrac{3t - 2} {t + 1} \overrightarrow{i} + \dfrac{-5t + 4} {t + 1} \overrightarrow{j} + \dfrac{8t + 7} {t + 1} \overrightarrow{k}$

This lies on the given plane

$\therefore \dfrac{3t - 2} {t + 1} (1) + \dfrac{-5t + 4} {t + 1} (- 2) + \dfrac{8t + 7} {t + 1} (3) = 17$

Solving, we get

$t = \dfrac{3} {10}$

The point on the line

$\frac{x - 2} {1} = \frac{y + 3} {-2} = \frac{z + 5} {-2}$ at a distance of 6 from the point

$\left ( 2, -3, -5 \right )$ is

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$\left ( 3, -5, -3 \right )$
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$\left ( 4, -7, -9 \right )$
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$\left ( 0,2, -1 \right )$
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$\left ( -3, 5, 3 \right )$

Explanation

b. Direction cosines of the given line are

$\frac{1} {3}, - \frac{2} {3}, -\frac{2} {3}$

Hence, the equation of line can be point in the form

$\frac{x - 2} {1/3} = \frac{y + 3} {-2/3} = -\frac{z + 5} {-2/3} = r$

Therefore, any point on the line is

$\left (2 + \frac{r} {3}, - 3 - \frac{2r} {3}, - 5 - \frac{2r} {3} \right )$ , where

$r = \pm 6$ .

Points are

$\left (4, -7, -9 \right )$ and

$\left (0, 1, -1 \right )$

The points

$A(1,2,-1),B(2,5,-2),C(4,4,-3)$ and

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

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collinear
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vertices of a rectangle
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vertices of a square
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vertices of a rhombus

Explanation

${=}$

$\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$
AB

${=}$

$\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$
AB

${=}$

$\sqrt{11}$
Similarly you find that BC

${=}$

$\sqrt{6}$ CD

${=}$

$\sqrt{11}$ and DA

${=}$

$\sqrt{6}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC

${=}$

$\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$
AC

${=}$

$\sqrt{17}$
similarly BD

${=}$

$\sqrt{17}$
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle

Let

$A \left ( 2\hat{i}+3\hat{j}+5\hat{k} \right )B\left ( -\hat{i}+3\hat{j}+2\hat{k} \right )$ and

$C \left ( \lambda \hat{i}+5\hat{j}+\mu\hat{k} \right )$ are vertices of a triangle and its median through

$A$ is equally inclined to the positive directions of the axes. The value of

$\lambda+\mu$ is equal to

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$-7$
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$2$
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$7$
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$17$

Explanation

We have,

$\vec{AD} = \dfrac 12 \left( \vec b + \vec c - 2 \vec a \right)$

$\vec{AD} = \dfrac{ \lambda -5 }{2} \hat i + \hat j + \dfrac{\mu-8}{2} \hat{k}$
The line is equally inclined to the axes. Hence, its direction ratios must be of the form

$1:1:1$ .
Hence,

$\lambda= 7$ and

$\mu = 10$

$(0, b, 0)$ is the centroid of the triangle formed by the points

$(4, 2, -3)$ ,

$({a}, -5, 1)$ and

$(2, -6, 2)$ . If

$a ,b$ are the roots of the quadratic equation

$x^2+px+q = 0$ , then

$p,q$ are

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$9,18$
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$-9,-18$
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$3,-18$
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$-3,18$

Explanation

Since

$a,b$ are the roots of the equation

$x^{2}+px+q=0$

$\Rightarrow a+b=-p$ and

$ab=q$

Centroid of triangle is

$\displaystyle \left(\dfrac{a+6}{3},-3,0\right)$
Given, centroid

$(0,b,0)$

Comparing, we get

$b=-3$
and

$\dfrac{a+6}{3}=0$

$\Rightarrow a=-6$
Hence,

$p=9, q=18$

Hence, option A.

$OABC$ is a tetrahedron such that the

$OA^{2}+BC^{2}=OB^{2}+CA^{2}=OC^{2}+AB^{2}$ , then which of the following is/are correct

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$AB\perp OC$
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$OB\neq CA$
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$OC=AB$
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$AB\perp BC$

Explanation

$OABC$ is a tetrahedron such that

$OA^{2}+BC^{2}=OB^{2}+CA^{2}=OC^{2}+AB^{2}$
Let

$O(0,0,0),A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2}),C(x_{3},y_{3},z_{3})$ are coordinates of vertices.

$OA^{2}+BC^{2}$ =

$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+((x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2})$
=

$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{2}x_{3}+y_{2}y_{3}+z_{2}z_{3})$
Similarly,

$OB^{2}+CA^{2}$ =

$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{1}x_{3}+y_{1}y_{3}+z_{1}z_{3})$

$OC^{2}+AB^{2}$ =

$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2})$

$\therefore$

$OA^{2}+BC^{2}=OB^{2}+CA^{2}\Rightarrow x_{2}x_{3}+y_{2}y_{3}+z_{2}z_{3} = x_{1}x_{3}+y_{1}y_{3}+z_{1}z_{3}$

$\Rightarrow (x_{1}-x_{2})x_{3}+(y_{1}-y_{2})y_{3}+(z_{1}-z_{2})z_{3} = 0$

$\Rightarrow (x_1-x_2)(x_3-0) + (y_1-y_2)(y_3-0)+(z_1-z_2)(z_3-0) = 0$

$\Rightarrow \vec{AB}\cdot\vec{OC} = 0$
Hence,

$OC$ is perpendicular to

$AB.$

The plane

$\displaystyle ax + by + cz + (-3) = 0$ meet the co-ordinate axes in

$A, B, C$ . The centroid of the triangle is

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$\displaystyle (3a, 3b, 3c)$
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$\displaystyle \left(\frac {3}{a}. \frac {3}{b}, \frac {3}{c}\right)$
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$\displaystyle \left(\frac {a}{3}. \frac {b}{3}, \frac {c}{3}\right)$
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$\displaystyle \left(\frac {1}{a}. \frac {1}{b}, \frac {1}{c}\right)$

Explanation

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

$\left(\dfrac{3}{a},0,0\right)$

similarily, the point on y axis becomes

$\left(0,\dfrac{3}{b},0\right)$

And that on z axis becomes

$\left(0,0,\dfrac{3}{c}\right)$

The centroid of the triangle formed by these points would be

$\left(\dfrac {\dfrac{3}{a}+0+0}{3} , \dfrac {{0}+\dfrac{3}{a}+0}{3} ,\dfrac{0+0+\dfrac{3}{c}}{3}\right)$

$\therefore \left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)$

$ABCD$ is a parallelogram.

$L$ is a point on

$BC$ which divides

$BC$ in the ratio

$1 : 2$ .

$AL$ intersects

$BD$ at

$P$ .

$M$ is a point on

$DC$ which divides

$DC$ in the ratio

$1 : 2$ and AM intersects

$BD$ in

$Q$ .

Point

$P$ divides

$AL$ in the ratio

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

Explanation

Given,

$BL:LC=1:2$ and

$DM:MC=1:2$

Now,

$BL:BC=1:3$

Consider

$\triangle APD$ and

$\triangle LPB$

$\angle ADP=\angle LBP$

$\angle APD=\angle BPL$

By AA corollary,

$\triangle APD \sim \triangle LPB$

$\Rightarrow$ Corresponding parts are proportional.

$\Rightarrow \dfrac{AP}{LP}=\dfrac{AD}{BL}$

$\Rightarrow \dfrac{AP}{LP}=\dfrac{BC}{BL}$ (

$\because AD=BC$ )

$\Rightarrow \dfrac{AP}{LP}=3:1$

If the vertices of a triangle are

$(-1,6,-4),(2,1,1)$ and

$(5,-1,0)$ then the centroid of the triangle is

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$(6,6,-3)$
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$(2,2,-1)$
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$\left ( 3,3,-\displaystyle \frac{3}{2} \right )$
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none of these

Explanation

the centroid of the triangle is

$(\dfrac {x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3})$

Thus by substituting the vertices we get

$=(\dfrac{-1+2+5}{3},\dfrac{6+1-1}{3},\dfrac{-4+1+0}{3})$

$=(\dfrac{6}{3},\dfrac{6}{3},\dfrac{-3}{3})$

$\therefore$ the centroid of the triangle is

$(2,2,-1)$

The coordinates of a point which is equidistant from the point

$(0,0,0),(a,0,0),(0,b,0)$ and

$(0,0,c)$ are given by

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$\displaystyle \left( \frac { a }{ 2 } ,\frac { b }{ 2 } ,\frac { c }{ 2 } \right)$
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$\displaystyle \left( \frac { -a }{ 2 } ,\frac { -b }{ 2 } ,\frac { c }{ 2 } \right)$
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$\displaystyle \left( \frac { a }{ 2 } ,\frac { -b }{ 2 } ,\frac { -c }{ 2 } \right)$
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$\displaystyle \left( \frac { -a }{ 2 } ,\frac { b }{ 2 } ,\frac { -c }{ 2 } \right)$

Explanation

Let

$P(x,y,z)$ be the required point.

Then

$OP=PA=PB=PC$

Now

$OP=PA \Longrightarrow OP^2+PA^2 \Longrightarrow x^2+y^2+z^2$

$(x-a)^2+(y-0)^2+(z-0)^2 \Longrightarrow x=\dfrac{a}{2}$

Similarily,

$OP=PB=y=\dfrac{b}{2}$ and

$OP=PC=z=\dfrac{c}{2}$

hence the coordinate of the required point are

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

$A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C= \left ( \sqrt{2},\sqrt{2},0 \right )$ and

$D= \left ( \displaystyle \frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17} \right )$ , then

$ABCD$ is a

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rhombus
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square
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parallelogram
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none of these

Explanation

Given points are,

$A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C= \left ( \sqrt{2},\sqrt{2},0 \right )$ and

$D= \left ( \displaystyle \frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17} \right )$

Length

$AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$
Length

$BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$
Length

$CD = \sqrt{\left( \dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$
Length

$AD = \sqrt{\left( \dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$

Angle between two vectors

$\overline{AB} = p_{1}\hat{i}+q_{1}\hat{j}+r_{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$ and

$\overline{BC} = p_{2}\hat{i}+q_{2}\hat{j}+r_{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$ is

$cos\theta = 0 \Rightarrow \theta = 90^o$

All sides are equal and angle between

$AB$ and

$BC$ is

$90^o$ , Hence, its a square.

The equation of motion of a rocket are:

$x=2t,y=-4t,z=4t,$ where the time

$t$ is given in seconds and the coordinate of a moving point in kilometers. At what distance will the rocket be from the starting point

$O(0,0,0)$ in

$10$ seconds ?

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$60$ km
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$30$ km
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$45$ km
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None of these

Explanation

Eliminating t from the given equation, we get the equation of the path

$\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t$

Thus the path of the Rocket represents a straight line passing through the origin for

$t=10sec.$

we have

$x=20,y=-40,z=40$

Let

$\vec r=x\vec i+y\vec j+z\vec k$

$\Longrightarrow |\vec r|=\sqrt{{x^2}+{y^2}+{z^2}}=\sqrt{400+1600+1600}=60km$

What is the distance in space between

$(1,0,5)$ and

$(-3,6,3)$ ?

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$4$
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$6$
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$2\sqrt { 11 }$
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$2\sqrt { 14 }$
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$12$

Explanation

The distance between the points

$(1,0,5)$ and

$(-3,6,3)$ is given below:

$D=\sqrt { \left( 3-5 \right) ^{ 2 }+\left( 6-0 \right) ^{ 2 }+\left( -3-1 \right) ^{ 2 } }$

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

$=\sqrt { 4+36+16 } =\sqrt { 56 }$

$=2\sqrt { 14 }$

A triangle

$ABC$ is placed so that the midpoints of its sides are on the

$x, y$ and

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

$\displaystyle \alpha ,\beta ,\gamma$ , then the coordinates of the centroid of the triangle

$ABC$ are

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$\displaystyle \left (- \dfrac {\alpha }{3}, \dfrac {\beta }{3}, \dfrac {\gamma }{3} \right )$
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$\displaystyle \left (\dfrac {\alpha }{3},- \dfrac {\beta}{3}, \dfrac {\gamma}{3}\right)$
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$\displaystyle \left ( \dfrac {\alpha }{3},\dfrac { \beta }{3},-\dfrac {\gamma }{3} \right )$
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$\displaystyle \left ( \dfrac {\alpha }{3}, \dfrac {\beta }{3},\dfrac {\gamma }{3} \right )$

Explanation

Equation of the plane containing the triangle

$ABC$ is

$\displaystyle \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$
which meets the axes in

$\displaystyle \left ( \alpha ,0,0 \right ),\left ( 0,\beta ,0 \right )$ and

$\displaystyle \left ( 0,0,y \right )$ .
Let the coordinates of

$A$ be

$\displaystyle \left (x_1,y_1,z_1 \right )$
Since the middle point of

$AB$ lies on the

$z$ -axis it is

$\displaystyle \left (0,0,\gamma \right )$ and thus the coordinates of

$B$ are

$\displaystyle \left (-x_1,-y_1,2\gamma -z_1 \right )$
Similarly the coordinates of

$C$ are

$\displaystyle \left (-x_1,2\beta -y_1,-z_1 \right )$
So that the middle point of

$\displaystyle BC=\left ( -x_1,\beta -y_1,\gamma -z_1 \right ) \displaystyle =\left ( \alpha ,0,0 \right )$

$\displaystyle \Rightarrow x_1=-\alpha ,y_1=\beta ,z_1=\gamma$ .
And thus the coordinates of

$A$ are

$\displaystyle \left ( -\alpha ,\beta ,\gamma \right )$
Similarly the coordinates of

$B$ are

$\displaystyle \left ( \alpha ,-\beta ,\gamma \right )$ and those of

$C$ are

$\displaystyle \left ( \alpha ,\beta ,-\gamma \right )$ .
Hence, the coordinates of the centroid of the triangle

$ABC$ are

$\displaystyle \left ( \dfrac {\alpha }{3},\dfrac {\beta }{3}, \dfrac {\gamma }{3} \right )$

Point A is

$\displaystyle a+2b,$ and a divides AB in the ratio 2 :The position vector of B is

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$\displaystyle 2a-b$
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$\displaystyle b-2a$
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$\displaystyle a-3b$
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b

Explanation

Let us consider x be the position vector of B,

then a divides AB in the ratio

$2 : 3$

$a$ =

$\dfrac {2x +3(a +2b)}{2+3}$

$\Rightarrow x= a - 3b$

$D(2, 1, 0), E(2, 0, 0), F(0, 1, 0)$ are mid point of the sides

$BC, CA, AB$ of

$\Delta$

$ABC$ respectively, The the centroid of

$\Delta$ ABC is

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$\left ( \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$
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$\left ( \displaystyle \frac{4}{3},\, \displaystyle \frac{2}{3},\, 0 \right )$
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$\left (- \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$
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$\left ( \displaystyle \frac{2}{3},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{1}{3} \right )$

Explanation

Centroid of triangle coincide with the centroid of triangle formed by joing the mid-point of sides of triangle

So, centroid of

$\triangle ABC =$ centroid of

$\triangle DEF$

$=\left(\dfrac{2+2+0}{3},\dfrac{1+0+1}{3},\dfrac{0+0+0}{3}\right)$

$=\left(\dfrac{4}{3},\dfrac{2}{3},0\right)$

$P(x,y,x)$ is a point on the line segment joining

$Q(2,2,4)$ and

$R(3,5,6)$ such that the projection of

$\overline { OP }$ on the axis are

$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 } ,\frac { 26 }{ 5 } ,$ respectively, then

$P$ divides

$QR$ in the ratio

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

Explanation

Since

$OP$ has projection

$\displaystyle \frac { 13 }{ 5 } ,\frac { 19 }{ 5 }$ and

$\displaystyle \frac { 26 }{ 5 }$ on the coordinate axis,

therefore

$\displaystyle OP=\frac { 13 }{ 5 }\vec i+\frac { 19 }{ 5 } \vec j+\frac { 26 }{ 5 } \vec k$ .
Suppose

$P$ divides the join of

$Q(2,2,4)$ and

$R(3,5,6)$ in the ratio

$\lambda :1$ .
Then, the position vector of

$P$ is

$\displaystyle \left( \frac { 3\lambda +2 }{ \lambda +1 } \right)\vec i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \vec j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \vec k$

$\displaystyle \therefore \frac { 13 }{ 5 } \vec i+\frac { 19 }{ 5 }\vec j+\frac { 26 }{ 5 }\vec k=\left( \frac { 3\lambda +2 }{ \lambda +1 } \right) \vec i+\left( \frac { 5\lambda +2 }{ \lambda +1 } \right) \vec j+\left( \frac { 6\lambda +4 }{ \lambda +1 } \right) \vec k$

$\displaystyle \Rightarrow \frac { 3\lambda +2 }{ \lambda +1 } =\frac { 13 }{ 5 } ;\frac { 5\lambda +2 }{ \lambda +1 } =\frac { 19 }{ 5 } ;\frac { 6\lambda +4 }{ \lambda +1 } =\frac { 26 }{ 5 }$

$\displaystyle \Rightarrow 2\lambda =3$

$\Rightarrow \lambda =\dfrac { 3 }{ 2 }$

There are three points with position vectors

$-2a+3b+5c, a+2b+3c$ and

$7a-c$ . What is the relation between the three points?

0%
Collinear
0%
Forms a triangle
0%
In different plane
0%
None of the above

The plane

$ax+by+cz+d=0$ divides the line joining the points

$\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 } \right)$ in the ratio

0%
$\displaystyle \frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) }$
0%
$\displaystyle \frac { \left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) }$
0%
$\displaystyle \frac { a{ x }_{ 1 }{ x }_{ 2 }+b{ y }_{ 1 }{ y }_{ 2 }+c{ z }_{ 1 }{ z }_{ 2 } }{ { d }^{ 2 } }$
0%
None of these

Explanation

Let the given plane meet the line joining the given points in

$\left( { x }_{ 3 },{ y }_{ 3 },{ z }_{ 3 } \right)$ .
Then

$a{ x }_{ 3 }+b{ y }_{ 3 }+c{ z }_{ 3 }+d=0$ ...(1)
Also let the point

$\left( { x }_{ 3 },{ y }_{ 3 },{ z }_{ 3 } \right)$ divide the line joining the given points in the ratio

$m:n$
Then

$\displaystyle { x }_{ 3 }=\frac { m{ x }_{ 1 }+n{ x }_{ 2 } }{ m+n } ,{ y }_{ 3 }=\frac { m{ y }_{ 1 }+n{ y }_{ 2 } }{ m+n }$ and

$\displaystyle { z }_{ 3 }=\frac { m{ z }_{ 1 }+n{ z }_{ 2 } }{ m+n }$

Substituting these values in (1), we get

$\displaystyle a\left( \frac { m{ x }_{ 1 }+n{ x }_{ 2 } }{ m+n } \right) +b\left( \frac { m{ y }_{ 1 }+n{ y }_{ 2 } }{ m+n } \right) +c\left( \frac { m{ z }_{ 1 }+n{ z }_{ 2 } }{ m+n } \right) +d=0$

$\Rightarrow a\left( m{ x }_{ 1 }+n{ x }_{ 2 } \right) +b\left( m{ y }_{ 1 }+n{ y }_{ 2 } \right) +c\left( m{ z }_{ 1 }+n{ z }_{ 2 } \right) +d\left( m+n \right) =0$

$\Rightarrow m\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) +n\left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) =0$

$\displaystyle\Rightarrow \frac { n }{ m } =\frac { -\left( a{ x }_{ 1 }+b{ y }_{ 1 }+c{ z }_{ 1 }+d \right) }{ \left( a{ x }_{ 2 }+b{ y }_{ 2 }+c{ z }_{ 2 }+d \right) }$

Hence, option A.

The set of points in space 4 inches from a given line and 4 inches from a given point on this line is ______ , if given point lies on the given line.

0%
a set consisting of two points
0%
a set consisting of four points
0%
a set consisting of two circles
0%
the empty set

Let

$\displaystyle a,b,c \epsilon R$ such that

$abc = p$ and

$qa-b = 0$ , where

$p$ and

$q$ are fixed positive number, then minimum distance of the point

$(a, b, c)$ from origin in the three dimensional coordinate system is:

0%
$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{2q} \right )^{1/3}$
0%
$\displaystyle \sqrt{3}\left ( \frac{p(q^{2}+1)}{q} \right )^{1/3}$
0%
$\displaystyle \sqrt{3}(p)^{1/3}$
0%
$\displaystyle \sqrt{2}\left ( \frac{p}{q} \right )^{1/2}$

Explanation

We know that the distance between point

$(a,b,c)$ and origin

$(0,0,0)$ is:

$d= \sqrt({a^2+b^2+c^2)}$ .

Using the inequality:

$AM\geq GM$ , for terms

$a^2, b^2$ and

$c^2$ :

$\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge \sqrt [ 3 ]{ { (abc) }^{ 2 } }$ .

$abc = p$ (Given)

$\implies \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } \ge { (p) }^{ \frac{2}{3} }$ .

Taking square root on both sides of the inequality:

$\sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } \ge \sqrt { 3 } { p }^{ 1/3 }$ .

Hence, Option C is correct.

CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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