Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
Find the incentre of the triangle whose vertices are $ O (0,0,0), \quad A(4,0,0) \text { and } \quad B (0,0,3) $.
- $(-1, 0, -1) $
- $(1, 0, -1) $
- $ (1, 0, 1) $
- $(-1, 0, 1) $
Correct

Incorrect

Hint

Given, $O (0,0,0), \quad A(4,0,0), \quad B (0,0,3)$
$OA=\sqrt{(0-4)^2+0^2+0^2}=4=b$
$OB=\sqrt{0^2+0^2+(0-3)^2}=3=a$
$AB=\sqrt{(4-0)^2+0^2+(0-3)^2}=\sqrt{4^2+3^2}=5=0$
$\begin{aligned} I&=\left( \dfrac{ax_1+bx_2+cx_3}{a+b+c},\dfrac{ay_1+by_2+cy_3}{a+b+c},\dfrac{az_1+bz_2+cz_3}{a+b+c} \right)\\&=\left( \dfrac{3\times 4+0+0}{3+4+5},\dfrac{3\times 0+4\times0+5\times0}{3+4+5},\dfrac{3\times 0+4 \times 3+5\times 0}{3+4+5} \right)\\&=\left( \dfrac{12}{12},0,\dfrac{12}{12} \right)=(1,0,1) \end{aligned}$
Question 2 of 25

2. Question
1 point(s)
If $ A(0,0,0), \quad B (0,12,0) \text { and } C (0,0,5) $ are vertices of a triangle $ ABC $, then find the incentre of $\triangle ABC $.
- $(0,2,-2) $
- $(0,2,2) $
- $ (0,-2,2) $
- $ (0,-2,-2) $
Correct

Incorrect

Hint

$AB=\sqrt{(0-0)^2+(0-12)^2+(0-0)^2}=12=c$
$BC=\sqrt{(0-0)^2+(12-0)^2+(0-5)^2}=13=a$
$CA=\sqrt{(0-0)^2+(0-0)^2+(0-5)^2}=5=b$
$\therefore a=13, \quad b=5, \text { and } c=12$
$\begin{aligned} I&=\left( \dfrac{ax_1+by_1+cz_1}{a+b+c},\dfrac{ax_2+by_2+cz_2}{a+b+c},\dfrac{ax_3+by_3+cz_3}{a+b+c} \right)\\&=\left( \dfrac{13 \times0+5 \times 0+12\times0}{13+5+12},\dfrac{13 \times 0+5\times 12+12 \times 0}{13+5+12}, \dfrac{13 \times 0+5 \times 0+12\times 5}{13+5+12} \right)\\&=\left( 0, \dfrac{60}{30}, \dfrac{60}{30} \right)=(0,2,2) \end{aligned}$
Question 3 of 25

3. Question
1 point(s)
Find the excentre with respect to $ A $ where $A (0,0,0), \quad B (0,12,0) \text { and } C (0,0,5) $ are coordinates of $ \triangle ABC$.
- $(0 ,15,15) $
- $(0,15,-15) $
- $(0,-15,-15) $
- $ (0,-15,15) $
Correct

Incorrect

Hint

$AB=\sqrt{0^2+12^2+0^2}=12=c$
$BC=\sqrt{0^2+12^2+(-5)^2}=13=a$
$AC=\sqrt{0^2+0^2+5^2}=5=b$
Excentre with respect to $A$ is
$\begin{aligned} I_1&=\left( \dfrac{-ax_1+by_1+cz_1}{-a+b+c},\dfrac{-ax_2+by_2+cz_2}{-a+b+c},\dfrac{-ax_3+by_3+cz_3}{-a+b+c} \right)\\&=\left( \dfrac{-13\times 0+5\times 0+12 \times 0}{-13+5+12},\dfrac{-13 \times 0+5\times 12+12 \times 0}{-13+5+12},\dfrac{-13\times 0+0 \times 12+12\times 5}{-13+5+12} \right)\\&=\left( 0, \dfrac{60}{4}, \dfrac{60}{4} \right) =(0,15,15) \end{aligned}$
Question 4 of 25

4. Question
1 point(s)
If $A(0,0,0), B(0,12,0) \text { and } C (0,0,5) $ are vertices of $ \triangle ABC $, then find the excentre w.r.to $ B $.
- $ (0,3,3) $
- $ (0,-3,3) $
- $ (0,3,-3) $
- $ (0,-3,-3) $
Correct

Incorrect

Hint

$AB=\sqrt{0^2+12^2+0^2}=12=c$
$AB=\sqrt{0^2+12^2+(-5)^2}=13=a$
$AC=\sqrt{0^2+0^2+5^2}=5=b$
$\begin{aligned} I_2&=\left( \dfrac{ax_1-by_1+cz_1}{a-b+c}, \dfrac{ax_2-by_2+cz_2}{a-b+c},\dfrac{ax_3-by_3+cz_3}{a-b+c} \right)\\&=\left( \dfrac{13 \times 0-5 \times 0+12 \times 0}{13-5+12}, \dfrac{13 \times 0-5 \times 12+12 \times 0}{13-5+12},\dfrac{13 \times 0- 0\times 12+12 \times 5}{13-5+12} \right) \\&= \left( 0, \dfrac{-60}{20}, \dfrac{60}{20} \right)=(0,-3,3) \end{aligned}$
Question 5 of 25

5. Question
1 point(s)
Find the excentre with respect to $ C $ where $ A (0,0,0), B (0,12,0) \text { and } C (0,0,5) $ are vertices of $ \triangle ABC $.
- $ (0,10,10) $
- $ (0,-10,-10) $
- $(0,-10,10) $
- $(0,10,-10) $
Correct

Incorrect

Hint

$AB=12 =c$
$BC=13=a$
$AC=5=b$
$\therefore a=13,\quad b=5,\quad c=12$
$\begin{aligned} I_3&=\left( \dfrac{ax_1+by_1-cz_1}{a+b-c},\dfrac{ax_2+by_2-cz_2}{a+b-c},\dfrac{ax_3+by_3-cz_3}{a+b-c} \right)\\&=\left( \dfrac{13 \times 0+5 \times 0-12 \times 0}{13+5-12},\dfrac{13 \times 0+5 \times 12-12 \times0}{13+5-12}, \dfrac{13 \times 0+0 \times 12-12\times 5}{13+5-12} \right)\\&=\left( 0, \dfrac{60}{6}, \dfrac{-60}{6} \right)=(0, 10,-10) \end{aligned}$
Question 6 of 25

6. Question
1 point(s)
In an square $ ABCD $, $ A (13,5,-2) \text { and } C (-3,7,-6) $ are opposite vertices. Find the coordinate of the intersection of two diagonals.
- $(5,6,-4) $
- $(5,6,4) $
- $ (5,-6, 4) $
- $(5,-6,-4) $
Correct

Incorrect

Hint

Given, $A= (13,5,-2) \text { and } C =(-3,7,-6)$
Let $'O'$ be the intersecting point of $\overline{AC} \text { and } \overline{BD}$ .
$\begin{aligned} \therefore O&=\left( \dfrac{13-3}{2},\dfrac{5+7}{2},\dfrac{-2-6}{2} \right) \\&=(5,6,-4)\end{aligned}$
Question 7 of 25

7. Question
1 point(s)
If $ O (0,0,0) $ is the intersecting point of a diagonal $ ABCD $, where $A (5,7,a) \text { and } C (b,-7,10) $ are opposite vertices, then find $ a \text { and } b $.
- $ a=5,\quad b=10 $
- $ a=10, \quad b=5 $
- $a=-5, \quad b=-10 $
- $a=-10, \quad b=-5 $
Correct

Incorrect

Hint

$O=\left( \dfrac{5+b}{2}, \dfrac{7-7}{0}, \dfrac{a+10}{2} \right)$
$\implies \dfrac{5+b}{2}=0 \implies b=-5$
$\dfrac{a+10}{2}=0 \implies a=-10$
Question 8 of 25

8. Question
1 point(s)
If $ A(5,-7,8) \text { and } C (2,-3,3) $ are opposite vertices of a square $ ABCD $, then find the length of its side.
- 2
- 7
- 5
- $5 \sqrt {2} $
Correct

Incorrect

Hint

Given, $A= (5,-7,8), \quad C=(2,-3,3)$
$\begin{aligned} AC&=\sqrt{(5-2)^2+(-7+3)^2+(8-3)^2} \\&=\sqrt{3^2+4^2+5^2}=\sqrt{50}=5 \sqrt{2} \end{aligned}$
$AB=BC=CD=AD=\dfrac{AC}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5$
$\therefore$ The length of its side is $5$ .
Question 9 of 25

9. Question
1 point(s)
The point $ (-2,3,4) $ lies in what octant?
- $ 2 \text {nd} $
- $ 4 \text {th} $
- $7 \text {th} $
- $ 5 \text {th} $
Correct

Incorrect

Hint

In $2 \text { nd}$ octant $\left( x < 0, y > 0, z >0 \right)$
Question 10 of 25

10. Question
1 point(s)
The point $ (3,-5,7) $ lies in what octant?
- $ 2 \text {nd} $
- $ 4 \text {th} $
- $5 \text {th} $
- $6 \text {th} $
Correct

Incorrect

Hint

In $4 \text {th}$ octant, $\left( x > 0, y < 0, z >0 \right)$
Question 11 of 25

11. Question
1 point(s)
The point $ (3,-7,-1) $ lies on what octant?
- $ 2 \text {nd} $
- $ 5 \text {th} $
- $ 8 \text {th} $
- $ 7 \text {th} $
Correct

Incorrect

Hint

In $8 \text {th}$ octant, $\left( x > 0, y < 0, z < 0 \right)$
Question 12 of 25

12. Question
1 point(s)
The point $(-7,5,-9) $ lies in what octant?
- $ 2 \text {nd} $
- $ 6 \text {th} $
- $ 5 \text {th} $
- $ 8 \text {th} $
Correct

Incorrect

Hint

In $6 \text {th}$ octant, $\left( x < 0, y > 0, z < 0 \right)$
Question 13 of 25

13. Question
1 point(s)
If the point $ P $ lies on $z $-axis, then the coordinates of $P $ are the form______
- $(0,0,z) $
- $ (x,y,z) $
- $ (x,y,0) $
- None of these
Correct

Incorrect

Hint

On $z$ -axis, $x=0, \quad y=0$ .
So, the required cooridnates of the form are $(0,0,z)$
Question 14 of 25

14. Question
1 point(s)
A point $ P $ lie on $x $-axis, then find the coordinates of $ P $ are of the form_____
- $(x,y,z) $
- $(0,y,z) $
- $(x,0,0) $
- None of these
Correct

Incorrect

Hint

On $x$ -axis, $y=0, \quad z=0$ .
So, the required cooridnates of the form are $(x,0,0)$
Question 15 of 25

15. Question
1 point(s)
A point $ P $ lie on $y $-axis, then find the coordinates of $ P $ are of the form_____
- $(x,0,z) $
- $(0,y,0) $
- $(x,y,z) $
- None of these
Correct

Incorrect

Hint

On $y$ -axis, $x=0, \quad z=0$ .
So, the required cooridnates of the form are $(0,y,0)$
Question 16 of 25

16. Question
1 point(s)
The plane $ x=a $ represent a plane parallel to_____
- $xy $ -plane
- $ yz $ -plane
- $zx $-plane
- None of these
Correct

Incorrect

Hint

On $yz$ -plane, $x=0$ .
Hence $x=0$ and $x=c$ are parallel to each other.
Question 17 of 25

17. Question
1 point(s)
The plane $ y=b $ represent a plane parallel to_____
- $xy $ -plane
- $ yz $ -plane
- $zx $-plane
- None of these
Correct

Incorrect

Hint

On $zx$ -plane, $y=0$ .
Hence $y=0$ and $y=b$ are parallel to each other.
Question 18 of 25

18. Question
1 point(s)
The plane $ z=c $ represents a plane parallel to_____
- $xy $ -plane
- $ yz $ -plane
- $zx $-plane
- None of these
Correct

Incorrect

Hint

On $xy$ -plane, $z=0$ .
$\therefore z=0 \text { and } z=c$ are parallel to each other.
Question 19 of 25

19. Question
1 point(s)
A line is parallel to $ x $-axis, if all the points on the line have equal______
- $ x \text { and } y $ coordinates
- $ y \text { and } z $ coordinates
- $ x \text { and } z $ coordinates
- None of these
Correct

Incorrect

Hint

On $x$ -axis, $y$ and $z$ coordinates are equal $(0)$ .
So any line parallel to $x$ -axis have equal $y$ and $z$ coordinates.
Question 20 of 25

20. Question
1 point(s)
A line is parallel to $ y $-axis, if all the points on the line have equal______
- $ x \text { and } y $ coordinates
- $ y \text { and } z $ coordinates
- $ z \text { and } x $ coordinates
- None of these
Correct

Incorrect

Hint

On $y$ -axis, $x$ and $z$ coordinates are zero (equal). So any line parallel to $y$ -axis have equal $x$ and $z$ coordinates.
Question 21 of 25

21. Question
1 point(s)
A line is parallel to $ z $-axis, if all the points on the line have equal____
- $ x \text { and } y $ coordinates
- $ y \text { and } z $ coordinates
- $ z \text { and } x $ coordinates
- None of these
Correct

Incorrect

Hint

On $z$ -axis, $x$ and $y$ coordinates are zero i.e., equal. So any line parallel to $z$ -axis have equal $x$ and $y$ cooridnates.
Question 22 of 25

22. Question
1 point(s)
The plane parallel to $ yz $-plane is perpendicular to ________
- $x $ -axis
- $ y $ -axis
- $z $ -axis
- None of these
Correct

Incorrect

Hint

In space $yz$ -plane is perpendicular to $x$ -axis.
So every plane parallel to $yz$ -plane also perpendicular to $x$ -axis.
Question 23 of 25

23. Question
1 point(s)
The plane parallel to $ xy $-plane is perpendicular to ________
- $x $ -axis
- $ y $ -axis
- $z $ -axis
- None of these
Correct

Incorrect

Hint

In space $z$ -axis is perpendicular to $xy$ -plane. So every plane parallel to $xy$ -plane also perpendicular to $x$ -axis.
Question 24 of 25

24. Question
1 point(s)
The plane parallel to $ zx $-plane is perpendicular to________
- $x $ -axis
- $ y $ -axis
- $z $ -axis
- None of these
Correct

Incorrect

Hint

In space, $y$ -axis is perpendicular to $zx$ -plane. So every plane parallel to $zx$ -plane also is perpendicular to $y$ -axis.
Question 25 of 25

25. Question
1 point(s)
If $ A (0,-1,-7), \quad B (2,1,-9) \text { and } C (6,5,-13)$ are three points on a line, then find the ratio in which $ B $ divides $ \overline{AC} $.
- $ 2:1 $
- $ 1:2 $
- $ 1:3 $
- $3:1 $
Correct

Incorrect

Hint

$\begin{aligned} AB&=\sqrt{(2-0)^2+(1+1)^2+(-9+7)^2}\\&=\sqrt{4+4+4}=2\sqrt{3} \end{aligned}$
$\begin{aligned} BC&=\sqrt{(6-2)^2+(5-1)^2+(-13+9)^2}\\&=\sqrt{16+16+16}=4\sqrt{3} \end{aligned}$
$\dfrac{AB}{BC}=\dfrac{2\sqrt{3}}{4\sqrt{3}}=\dfrac{2}{4}=\dfrac{1}{2}$
$\therefore B$ divides $AC$ in $1:2$ ratio.