MATHS M.C.Q (1 Marks)

3. -3 : 7

Solution:

Let the plane divide the line in the ratio p : 1

A point that divides the line joining these 2 points in the ratio p : 1 is

given by $\Big(\frac{6\text{p}+2}{\text{p}+1},\frac{7\text{p}+3}{\text{p}+1},\frac{\text{p}+1}{\text{p}+1}\Big)$

Since, this point has to lie on the zx plane, so, 7p + 3 = 0 

$\Rightarrow\text{p}=\frac{-3}{7}$

1. $\frac{1}{3},\frac{1}{6},1$

2. $\sqrt{34}$

3. $\sqrt{(\text{b}^2 + \text{c}^2)}$

Solution:

The coordinate of the foot of the perpendicular from P on x-axis are (a, 0, 0).

So, the required distance $= \sqrt{{(\text{a – a})^2 + (\text{b – 0})^2 + (\text{c – 0})^2}}$

$=\sqrt{(\text{b}^2 + \text{c}^2)}$

1. $\mid{\text{r}}\mid = 5$

4. a + 3b - 4c

Solution:

a - 2b + 3c 

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

$\overrightarrow{\text{c}}=\frac{2\overrightarrow{\text{b}}+\overrightarrow{\text{a}}}{3}$

$\overrightarrow{\text{b}}=\frac{3\overrightarrow{\text{c}}-\overrightarrow{\text{a}}}{2}$

$=\frac{\big(3\overrightarrow{a}+4\overrightarrow{b}-5\overrightarrow{c}\big)-\big(\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}\big)}{2}$

$=\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$

4. $\frac{1}{3}$

Solution:

The plane XOZ divides the join of (1, -1, 5) and (2, 3, 4) in the ratio $\lambda:1$ i.e. y = 0 divide the join of (1, -1, 5) and (2, 3, 4) in the ratio.

$\lambda:1\therefore\frac{3\lambda−1}{\lambda+1}=0$

$\Rightarrow\lambda=\frac{1}{3}$

4. $\frac{-11}{3}$

Solution:

Given, C is a point of trisection of AB.

C either divides AB in the ratio 2 : 1 or 1 : 2

Case 1: C divides in the ratio 2 : 1

The coordinates of C will be $\Big(\frac{8}{3},-\frac{11}{3},\frac{13}{3}\Big)$

Case 2: C divides in the ratio 1 : 2 

The coordinates of C will be $\Big(\frac{7}{3},-\frac{10}{3},\frac{8}{3}\Big)$

either C_y = $\frac{-11}{3}\text{ or}-\frac{10}{3}$

4. 3 : 2

Solution:

B divides AC in the ratio is x₁​ - x₂ ​: x₂ ​- x_3​

_{$3-\frac{33}{5}:\frac{33}{5}-9$}

$3:2$

4. $\frac{\pi}{2}$

Solution:

PQ² = (4 - 0)² + (-2 - 1)² + (1 - 2)² = 16 + 9 + 1 = 26

OP² = (0 - 0)² + (1 - 0)² + (2 - 0)² = 0 + 1 + 4 = 5

QO² = (0 - 4)² + (0 + 2)² + (0 - 1)² = 16 + 1 + 4 = 21

Since, PQ² = OP² + QO²

Hence, $\angle\text{POQ}=\frac{π}{2}$

2. $\Big(\frac{6}{4},1,\frac{2}{4}\Big)$

Solution:

Angular points of tetrahedron OABC are.

(0, 0, 0), (0, 0, 2), (0, 4, 0), (6, 0, 0) To find the coordinates of the centroid of the tetrahedron whose vertices are

(x₁​, y₁​, z₁​), (x₂​, y₂​, z₂​), (x₃​, y₃​, z₃​) and (x₄​, y₄​, z_4​) the centroid is

$\Big(\frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}+\text{x}_{4}}{4}\Big),\Big(\frac{\text{y}_{1}+\text{y}_{2}+\text{y}_{3}+\text{y}_{4}}{4}\Big),\Big(\frac{\text{z}_{1}+\text{z}_{2}+\text{z}_{3}+\text{z}_{4}}{4}\Big)$

Now, substituting the values we get

$\Big(\frac{0+0+0+6}{4}\Big),\Big(\frac{0+0+4+0}{4}\Big),\Big(\frac{0+2+0+0}{4}\Big)$

$\therefore$ The coordinates of the centroid are $\Big(\frac{6}{4},1,\frac{2}{4}\Big)$

2. VII

Solution:

Here all the three x, y, z coordinate are negative of the given point.

$\therefore$ it will lie in the seventh Octant.

2. 3

Solution:

$\text{d}^2=(−1−3)^2+(2−4)^2+(3+1)^2$

$\Rightarrow\text{d}^2=(−4)^2+(−2)^2+(4)^2$

$\Rightarrow\text{d}^2=16+4+16$

$\Rightarrow\text{d}^2=36$

$\Rightarrow\text{d}^2=6$

Hence, radius of the sphere is 3 units.

1. $\Big(\frac{13}{5},0,-2\Big)$

2. $\frac{11}{3}$

1. $\frac{8\sqrt{21}}{3}$

Solution:

Given, A = (1, 2, -3), G (-3, 4, 5)

$\therefore\text{AG}=\sqrt{(-3-1)^2+(4-2)^2+(5-(-3))^2}$

$\text{and AG}=\sqrt { 84 } =2\sqrt { 21 }$

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 AP = AG + GP

⇒ AP = 3x + x

⇒ AP = 4x

$\Rightarrow\text{AP}=4\Big(\frac{2\sqrt2}{3}\Big)=\frac{8\sqrt21}{3}$

2. -2

Solution:

Coordinate 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

$\Rightarrow2\times\frac{29}{20}-3\times\frac{31}{20}+5\times\frac{9\text{k}+33}{20}-2=0$

$\Rightarrow\text{k}=-2$

2. (2, 2, -1)

Solution:

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)

4. None of these

Solution:

Suppose:

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.

1. $\sqrt{\text{a}^2+\text{b}^2}$

Solution:

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)

4. $\frac{4}{3}$

Solution:

Given, points are (1, -1, 5) and (2, 3, 4) since ZX-plane divides the line segment in the ratio p : 1, y-coordinate will be 0 the y-coordinate of the point dividing the line segment will be.

$=\frac{3\text{p}-1}{\text{p} + 1}=0,\text{ p}=\frac{1}{3}\text{ p}+1=\frac{1}{3}+1=\frac{4}{3}$

2. $\sqrt{77}$

Solution:

Origin is O(0, 0, 0) and given point is A(4, 5, 6)

So, distance $=\sqrt{(4-0)^2+(5-0)^2+(6-0)^2}$

$=\sqrt{4^2+5^2+6^2}=\sqrt{77}$

2. Octant II

Solution:

Given, (-5, 4, 3) is the point

Here, the x-coordinate is negative but y and z coordinates are positive

$\therefore$ (-5, 4, 3) lie in octant II.