Class 11th

10. i. Let P(x, y, z) be the point which divides line segment joining (–2, 3, 5) and (1, –4, 6) internally in the ratio 2 : 3

Therefore,

x = $\frac{2\left(1\right) + 3\left(-2\right)}{2 + 3}$ = $\frac{2 - 6}{5}$ = $\frac{-4}{5}$

y = $\frac{2\left(-4\right) + 3\left(3\right)}{2 + 3}$ = $\frac{-8 + 9}{5}$ = $\frac{1}{5}$

z = $\frac{2\left(6\right) + 3\left(5\right)}{2 + 3}$ = $\frac{12 + 15}{5}$ = $\frac{27}{5}$

Thus, the required points are $\left(\frac{-4}{5},\frac{1}{5},\frac{27}{5}\right)$

ii. Let P(x, y, z) be the point which divides line segment joining (–2, 3, 5) and (1, –4, 6) externally in the ratio 2 : 3

Therefore,

x = $\frac{2\left(1\right) - 3\left(-2\right)}{2 - 3}$ = $\frac{2 + 6}{-1}$ = –8

y = $\frac{2\left(-4\right) - 3\left(3\right)}{2 - 3}$ = $\frac{-8 - 9}{-1}$ = 17

z = $\frac{2\left(6\right) - 3\left(5\right)}{2 - 3}$ = $\frac{12 - 15}{-1}$ = 3

Thus, the required points are (–8, 17, 3).

9. Let P have the co-ordinates (x, y, z).

Given that,

PA + PB = 10

8. Let P(x, y, z) be the point equidistant from the given points (1, 2, 3) say A and (3, 2, –1) say B.

So, PA = PB

=>  ${\left(1-x\right)}^2+{\left(2-y\right)}^2+{\left(3-z\right)}^2$ = ${\left(3-x\right)}^2+{\left(2-y\right)}^2+{\left(-1-z\right)}^2$

Squaring both sides,

=> ${\left(1-x\right)}^2+{\left(2-y\right)}^2+{\left(3-z\right)}^2$ = ${\left(3-x\right)}^2+{\left(2-y\right)}^2+{\left(-1-z\right)}^2$

=> ${\left(1-x\right)}^2+{\left(3-z\right)}^2={\left(3-x\right)}^2+[{\left(-\right)}^2{\left(1+z\right)}^2$

=> $\left(1^2+x^2-2x\right)+\left(3^2+z^2-2.3.z\right)=\left(3^2+x^2-2.3.x\right)+(1^2+z^2+2z)$

=> $1+x^2-2x+9+z^2-6z=9+x^2-6x+1+z^2+2z$

=> $6x-2x-6z-2z=0$

=> $4x-8z=0$

=> $x-2z=0$

Therefore, the required equation of point is $x-2z=0$

5. i. The distance between points (2, 3, 5) and (4, 3, 1) is

4. i. The x-axis and y-axis taken together determine a plane known as XY plane.

ii. The coordinates of points in the XY-plane are of the form (x, y, 0).

iii. Coordinate planes divide the space into 8 (eight) octants.

3. 

(1, 2, 3) lies in octant I.

(4, –2, 3) lies in octant IV.

(4, –2, –5) lies in octant VIII.

(4, 2, –5) lies in octant V.

(–4, 2, –5) lies in octant VI.

(–4, 2, 5) lies in octant II.

(–3, –1, 6) lies in octant III.

(–2, –4, –7) lies in octant VII.

2. When a point lies on XZ-plane, its y-coordinate is zero.

1. When a point lies on x-axis, its y-coordinate and z-coordinate are zero.

(c) Given, f(x)=(x-a)(x-b)

where a and b are constants.

So,

$f^{\prime }(x)=(x-a)\frac{d}{dx}(x-b)+(x-b)\frac{d}{dx}(x-a)$

=(x-a)+(x-b)

= 2x– a- b. $$

(ii) Given f(x)= ${\left(a{x}^2+b\right)}^2.$ where ab are constant

So, $f^{\prime }(x)=\frac{d}{dx}{\left(a{x}^2+b\right)}^2$

$=\frac{d}{dx}\left(a^2{x}^4+b^2+2a{x}^{2}b\right)$

$=\frac{d}{dx}{a}^2{x}^4+\frac{d}{dx}{b}^2+\frac{d}{dx}2a{x}^{2}b$

= $4a^2{x}^3+0+4axb$

= $4ax\left(a{x}^2+b\right).$

(iii) Given, f(x)= $\frac{x-9}{x-b}$ where a and bare constants

So, $f(x)=\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{{(x-b)}^2}$

$=\frac{(x-b)-(x-a)}{{(x-b)}^2}$

$=\frac{a-b}{{(x-b)}^2}$