Class 11th

13. Let P divides AB in ratio k : 1. Then co-ordinates of point P are

$(\frac{k(-1)+1(2)}{k+1},\frac{k(2)+1(-3)}{k+1},\frac{k(1)+1(4)}{k+1})$

= $\left(\frac{-k+2}{k+1},\frac{2k-3}{k+1},\frac{k+4}{k+1}\right)$

Let us examine whether the value of k, the point P coincides with point C

Putting $\frac{-k+2}{k + 1}=0$

=> $-k+2=0$

=> $k=2$

Put $k=2$ in

$\frac{2k - 3}{k + 1}$

= $\frac{2\left(2\right) - 3}{2 + 1}$

= $\frac{4 - 3}{3}$

= $\frac{1}{3}$

And put $k=2$ in

$\frac{k + 4}{k + 1}$

= $\frac{2 + 4}{2 + 1}$

= $\frac{6}{3}$

= 2

Therefore, C $\left(0,\frac{1}{3},2\right)$ is a point which divides AB internally in ratio 2 : 1 and is same as P. Hence A, B and C are collinear.

12. Let YZ-plane divides the line segment joining A (–2, 4, 7) and B (3, –5, 8) at point P (x, y, z) in the ratio k : 1.

Then the co-ordinates of P are.

$\left(\frac{k(3)+1(-2)}{k+1},\frac{k(-5)+1(4)}{k+1},\frac{k(8)+1(7)}{k+1}\right)$

$\left(\frac{3k-2}{k+1},\frac{-5k+4}{k+1},\frac{8k+7}{k+1}\right)$

As P lies on YZ-plane its x-coordinate is zero.

i.e. $\frac{3k-2}{k + 1}=0$

=> $3k-2=0$

=> $k=\frac{2}{3}$

Hence the YZ-plane divides AB internally in ratio.

$\frac{2}{3}$ : 1 = 2 : 3

11. Let point Q divides PR in the k : 1. Then co-ordinate of Q will be

$(\frac{k(9)+1(3)}{k+1},\frac{k(8)+1(2)}{k+1},\frac{k(-10)+1(-4)}{k+1})$

=> (5, 4, –6) = $·\left(\frac{9k+3}{k+1}·,·\frac{8k+2}{k+1}·,·\frac{-10k-4}{k+1}·\right)$

Equating the co-ordinates we get,

$\frac{9k+3}{k + 1}$ = 5

=> $9k+3=5\left(k+1\right)$

=> $9k+3=5k+5$

=> $9k-5k=5-3$

=> $4k=2$

=> $k=\frac{2}{4}$

=> $k=\frac{1}{2}$

Putting $k=\frac{1}{2}$ in y-coordinate and z-coordinate

$\frac{8k + 2}{k + 1}$

= $\frac{\left(8 x\frac{1}{2}\right) + 2}{\frac{1}{2} + 1}$

= (4 + 2) ÷ $\frac{3}{2}$

= 6 x $\frac{2}{3}$

= 4

And

$\frac{\left(-1\mathrm{0x}\frac{1}{2} \right) - 4}{\frac{1}{2} + 1}$

= (–5 – 4) ÷ $\frac{3}{2}$

= – 9 x $\frac{2}{3}$

= – 6

Which is matching with the given co-ordinates of Q.

Hence, the ration in which Q divides PR is k : 1

= $\frac{1}{2}$ : 1

= 1 : 2

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.