MATHS 2 Marks Questions

 Here points are A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7).

$\text{AB}=\sqrt{(3-0)^2+(3-6)^2+(3-3)^2}$

$=\sqrt{9+9}$

$=3\sqrt{2}\text{ units}$

$\text{BC}=\sqrt{(0-1)^2+(6-7)^2+(3-7)^2}$

$=\sqrt{1+1+16}$

$=3\sqrt{2}\text{ units}$

$\text{AC}=\sqrt{(3-1)^2+(3-7)^2+(3-7)^2}$

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

$=6\text{ units}$

$\text{BD}=\sqrt{(0-4)^2+(6-4)^2+(3-7)^2}$

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

$=6\text{ units}$

$\text{CD}=\sqrt{(1-4)^2+(7-4)^2+(7-7)^2}$

$=\sqrt{9+9}$

$=3\sqrt{2}\text{ units}$

$\text{AD}=\sqrt{(3-4)^2+(3-4)^2+(3-7)^2}$

$=\sqrt{1+1+16}$

$=3\sqrt{2}\text{ units}$ 

Since, AB = BC = CD = DA

And AC = BC

So,

A, B, C, D are vertices of a square.

Tri section points are those which divide line in ratio 1 : 2 or 2 : 1

P(4, 2, -6) and Q(10, -16, 6)

Consider 1 : 2 case, we get

$\Big(\frac{10+8}{3},\frac{-16+4}{3},\frac{6–12}{3}\Big)= (6, -4, -2)$

Consider 2 : 1 case, we get

$\Big(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3}\Big)=(8, -10, 2)$

(6, -4, -2) and (8, -10, 2) are tri section points.

(4, 8, 10) and (6, 10, -8) is divided by the yz-plane.

Equation of yz-plane is x = 0

assume ratio is m : n

Equating x-term, we get

$0=\frac{\text{6m+4n}}{\text{m+n}}$

m : n=-2 : 3

So YZ plane divides the line segment in ratio 2 : 3 externally

Distance between points P and Q

$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

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

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

$=\sqrt{16+4+1}$

$\text{PQ}=\sqrt{21}\text{ units}$

Distance between points A and B

$\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

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

$=\sqrt{(4)^2+(3)^2+(0)^2}$

$=\sqrt{16+9+0}$

$=\sqrt{25}$

$\text{AB = 5 units}$

Distance between points P and Q

$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

$=\sqrt{(1-2)^2+(-1-1)^2+(0-2)^2}$

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

$=\sqrt{1+4+4}$

$\text{PQ = 3 units}$