MATHS 5 Marks Questions

Let R(x, 0, z) be the required point.

So

(AR)² = (BR)² ⇒ (1 - x)² + (-1 - 0)² + (0 - z)² = (2 - x) + (1 - 0)² + (2 - z)² 

⇒ 1 + x² - 2x + 1 + z² = 4 + x² - 4x + 1 + z² + 4z

⇒ 2x + 4z = 7 ... (i)

(BR)² = (CR)² ⇒ (z - z)² + (1 - 0)² + (2 - z)² = (3 - x)² + (2 - 0)² (-1 - z)²

⇒ 4 + x² - 4x + 4 + z² - 4z = 9 + x² - 6x + 4 + 1 + z² + 2z

⇒ 2x - 6z = 5 ... (ii)

(AR)² = (CR)² ⇒ (1 - x)² + (1 - 0)² + (0 - z)² = (3 - x)² + (2 - 0)² + (-1 - z)²

⇒ 1 + x² - 2x + 1 + z² = 9 + 6x + 4 + 1 + z²

⇒ 4x - 2z = 12 ... (iii)

Solving equation (i) and (ii) we get

$\text{z}=\frac{1}{5},\ \text{x}=\frac{31}{10}$

Put the value of x and z in equation (iii)

4x - 2z = 12

$4\Big(\frac{31}{10}\Big)-2\Big(\frac{1}{5}\Big) = 12$

$\frac{124}{10}+\frac{2}{10}=12$

$\frac{120}{10}=12$

$12=12$

LHS = RHS.

so,

$\text{x}=\frac{31}{10},\ \text{z}=\frac{1}{5}$

Required point $=\Big(\frac{31}{10},\ 0,\ \frac{1}{5}\Big)$

A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

$\text{AB}=\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}$

$=\sqrt{(4)^2+(-5)^2+(-7)^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}\text{ units}$

$\text{BC}=\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}$

$=\sqrt{(-8)^2+(10)^2+(14)^2}$

$=\sqrt{64+100+196}$

$=\sqrt{360}$

$=6\sqrt{10}\text{ units}$

$\text{CA}=\sqrt{(3-7)^2+(-5+10)^2+(1+6)^2}$

$=\sqrt{(-4)^2+(5)^2+(7)^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}\text{ units}$

Since, AB + AC = BC

so, A, B and C are collinear.

 Let A = (0, 7, 10), B = (-1, 6, 6) and C = (-4, 9, 6)

$\text{AB}=\sqrt{(0+1)^2+(7-6)^2+(10-6)^2}$

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

$=\sqrt{18}$

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

$\text{BC}=\sqrt{(-1+4)^2+(6-9)^2+(6-6)^2}$

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

$=\sqrt{18}$

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

$\text{AC}=\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}$

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

$=\sqrt{36}$

$=6\text{ units}$

$(\text{AB})^2+\text{(BC)}^2$

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

$=18+18$

$=36$

$=\text{(AC)}^2$

Also (AB) = (BC)

Hence (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle. 

A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

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

$=\sqrt{(4-5)^2+(-3+7)^2+(-1-6)^2}$

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

$=\sqrt{1+16+49}$

$=\sqrt{66}\text{ units}$

$\text{BC}=\sqrt{(5-3)^2+(-7-1)^2+(6+8)^2}$

$=\sqrt{(2)^2+(-8)^2+(14)^2}$

$=\sqrt{4+64+196}$

$=\sqrt{264}$

$=2\sqrt{66}\text{ units}$

$\text{AC}=\sqrt{(4-3)^2+(-3-1)^2+(-1+8)^2}$

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

$=\sqrt{1+16+49}$

$=\sqrt{66}\text{ units}$

Since, AC + AB = BC

so, A, B, C are collinear.

Assume ratio is $\lambda:1$

Plane is ax + by + cz + d = 0

points (x₁, y₁, z₁) and (x₂, y₂, z₂)

Assume point of intersection of line and plane is D

$\text{D}=\Big(\frac{\lambda\text{x}_2+\text{x}_1}{\lambda+1},\frac{\lambda\text{y}_2+\text{y}_1}{\lambda+1},\frac{\lambda\text{z}_2+\text{z}_1}{\lambda+1}\Big)$

As D lies on plane, substitute D in plane equation, we get

$\lambda(\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d})+\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}=0$

$\Rightarrow\lambda=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$

Let Q(0, y, z) be the required point.

So

(AQ)² = (BQ)² ⇒ (0 - 1)² + (y + 1)² - (z - 0)² = (0 - 2) + (y + 1)² + (z - 2)² 

⇒ 1 + y² + 1 + 2y + z² = 4 + y² + 1 - 2y + z² + 4 - 42

⇒ 4y + 4z = 7 ... (i)

(BQ)² = (CQ)² ⇒ (0 - z)² + (y - 1)² + (z - 2)² = (0 - 3)² + (y - 2)² (2 + 1)²

⇒ 4 + y² + 1 - 2y + z² + 4 - 4z - 9 + y² + 4 - 4y + z² + 1 + 2z

⇒ 2y - 6z = 5 ... (ii)

(AQ)² = (CQ)² ⇒ (0 - 1)² + (y + 1)² + (z - 0)² = (0 - 3)² + (y - 2)² (z + 1)²

⇒ 1 + y² + 2y + 1 + z² = 9 + y² - 4y + 4 + z² + 1 +2z

⇒ 6y - 2z = 12 ... (iii)

Solving equation (i) and (ii) we get

$\text{z}=\frac{-3}{16}$ and $\text{y} = \frac{31}{16}$

Put the value of y and z in equation (iii)

6y - 2z = 12 = 12

$6\Big(\frac{31}{16}\Big)-2\Big(\frac{-3}{16}\Big) = 12$

$\frac{186}{16}+\frac{6}{16}=12$

$\frac{192}{16}=12$

$12=12$

LHS = RHS.

so,

$\text{y}=\frac{31}{16},\ \text{z}=\frac{13}{16}$

Required point $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$

IMAGE

Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.

$\therefore$ PA = Distance between planes PBEC and QDAF

= 5 - 3

= 2

PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and -2, respectively, from the zx-plane.

$\therefore$ PB = 0 - (-2)

= 2

PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.

$\therefore$ PC = 2 - 5

= -3

 Here

$\text{AB}=\sqrt{(1+5)^2+(3-5)^2+(0-2)^2}$

$=\sqrt{36+4+4}$

$=\sqrt{44}$

$=2\sqrt{11}\text{ units}$

$\text{BC}=\sqrt{(-5+9)^2+(5+1)^2+(2-2)^2}$

$=\sqrt{16+36}$

$=\sqrt{52}$

$=2\sqrt{13}\text{ units}$

$\text{CD}=\sqrt{(-9+3)^2+(-1+3)^2+(2-0)^2}$

$=\sqrt{36+4+4}$

$=2\sqrt{11}\text{ units}$

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

$=\sqrt{16+36}$

$=\sqrt{52}$

$=2\sqrt{13}\text{ units}$

$\text{AC}=\sqrt{(1+9)^2+(3+1)^2+(0-2)^2}$

$=\sqrt{150+16+4}$

$=\sqrt{120}$

$=4\sqrt{5}\text{ units}$

$\text{BD}=\sqrt{(-3+5)^2+(-3-5)^2+(0-2)^2}$

$=\sqrt{4+64+4} $

$=\sqrt{72}$

$=6\sqrt{2}\text{ units}$ 

Since,

AB = CD and BC = DA

⇒ ABCD is a parallelgram = BD

but, $\text{AC}\neq\text{BD}$

⇒ ABCD is not a rectangles.

P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

$\text{PQ}=\sqrt{(0-1)^2+(7-4)^2+(-7+5)^2}$

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

$=\sqrt{1+9+4}$

$=\sqrt{14}\text{ units}$

$\text{QR}=\sqrt{(1+1)^2+(4-10)^2+(-5+9)^2}$

$=\sqrt{(2)^2+(-6)^2+(4)^2}$

$=\sqrt{4+36+16}$

$=2\sqrt{14}\text{ units}$

$\text{PR}=\sqrt{(0+1)^2+(7-10)^2+(-7+9)^2}$

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

$=\sqrt{1+9+4}$

$=\sqrt{14}\text{ units}$

Since, PQ + PR = QR

so, P, Q, R are collinear.

Here, 0(0, 0, 0), A(2, 0, 0), B(0, 3, 0), C(0, 0, 8)

Since, (OP)²  = (PA)²

(x - 0)² + (Y - 0)² + (z - 0)² = (x - 2)² + (y - 0)² + (z- 0)²

x² + y² + z² = x² - 4x + 4 + y² + z²

4x = 4

X = 1

(OP)² = (PB)²

(x - 0)² + (y - 0)² + (z - 0)² = (x - 0)² + (y - 3)² + (z - 0)²

x² + y² + z² = x² + y² - 6y + 9 + z²

6y = 9

$\text{y}=\frac{3}{2}$

(OP)² = (PC)²

(x - 0)² + (y - 0)² + (z - 0)² = (x - 0)² + (Y - 0)² + (z - 8)²

x² + y² + z² = x² + y² + z² - 16z + 64

16z = 64

z = 4

The required point $=\Big(1,\ \frac{3}{2},\ 4\Big)$ 

Let the point on xy-plane be P(x, y, 0).

Now P is equidistance from A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).

So, AP - BP - CP

Now,

(AP)² = (x - 1)² + (y + 1)² + (0 - 0)²

(BP)² = (x - 2)² + (y - 1)² + (0 - 2)²

(CP)² = (x - 3)² + (y - 2)² + (0 + 1)²

(AP)² = (BP)² ⇒ (x - 1)² + (y + 1)² - (x - 2)² + (y - 1)² + 4

⇒ x² + 1 - 2 + y² + 1 + 2y + z² - x² + 4 - 4x + y² + 1 - 2y + 4

⇒ 2x + 4y = 7 ... (i)

(BP)² = (CP)² ⇒ (x - 2)² + (y - 1)² + 4 = (x - 3)² + (y - 2)² + 1

⇒ x² + 4 - 4x + y² + 1 - 2y + z² + 4 = x² + 9 - 6x + y² + 4 - 4y +1

⇒ 2x + 2y = 5 ... (ii)

(AP)² = (CP)² ⇒ (x - 1)² + (y + 1)² + (x - 3)² + (y - 2)² + 1

⇒ x² + 1 - 2x + y² + 1 + 2y = x² + 9 - 6x + y² + 4 - 4y + 1

⇒ 4x + 5y = 12 ... (iii)

Solving equation (i) and (ii) we get

y = 1, $\text{x} = \frac{3}{2}$

Put x and y in equation (iii)

$4\Big(\frac{3}{2}\Big) + 6(1) = 12$

$12-12$

So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$