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Introduction To 3D Coordinate Geometry — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSIntroduction To 3D Coordinate Geometry5 Marks
Question
Determine the points zx-plane equidistant from the points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).
✓

Answer

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

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