Important Questions: Three Dimensional Geometry | Mathematics (Maths) Class 12 - JEE PDF Download

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Q1: Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are collinear.
Ans: We know that the direction ratios of the line passing through two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are given by:
x₂ – x₁, y₂ – y₁, z₂ – z₁ or x₁ – x₂, y₁ – y₂, z₁ – z₂
Given points are A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11).
Direction ratios of the line joining A and B are:
1 – 2, – 2 – 3, 3 + 4
i.e. – 1, – 5, 7.
The direction ratios of the line joining B and C are:
3 –1, 8 + 2, – 11 – 3
i.e., 2, 10, – 14.
From the above, it is clear that direction ratios of AB and BC are proportional.
That means AB is parallel to BC. But point B is common to both AB and BC.
Hence, A, B, C are collinear points.

Q2: Find the angle between the pair of lines given by

Ans:
From the given,

Let θ be the angle between the given pair of lines.

Q3: Find the distance between the lines l₁ and l₂ given by:

 
Ans: Given two lines are parallel.

The distance between the two given lines is


Q4: Find the intercepts cut off by the plane 2x + y – z = 5.
Ans: Given plane is 2x + y – z = 5 ……(i)
Dividing both sides of the equation (i) by 5,
(⅖)x + (y/5) – (z/5) = 1

We know that,
The equation of a plane in intercept form is (x/a) + (y/b) + (z/c) = 1, where a, b, c are intercepts cut off by the plane at x, y, z-axes respectively.
For the given equation,
a = 5/2, b = 5, c = -5
Hence, the intercepts cut off by the plane are 5/2, 5 and -5.

Q5: Find the direction cosines of the line passing through the two points (– 2, 4, – 5) and (1, 2, 3).
Ans: We know that the direction cosines of the line passing through two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are given by

Using the distance formula,

Hence, the direction cosines of the line joining the given two points are


Q6: If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
Ans: Let the direction cosines of the line be l, m, and n.
l = cos 90° = 0
m = cos 135° = -1/√2
n = cos 45° = 1/√2
Hence, the direction cosines of the line are 0, -1/√2, and 1/√2.

Q7: Find the angle between the pair of lines given below.
(x + 3)/3 = (y -1)/5 = (z + 3)/4
(x + 1)/1 = (y – 4)/1 = (z – 5)/2
Ans: Given,
(x + 3)/3 = (y -1)/5 = (z + 3)/4
(x + 1)/1 = (y – 4)/1 = (z – 5)/2
The direction ratios of the first line are:
a₁ = 3, b₁ = 5, c₁ = 4
The direction ratios of the second line are:
a₂ = 1, b₂ = 1, c₂ = 2

Hence, the required angle is


Q8: Show that the lines (x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
Ans: Given lines are:
(x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3
The direction ratios of the given lines are 7, -5, 1 and 1, 2, 3, respectively.
We know that,
Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular to each other if a₁a₂ + b₁b₂ + c₁c₂ = 0
Therefore, 7(1) + (-5) (2) + 1 (3)
= 7 – 10 + 3
=0
Hence, the given lines are perpendicular to each other.

Q9: Find the equations of the planes that passes through three points (1, 1, 0), (1, 2, 1), and (– 2, 2, – 1).
Ans: Given points are (1, 1, 0), (1, 2, 1), and (– 2, 2, – 1).

Therefore, the plane will pass through the given three points.
We know that,
The equation of the plane through the points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is

(x – 1)(-2) -(y – 1) (3) + z (3) = 0
-2x + 2 – 3y + 3 + 3z = 0
-2x – 3y + 3z + 5 = 0
-2x – 3y + 3z = -5
Therefore, 2x + 3y – 3z = 5 is the required Cartesian equation of the plane.