Equation of a Line MCQ Quiz - Objective Question with Answer for Equation of a Line - Download Free PDF

Explanation:

Given:

L = x + y + z + 4 = 0 = 2x – y – z + 8 and

P = x + 2y + 3z + 1 = 0 is a plane.

To find the point of intersection of L and P,

The point must satisfy both line L and plane P and Option (d) (–4, –3, 3) will satisfy both line L and plane P.

So option (d) is the correct answer.

Explanation:

Given:

L = x + y + z + 4 = 0 = 2x – y – z + 8 and

P = x + 2y + 3z + 1 = 0 is a plane. 

⇒ Let direction ratios of line (a, b, c)

Since line is obtained by the intersection of planes.

⇒ x + y + z + 4 = 0 and 2x – y – z + 8 = 0

So a + b + c = 0....(i)

⇒ 2a – b – c = 0...(ii)

Solve (i) and (ii)

⇒

⇒ a = 0, b = 3λ  and c = –3λ 

⇒ Direction ratios of line is (0, 3, –3) or (0, 1, –1).

∴ Option (c) is correct.

Calculation

Equation of line through point (–1, 2,1) is →

⇒ 

So, 

By (1) → 

So, 

For intersection point ‘P’

x = 2λ – 1 = 3μ – 2

y = 3λ + 2 = 2μ + 3 

z = 4λ + 1 = µ + 4 

So, point P(x, y, z) = (1, 5, 5)

& Q(4, – 5, 1) 

∴ PQ = 

= 

Hence option 4 is correct

Concept:

Equation of line joining two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given as

A point lies on the line only when its coordinates satisfy the equation of the line.

Calculation:

Given:

P = (5,y,z)

The equation of line joining (3,2,-1) and (6,-4,-2) is 

so if point P lies on the line then it must satisfy the above equation

⇒ - 4 = y - 2

⇒ y =  - 2

Hence y coordinate of P is -2

Concept:

 The equation of a line with direction ratio (a, b, c) that passes through the point (x1, y1, z1) is given by the formula: 

If two planes intersect each other, the intersection will always be a line.

Calculation:

Given:

x = 3z + 4 and y = 2z - 3

⇒ x - 4 = 3z and y + 3 = 2z

If two planes intersect each other, the intersection will always be a line.

Now,  direction ratios of the line are ⟨3, 2, 1⟩

Concept:

Equation of a line in cartesian form:- 

Equation of a line in vector form:-  

where,  and 

Calculation:

We have,

Cartesian equation:- 

⇒       ------- equation (1)

⇒            -------- equation (2)

Comparing (1) and (2)

⇒ x₁ = -5, y₁ = 7, z₁ = -3

⇒ a = 3, b = 2, c = 2

⇒ 

⇒ 

Equation of line in vector form:- 

⇒ 

∴ The vector equation for the line is  

Concept: 

1. Equation of line of slope m passing through (x₁, y₁) is given by

(y - y₁) = m(x - x₁)

2. If two line with slope m₁ and m₂ are perpendicular to each other, then m₁m₂ = -1

Calculation:

Slope of line passes through (0, 0) and (3, -4) is 

Let slope of perpendicular at (3, -4) is m₂, then

m₁m₂ = -1 ⇒ m₂ = 3/4

We know that, 

Equation of line of slope m passing through (x1, y1) is given by

(y - y1) = m(x - x1)

Hence, equation of line passing through (3, -4) and slope 3/4 is given by

⇒ 3x - 4y = 25

Hence, option 3 is correct.

Given:

 The straight line x + y + 3z - 2 = 0 = 2x + y - z - 3 is parallel to the plane 3x + 2y + kz - 4 = 0

Concept:

If a line parallel to the plane then dot product of direction ratio of line and normal to plane must be 0 

Calculation:

 The straight line x + y + 3z - 2 = 0 = 2x + y - z - 3 is parallel to the plane 3x + 2y + kz - 4 = 0

So, the direction ratios of normals to the planes are ( 1, 1, 3 ) and ( 2, 1, -1 ) repectively .

So direction ratio of the required line is the cross product of these normals

i.e.  

= i(-1 - 3) - j(-1 - 6) + k(1 - 2)

= - 4i + 7j - k

Therefore, direction ratio of line is ( -4, 7, -1)

Now line is parallel to 3x + 2y + kz - 4 = 0, So dot product of direction ratio of line and normal to plane must be 0 

∴ 3(-4) + 2(7) + k(-1) = 0

⇒ 2 - k = 0

⇒ k = 2

Hence the option (2) is correct.

Given:

The plane : 4x - 2y + 3z - 6 = 0

Concept:

For a plane having equation ax + by + cz = d(a, b, c) are direction ratios of the normal to the plane.

Equation of a line passing through (x₁, y₁, z₁) and having direction ratios a, b, c in the cartesian form :

Calculation:

Direction ratios of normal to the plane are 4, -2, and 3. 

Equation of a line passing through the origin (0, 0, 0) and having direction ratios 4, -2, 3 is:

⇒ x = 4λ , y = -2λ and z = 3λ

Satisfying equation of the plane with above coordinates :

⇒ 4(4λ) - 2(-2λ) + 3(3λ) - 6 = 0

⇒ λ = 6/29

∴ Foot of the perpendicular,

⇒ x = 24/29, y = -12/29 and z = 18/29

Concept:

Equation of the line in 3D:

If a line passes through a point  and having direction cosines  then the equation of the line is given by:

Direction cosines of the line equally inclined to co-ordinate axes:

If a line is equally inclined to the co-ordinate axes then the direction cosines of the line are 

Calculation:

It is given that the line is equally inclined to the co-ordinate axes.

Therefore, the direction cosines of the line are 

It is given that the line passes through (-3, 2, -5).

Therefore, we have  and .

Thus, the equation of the line is given as follows:

CONCEPT:

• The direction ratios of the line joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by: a = x₂ - x₁, b = y₂ - y₁ and c = z₂ - z₁
• If a, b, c are the direction ration ratios of a line passing through the point (x₁, y₁, z₁), then the equation of line is given by: 

CALCULATION:

Given: The line joining the points (k, 1, 3) and (1, -2, k + 1) also passes through the point (15, 2, -4)

As we know that, the direction ratios of the line joining the points (x1, y1, z1) and (x2, y2, z2) is given by: a = x2 - x1, b = y2 - y1 and c = z2 - z₁

So, the direction ratios of the line joining the points (k, 1, 3) and (1, -2, k + 1) is: a = 1 - k, b = - 3 and c = k - 2

As we know, if a, b, and c are the direction ratio ratios of a line passing through the point (x1, y1, z1), then the equation of a line is given by: 

So, the equation of the line with direction ratios a, b, c and passing through the point (k, 1, 3) is given by: 

∵ It is given that, the line represented by  also passes through the point (15, 2, -4)

So, substitute x = 15, y = 2 and z = - 4 in the equation 

⇒ 

⇒ 

⇒ - 45 + 3k = 1 - k

⇒ k = 23/2

⇒ 

⇒ k - 2 = 21

⇒ k = 23

So, there are two possible values of k as shown above.

Hence, correct option is 3

Calculation:

Let (h, k) be the coordinates of the foot of the perpendicular from the point (2, 3) on the line 4x - 5y + 8 = 0.

The slope of the perpendicular line will be 

The slope of the line 4x - 5y + 8 = 0

⇒ 4x - 5y + 8 = 0

⇒ -5y = -4x - 8

⇒ y = 4/5x + 8/5

On comparing the above equation with y = mx + c we get m = 4/5

∴ the slope of the line is 4/5

Using the condition of perpendicularity of lines,

4k - 12 = - 5h + 10

4k + 5h = 22       ----(i)

Since (h, k) lies on the given line 4x - 5y + 8 = 0,

⇒ 4h - 5k + 8 = 0

⇒ 4h - 5k = - 8       ----(ii)

on solving the equations (i) and (ii) using elimination method, we get h = 78/41 and k = 128/41

Therefore, are the coordinates of the foot of the perpendicular from the point (2, 3) on the line 4x - 5y +8 = 0.

Concept:

The cartesian equation of a line passing through A (x1, y1, z1) and B (x2, y2, z2) is given by: 

Calculation:

Given: A (2, - 3, 1) and B (3, - 4, - 5) are two points

Let x₁ = 2, y₁ = - 3, z₁ = 1, x₂ = 3, y₂ = - 4 and z₂ = - 5.

As we know that, equation of a line passing through A (x1, y1, z1) and B (x2, y2, z2) is given by: 

So, the required equation of line is: