Coplanar Lines MCQ Quiz - Objective Question with Answer for Coplanar Lines - Download Free PDF

Concept:

If the 2 lines  and  are coplanar, then 

Calculation:

Given equations of the coplanar lines are 

 and 

x₁ = 1, y₁ = - 1, z₁ = 0 and x₂ = - 2, y₂ = 1, z₂ = 2

∵ The lines are coplanar so,

-3(6 + λ) - 2(4 + 1) + 2(2λ - 3) = 0

λ  - 34 = 0

λ = 34

Concept:

Co-planar Vectors and Scalar Triple Product:

• Three vectors are co-planar if their scalar triple product is zero. The scalar triple product of three vectors A, B, and C is given by:
  A · (B × C) = 0
• In this case, we have the vectors:
  A = 6i + 4j + k,
  B = i + 4j + 2k,
  C = 7i + Xj + 3k.
• To find the value of X, we compute the scalar triple product and set it equal to zero, since the vectors are co-planar.
• First, calculate the cross product of B and C:
  • B × C = |i j k|
    |1 4 2|
    |7 X 3|
  • Expand the determinant:
    B × C = (12 - 2X)i - (3 - 14)j + (X - 28)k = (12 - 2X)i + 11j + (X - 28)k.
• Now, calculate the dot product of A with B × C:
  • A · (B × C) = (6i + 4j + k) · [(12 - 2X)i + 11j + (X - 28)k]
  • A · (B × C) = 6(12 - 2X) + 4(11) + 1(X - 28)
  • After simplifying:
    A · (B × C) = 72 - 12X + 44 + X - 28 = 88 - 11X.
• For the vectors to be co-planar, A · (B × C) = 0:
  • 88 - 11X = 0
  • X = 88 / 11 = 8.

Calculation:

Given vectors A = 6i + 4j + k, B = i + 4j + 2k, and C = 7i + Xj + 3k.

First, calculate B × C, then compute A · (B × C). Set A · (B × C) = 0 and solve for X:

A · (B × C) = 88 - 11X = 0 → X = 8.

∴ The value of X is 8.

∴ Hence, the correct answer is: Option 2 (X = 8).

Calculation

Two planes are coplanar if

Applying C₂ → C₂ + C₁, C₃ → C3 + C₁

⇒ 1[2 + 2k - (k + 2)(1 − k)] = 0

⇒ 2 + 2k - (-k² - k + 2) = 0

k² + 3k = 0 ⇒ k(k + 3) = 0

k = 0 or k = -3 

Hence option 4 is correct

The correct answer is: 1.49

Calculation: To find the distance between two parallel lines of the form (ax + by + c₁= 0) and (ax + by + c₂ = 0), we use the formula:

• Given the two lines:
• We can rewrite them in the standard form:
• Here, (a = 2), (b = 5), (c₁ = -7), and (c₂ = -15).

Step-by-Step Calculation:

• Substitute the given values into the formula:
• Simplify the numerator: [ |-15 + 7| = |-8| = 8 ]
• Calculate the denominator:
• Compute the distance:
• Rationalize the denominator:
• Determine the approximate value by calculation:

Rounded off to 2 decimal places:

Concept:

The scalar triple product of three vectors is given by the determinant

Explanation:

This translates to finding the determinant of the matrix formed by the

components of these vectors

Hence,

=0

⇒  = 0 

⇒  = 0

⇒  = 0

⇒ 

So the correct option is option 2.

Concept:

If the 2 lines  and  are coplanar, then 

Calculation:

Given equations of the coplanar lines are 

 and 

x₁ = 1, y₁ = - 1, z₁ = 0 and x₂ = - 2, y₂ = 1, z₂ = 2

∵ The lines are coplanar so,

-3(6 + λ) - 2(4 + 1) + 2(2λ - 3) = 0

λ  - 34 = 0

λ = 34

Concept:

Let a pair of linear equation in two variable a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

The condition of parallel lines or inconsistent equations.

Calculation:

Given equation of lines

3x + 2ky = 2 and 2x + 5y + 1 = 0

a1 = 3; a2 = 2

b1 = 2k; b2 = 5

c1 = -2; c2 = 1

Here, given lines are parallel

So that,

∴ 

Additional Information

(I) If 

Then the graph will be a pair of lines interesting at a unique point. Which is the solution of the pair of equations.

(II) If 

 then

Then graph will be a pair of coincident lines

Concept:Coplanar:

 Lines are said to be coplanar if they lie in the same plane.

If the two lines  and  are coplanar, then 

∴ Option 2 is correct 

Concept:

Let two line represented by ax2 + 2hxy + by2 = 0 are

y = m1x

y = m2x

Where, 

      ----(i)

And       ----(ii)

Let θ be angle between two lines

∴ using (i) and (ii)

  ----(iii)

Calculation:

Given equation is:

y2 - xy - 6x2 = 0 

Compare it with standard equation:

ax2 + 2hxy + by2 = 0

b = 1, a = - 6, h = -1/2

From equation (1);

∴ θ = 45°

Concept:

The equation of a line with a direction ratio (1, 1, 2) that passes through the point (4, 2, k) is given by the formula

Equation of plane in 3-D: 2x - 4y + z  = 7,

Since the given line is lying on the given plane so the point through which the line is passing ie (4, 2, k) will also lie on the given plane.

Calculation:

Given:

Here, equation of plane 2x - 4y + z  = 7,

Equation of the straight line 

Put the point  (4, 2, k) on the given plane as it will also lie on this plane.

⇒ 2×4 - 4×2 + k  = 7

⇒ k = 7

So, the correct answer will be option 2.

Concept:

If the 2 lines  and  are coplanar, then 

Calculation:

Given equations of the coplanar lines are 

 and 

x₁ = 1, y₁ = - 1, z₁ = 0 and x₂ = - 2, y₂ = 1, z₂ = 2

∵ The lines are coplanar so,

-3(6 + λ) - 2(4 + 1) + 2(2λ - 3) = 0

λ  - 34 = 0

λ = 34

Concept:

Let a pair of linear equation in two variable a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

The condition of parallel lines or inconsistent equations.

Calculation:

Given equation of lines

3x + 2ky = 2 and 2x + 5y + 1 = 0

a1 = 3; a2 = 2

b1 = 2k; b2 = 5

c1 = -2; c2 = 1

Here, given lines are parallel

So that,

∴ 

Additional Information

(I) If 

Then the graph will be a pair of lines interesting at a unique point. Which is the solution of the pair of equations.

(II) If 

 then

Then graph will be a pair of coincident lines

Concept:Coplanar:

 Lines are said to be coplanar if they lie in the same plane.

If the two lines  and  are coplanar, then 

∴ Option 2 is correct 

Concept:

Let two line represented by ax2 + 2hxy + by2 = 0 are

y = m1x

y = m2x

Where, 

      ----(i)

And       ----(ii)

Let θ be angle between two lines

∴ using (i) and (ii)

  ----(iii)

Calculation:

Given equation is:

y2 - xy - 6x2 = 0 

Compare it with standard equation:

ax2 + 2hxy + by2 = 0

b = 1, a = - 6, h = -1/2

From equation (1);

∴ θ = 45°

Concept:

Co-planar Vectors and Scalar Triple Product:

• Three vectors are co-planar if their scalar triple product is zero. The scalar triple product of three vectors A, B, and C is given by:
  A · (B × C) = 0
• In this case, we have the vectors:
  A = 6i + 4j + k,
  B = i + 4j + 2k,
  C = 7i + Xj + 3k.
• To find the value of X, we compute the scalar triple product and set it equal to zero, since the vectors are co-planar.
• First, calculate the cross product of B and C:
  • B × C = |i j k|
    |1 4 2|
    |7 X 3|
  • Expand the determinant:
    B × C = (12 - 2X)i - (3 - 14)j + (X - 28)k = (12 - 2X)i + 11j + (X - 28)k.
• Now, calculate the dot product of A with B × C:
  • A · (B × C) = (6i + 4j + k) · [(12 - 2X)i + 11j + (X - 28)k]
  • A · (B × C) = 6(12 - 2X) + 4(11) + 1(X - 28)
  • After simplifying:
    A · (B × C) = 72 - 12X + 44 + X - 28 = 88 - 11X.
• For the vectors to be co-planar, A · (B × C) = 0:
  • 88 - 11X = 0
  • X = 88 / 11 = 8.

Calculation:

Given vectors A = 6i + 4j + k, B = i + 4j + 2k, and C = 7i + Xj + 3k.

First, calculate B × C, then compute A · (B × C). Set A · (B × C) = 0 and solve for X:

A · (B × C) = 88 - 11X = 0 → X = 8.

∴ The value of X is 8.

∴ Hence, the correct answer is: Option 2 (X = 8).