Equation of a plane Practice Questions - NDA Mathematics Question Bank

Question 1 of 10 Authentic UPSC NDA Question

Consider the following for the next three (03) items that follow: The plane 6x + ky + 3z-12 = 0 where k ≠ 0 meets the coordinate axes at A, B and C respectively.

The equation of the sphere passing through the

origin and A, B, C is x2 + y2 + z2 - 2x - 3y - 4z = 0. What is the value of k?

Step-by-Step Logic & Solution

Explanation:

Since, B(0, 12/k, 0) lies on sphere

∴ 0 + (12/k)² + 0 + 0 − 3(12/k) − 0 = 0

⇒ (12/k)[12/k − 3] = 0

⇒ k = 4

Question 2 of 10 Authentic UPSC NDA Question

For the following two (02) items:

A plane P is parallel to the line having direction ratios ⟨1, 3, 2⟩ and contains the line of intersection of the planes 6x + 4y - 5z = 2 and x - 2y + 3z = 0.

What is the equation of the plane P?

Step-by-Step Logic & Solution

Question 3 of 10 Authentic UPSC NDA Question

What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratio is (3, 2, 1)?

Step-by-Step Logic & Solution

Question 4 of 10 Authentic UPSC NDA Question

If p is the perpendicular distance from origin to the plane passing through (1, 0, 0), (0, 1, 0) and (0, 0, 1), then what is 3p² equal to?

Step-by-Step Logic & Solution

Question 5 of 10 Authentic UPSC NDA Question

If < l, m, n > are the direction cosines of a normal to the plane 2x – 3y + 6z + 4 = 0, then what is the value of 49(7l² + m² – n²)?

Step-by-Step Logic & Solution

Given the circle:

x² + y² + 2x + 6y + 1 = 0

To find the centre (a, b) and radius c, rewrite the equation in the form

(x − h)² + (y − k)² = r²

Completing the square:

(x² + 2x + 1) − 1 + (y² + 6y + 9) − 9 + 1 = 0

⇒ (x + 1)² − 1 + (y + 3)² − 9 + 1 = 0

⇒ (x + 1)² + (y + 3)² − 9 = 0

⇒ (x + 1)² + (y + 3)² = 9

⇒ [x − (−1)]² + [y − (−3)]² = 3²

Therefore, the centre is (−1, −3) and the radius is 3.

∴ a = −1, b = −3, c = 3

∴ a² + b² + c²
= (−1)² + (−3)² + 3²
= 1 + 9 + 9
= 19

Question 6 of 10 Authentic UPSC NDA Question

What is the equation of the plane which cuts an intercept 5 units on the z-axis and it parallel to xy-plane ?

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

The equation of a plane parallel to the xy-plane can be represented as:

\[ z = d \]

where \( d \) is the z-intercept. Given that the z-intercept is 5 units, we have:

\[ z = 5 \]

Thus, the equation of the plane is:

\[ z = 5 \]

📌 Hints / Properties Used:

Equation of a Plane: \( z = d \) for planes parallel to xy-axis.
Intercept on axis defines distance from origin.

📊 Visual Diagram Suggestion:

Illustrate a 3D Cartesian coordinate system highlighting the plane intersecting the z-axis at point (0, 0, 5) and parallel to the xy-plane.

📖 Factual Verification & Reference:

Verified against standard geometry principles related to planes and intercepts.

Question 7 of 10 Authentic UPSC NDA Question

A plane cuts intercepts 2, 2, 1 on the coordinate axes. What are the direction cosines of the normal to the plane ?

Step-by-Step Logic & Solution

Solution:

Plane cuts intercepts 2, 2, 1.

Equation of plane:

x/2 + y/2 + z/1 = 1

⇒ x + y + 2z = 2

Direction ratios of normal:

⟨1, 1, 2⟩

Magnitude:

√(1² + 1² + 2²)

= √6

Direction cosines:

⟨1/√6, 1/√6, 2/√6⟩

Hence option (c).

Hint:

Plane with intercepts a, b, c:

x/a + y/b + z/c = 1

Question 8 of 10 Authentic UPSC NDA Question

If p, q and r are the intercepts made by the plane on the coordinate axes respectively, then what is (p + q + r) equal to?

Step-by-Step Logic & Solution

Question 9 of 10 Authentic UPSC NDA Question

Consider the following for the next two (02) items that follow:

Let the plane 2x/k + 2y/3 + z/3 = 2 pass through the point (2, 3, −6).

What are the direction ratios of a normal to the plane?

Step-by-Step Logic & Solution

Question 10 of 10 Authentic UPSC NDA Question

The foot of the perpendicular drawn from the origin to the plane x + y + z = 3 is

Step-by-Step Logic & Solution

Equation of a plane Questions

Solution