3D points Practice Questions - NDA Mathematics Question Bank

Question 1 of 10 Authentic UPSC NDA Question

Consider the points A(2, 4, 6), B(-2, -4, -2), C(4, 6, 4) andD(8, 14, 12).

Which of the following statements is/are correct? 1. The points are the vertices of a rectangle ABCD.

2. The mid-point of AC is the same as that of BD.

Select the correct answer using the code given below:

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Question 2 of 10 Authentic UPSC NDA Question

The number of spheres of radius r touching the coordinate axes is

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Question 3 of 10 Authentic UPSC NDA Question

The xy-plane divides the line segment joining the points (-1, 3, 4) and (2, -5, 6)

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Question 4 of 10 Authentic UPSC NDA Question

The locus of a point P(x, y, z) which moves in such a way that z = 7 is a

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Explanation : Given, point z = 7

Question 5 of 10 Authentic UPSC NDA Question

What is the number of possible values of k for which the line joining the points (k, 1, 3) and (1, - 2, k + 1) also passes through the point (15, 2, - 4)??

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Question 6 of 10 Authentic UPSC NDA Question

What is the projection of the line segment joining (1, 7, -5) and B (-3, 4, -2) on y-axis?

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Question 7 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

P(−2, −3, 5), Q(4, −1, 5), R(6, −4, 8) and S(2, −6, 10) are four points.

What is (PQ² + 2QS² − 2PR²) equal to?

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

To solve for \(PQ^2 + 2QS^2 - 2PR^2\), we compute the distances first. The distance formula in 3D: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] Calculating \(PQ\): \[ PQ = \sqrt{(4 - (-2))^2 + (-1 - (-3))^2 + (5 - 5)^2} \] \[ = \sqrt{(6)^2 + (2)^2 + 0} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \] \[ PQ^2 = (2\sqrt{10})^2 = 40 \] Calculating \(QS\): \[ QS = \sqrt{(2 - 4)^2 + (-6 - (-1))^2 + (10 - 5)^2} \] \[ = \sqrt{(-2)^2 + (-5)^2 + (5)^2} = \sqrt{4 + 25 + 25} = \sqrt{54} = 3\sqrt{6} \] \[ QS^2 = (3\sqrt{6})^2 = 54 \] Calculating \(PR\): \[ PR = \sqrt{(6 - (-2))^2 + (-4 - (-3))^2 + (8 - 5)^2} \] \[ = \sqrt{(8)^2 + (-1)^2 + (3)^2} = \sqrt{64 + 1 + 9} = \sqrt{74} \] \[ PR^2 = 74 \] Now substituting into the expression: \[ PQ^2 + 2QS^2 - 2PR^2 = 40 + 2 \cdot 54 - 2 \cdot 74 \] \[ = 40 + 108 - 148 = 0 \] Thus, the final answer is \(0\).

📌 Hints / Properties Used:

3D Distance Formula.
Properties of squares in distances.

📊 Visual Diagram Suggestion:

A 3D coordinate system diagram showing points P, Q, R, and S, with lines connecting them, would clarify position relationships and distances.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Mathematics, Chapter 11 (Conic Sections).

Question 8 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

P(−2, −3, 5), Q(4, −1, 5), R(6, −4, 8) and S(2, −6, 10) are four points.

Consider the following statements:

I. PQ is parallel to RS.

II. PR is perpendicular to QS.

Which of the statements given above is/are correct?

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

To determine the correctness of statements I and II, we first check the slopes of the line segments. 1. For line segment PQ: - Coordinates: P(-2, -3) and Q(4, -1) - Slope (m_PQ):

\[ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 + 3}{4 + 2} = \frac{2}{6} = \frac{1}{3} \]

2. For line segment RS: - Coordinates: R(6, -4) and S(2, -6) - Slope (m_RS):

\[ m_{RS} = \frac{-6 + 4}{2 - 6} = \frac{-2}{-4} = \frac{1}{2} \] \end{aligned} \

Since \( m_{PQ} \neq m_{RS} \), segment PQ is not parallel to segment RS. 3. For line segment PR: - Coordinates: P(-2, -3) and R(6, -4) - Slope (m_PR):

\[ m_{PR} = \frac{-4 + 3}{6 + 2} = \frac{-1}{8} \]

4. For line segment QS: - Coordinates: Q(4, -1) and S(2, -6) - Slope (m_QS):

\[ m_{QS} = \frac{-6 + 1}{2 - 4} = \frac{-5}{-2} = \frac{5}{2} \] \end{aligned} \

To check for perpendicularity, we establish that the product of the slopes of perpendicular lines is -1: \[ m_{PR} \cdot m_{QS} = \left(-\frac{1}{8}\right) \cdot \frac{5}{2} = -\frac{5}{16} \neq -1 \] Thus, segment PR is not perpendicular to segment QS. Both statements are incorrect.

📌 Hints / Properties Used:

Slope of a line: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
Two lines are parallel if their slopes are equal.
Two lines are perpendicular if the product of their slopes is -1.

📊 Visual Diagram Suggestion:

Utilize a coordinate plane to plot points P, Q, R, and S. Draw line segments PQ and RS, indicating their slopes, and label angles to show potential perpendicularity.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Mathematics, Coordinate Geometry.

Question 9 of 10 Authentic UPSC NDA Question

For the following two (02) items:

Let A(1, –1, 0), B(-2, 1, 8) and C(-1, 2, 7) are three consecutive vertices of a parallelogram ABCD.

If angle BCD is θ, then what is cos²θequal to?

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Question 10 of 10 Authentic UPSC NDA Question

For the following two (02) items:

Let A(1, –1, 0), B(-2, 1, 8) and C(-1, 2, 7) are three consecutive vertices of a parallelogram ABCD.

What is the fourth vertex D?

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3D points Questions

Solution

Solution