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PUCET (Punjab) Surface Areas and Volumes — practice questions

Explore 28 free Surface Areas and Volumes questions in PUCET (Punjab) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Surface Areas and Volumes questions and exercises with answers

Showing questions 1–28 of 28.

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensional space is:Read the answer and worked solution →The direction cosines of a line are $l$, $m$, $n$. They satisfy:Read the answer and worked solution →A line passes through the points $(1, 2, 3)$ and $(4, 6, 9)$. The direction ratios of the line are:Read the answer and worked solution →In 3D, distance between points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):Read the answer and worked solution →Direction cosines (l, m, n) of a line satisfy:Read the answer and worked solution →Equation of line through point (x₀, y₀, z₀) with direction (a, b, c):Read the answer and worked solution →Two lines are parallel if their direction ratios are:Read the answer and worked solution →Normal vector to plane ax + by + cz = d is:Read the answer and worked solution →Distance from point (1, 2, 3) to plane x + y + z = 0:Read the answer and worked solution →Find equation of plane through (1, 0, 0), (0, 2, 0), (0, 0, 3):Read the answer and worked solution →Two planes are perpendicular if:Read the answer and worked solution →Vector form of line: r = a + λ b where:Read the answer and worked solution →Coplanar lines have:Read the answer and worked solution →Equation of sphere with center (h, k, l) and radius r:Read the answer and worked solution →Angle between line and plane: if direction of line is L and normal to plane is N, sin(angle line-plane):Read the answer and worked solution →Projection of vector a on b:Read the answer and worked solution →Skew lines in 3D are:Read the answer and worked solution →Distance between two parallel planes 2x + y + z = 5 and 2x + y + z = 10:Read the answer and worked solution →Find equation of plane perpendicular to (1, 2, 3) passing through (4, 5, 6):Read the answer and worked solution →Angle θ between two planes 2x + 3y + 4z = 0 and x + y + z = 0:Read the answer and worked solution →Three points (1,0,0), (0,1,0), (0,0,1) are vertices of a triangle. Area:Read the answer and worked solution →Plane passing through origin and perpendicular to line (x-1)/2 = (y-2)/3 = (z-3)/-1:Read the answer and worked solution →Foot of perpendicular from point (3, 1, 2) to plane x + y + z = 6:Read the answer and worked solution →Direction ratios of normal to plane 2x - y + 3z + 7 = 0:Read the answer and worked solution →For a line with direction cosines l, m, n, the angles it makes with axes are:Read the answer and worked solution →Coordinates of foot of perpendicular from point (1, 2, 3) to x-axis:Read the answer and worked solution →Equation of straight line passing through two points A(1,2,3) and B(4,5,6):Read the answer and worked solution →3D analog of Pythagoras: for a box with sides a, b, c, the body diagonal is:Read the answer and worked solution →

Worked practice questions

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensional space is:

  1. $\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2 + (z_2 + z_1)^2}$
  2. $(x_2 - x_1)(y_2 - y_1)(z_2 - z_1)$
  3. $|x_2 - x_1| + |y_2 - y_1| + |z_2 - z_1|$
  4. $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
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Correct answer: D. $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

3D distance formula extends Pythagoras into three dimensions: the distance is the square root of the sum of squared coordinate differences. Derivation: project the segment $PQ$ onto each axis to get the leg lengths $|x_2 - x_1|$, etc. Apply Pythagoras twice. Option B is the **Manhattan** (taxicab) distance, useful in some lattice problems but not the Euclidean answer.

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The direction cosines of a line are $l$, $m$, $n$. They satisfy:

  1. $l^2 + m^2 + n^2 = 1$
  2. $l \cdot m \cdot n = 1$
  3. $l + m + n = 1$
  4. $l^2 + m^2 + n^2 = 0$
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Correct answer: A. $l^2 + m^2 + n^2 = 1$

Direction cosines are the cosines of the angles that a line makes with the three coordinate axes: $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$. Key identity: $l^2 + m^2 + n^2 = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. This just says the unit vector along the line has unit length when projected onto the three axes.

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A line passes through the points $(1, 2, 3)$ and $(4, 6, 9)$. The direction ratios of the line are:

  1. $(4, 6, 9)$
  2. $(3, 4, 6)$
  3. $(5, 8, 12)$
  4. $(1, 2, 3)$
Show answer and solution

Correct answer: B. $(3, 4, 6)$

Direction ratios of the line joining two points are the differences in coordinates: $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. Here: $(4 - 1, 6 - 2, 9 - 3) = (3, 4, 6)$. Direction ratios are not unique. $(6, 8, 12)$ or $(1.5, 2, 3)$ would also be valid since multiplying by any non-zero scalar preserves the direction. The standard answer just uses the raw differences. Direction cosines (the normalised version) are $(3, 4, 6) / \sqrt{9 + 16 + 36} = (3, 4, 6) / \sqrt{61}$.

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