CBSE Class 11 Mathematics Three Dimensional Geometry Assignment Set 02

THREE DIMENSIONAL GEOMETRY

Any point on x-axis -> \((x, 0, 0)\)
Any point on y-axis -> \((0, y, 0)\)
Any point on z-axis -> \((0, 0, z)\)
Any point on XY-plane -> \((x, y, 0)\)
Any point on YZ-plane -> \((0, y, z)\)
Any point on ZX-plane -> \((x, 0, z)\)
Distance between two points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\) is:
\(|PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)
The coordinates of \(R\) which divides a line segment joining the points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\):
Internally in the ratio \(m : n\) is:
\(R\left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n}\right)\)
Externally in the ratio \(m : n\) is:
\(S\left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n}\right)\)
The coordinates of the centroid of the triangle whose vertices are \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\) and \((x_3, y_3, z_3)\) is:
\(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)\)

Extra Questions:

Question 1. Find the distance between (-3, 4, -6) and its image in the XY-plane.
Answer: Let the given point be \(P(-3, 4, -6)\).
The image of point \(P(x, y, z)\) in the XY-plane has the same x and y coordinates, but the z coordinate is negated.
Therefore, the image of \(P(-3, 4, -6)\) in the XY-plane is \(Q(-3, 4, 6)\).
Now, using the distance formula between \(P(-3, 4, -6)\) and \(Q(-3, 4, 6)\):
\(|PQ| = \sqrt{(-3 - (-3))^2 + (4 - 4)^2 + (6 - (-6))^2}\)
\(|PQ| = \sqrt{0 + 0 + (12)^2}\)
\(|PQ| = \sqrt{144} = 12 \text{ units}\)
In simple words: The reflection of a point across the XY-plane changes the sign of its z-coordinate, and the distance is simply twice the absolute value of the z-coordinate.

Exam Tip: Remember that the image in the XY-plane leaves x and y unchanged and changes z to -z. For the YZ-plane, x becomes -x, and for the ZX-plane, y becomes -y.

Question 2. Find the points on the y-axis which are at a distance of 3 units from the point (2, 3, -1).
Answer: Any point on the y-axis is of the form \(P(0, y, 0)\).
Let the given point be \(A(2, 3, -1)\).
Given that the distance between \(P\) and \(A\) is 3 units, so \(PA = 3\).
Squaring both sides, \(PA^2 = 9\).
Using the distance formula:
\((0 - 2)^2 + (y - 3)^2 + (0 - (-1))^2 = 9\)
\((-2)^2 + (y - 3)^2 + (1)^2 = 9\)
\(4 + (y - 3)^2 + 1 = 9\)
\(5 + (y - 3)^2 = 9\)
\((y - 3)^2 = 4\)
Taking square roots on both sides:
\(y - 3 = \pm 2\)
Case 1: \(y - 3 = 2 \implies y = 5\)
Case 2: \(y - 3 = -2 \implies y = 1\)
Therefore, the points on the y-axis are \((0, 1, 0)\) and \((0, 5, 0)\).
In simple words: Any point on the y-axis has x and z coordinates equal to zero. We set up the distance equation to find the possible y-values.

Exam Tip: When a point lies on a specific axis, immediately write its coordinates with the other two variables as zero (e.g., \((0, y, 0)\) for y-axis) to reduce variables.

Question 3. If A and B are the points (1, 2, 3) and (-1, 4, -3) respectively then find the locus of a point P such that \(PA^2 - PB^2 = 2k^2\).
Answer: Let the coordinates of the moving point \(P\) be \((x, y, z)\).
Given points are \(A(1, 2, 3)\) and \(B(-1, 4, -3)\).
Using the distance formula for \(PA^2\) and \(PB^2\):
\(PA^2 = (x - 1)^2 + (y - 2)^2 + (z - 3)^2\)
\(PB^2 = (x - (-1))^2 + (y - 4)^2 + (z - (-3))^2 = (x + 1)^2 + (y - 4)^2 + (z + 3)^2\)
Substitute \(PA^2\) and \(PB^2\) into the given condition \(PA^2 - PB^2 = 2k^2\):
\([(x - 1)^2 + (y - 2)^2 + (z - 3)^2] - [(x + 1)^2 + (y - 4)^2 + (z + 3)^2] = 2k^2\)
Expanding each term using algebraic identities:
\([x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9] - [x^2 + 2x + 1 + y^2 - 8y + 16 + z^2 + 6z + 9] = 2k^2\)
Simplifying by canceling common squared terms:
\((-2x - 4y - 6z + 14) - (2x - 8y + 6z + 26) = 2k^2\)
\(-2x - 4y - 6z + 14 - 2x + 8y - 6z - 26 = 2k^2\)
\(-4x + 4y - 12z - 12 = 2k^2\)
Dividing the entire equation by -2 and rearranging:
\(2x - 2y + 6z + 6 + k^2 = 0\)
In simple words: We express the square distances from point P to A and B algebraically, subtract them, and simplify to get the linear equation of the locus.

Exam Tip: Expanding squared binomials carefully and keeping track of negative signs during subtraction is crucial to avoid arithmetic errors.

Question 4. If the points A(1, 0, -6), B(-3, p, q) and C(-5, 9, 6) are collinear, find the values of p and q.
Answer: Since the points \(A\), \(B\), and \(C\) are collinear, the direction ratios of line segments \(AB\) and \(BC\) must be proportional.
First, find the direction ratios of \(AB\) \((a_1, b_1, c_1)\):
\(a_1 = -3 - 1 = -4\)
\(b_1 = p - 0 = p\)
\(c_1 = q - (-6) = q + 6\)
Next, find the direction ratios of \(BC\) \((a_2, b_2, c_2)\):
\(a_2 = -5 - (-3) = -2\)
\(b_2 = 9 - p\)
\(c_2 = 6 - q\)
Since the points are collinear, the ratios are equal:
\(\frac{-4}{-2} = \frac{p}{9 - p} = \frac{q + 6}{6 - q}\)
Simplify the first ratio:
\(\frac{-4}{-2} = 2\)
Now, equate the second ratio to 2 to find \(p\):
\(\frac{p}{9 - p} = 2\)
\(p = 2(9 - p)\)
\(p = 18 - 2p\)
\(3p = 18 \implies p = 6\)
Now, equate the third ratio to 2 to find \(q\):
\(\frac{q + 6}{6 - q} = 2\)
\(q + 6 = 2(6 - q)\)
\(q + 6 = 12 - 2q\)
\(3q = 6 \implies q = 2\)
Therefore, \(p = 6\) and \(q = 2\).
In simple words: Collinear points lie on the same straight line, meaning the slopes or direction ratios between consecutive pairs of points are equal.

Exam Tip: Alternatively, you can use the section formula or check that the sum of distances of any two segments equals the third, but equating direction ratios is the fastest method.

Question 5. Two vertices of a triangle are (2, -6, 4), (4, -2, 3) and its centroid is \(\left(\frac{8}{3}, -1, 3\right)\), find the third vertex.
Answer: Let the third vertex of the triangle be \(C(x_3, y_3, z_3)\).
The given vertices are \(A(2, -6, 4)\) and \(B(4, -2, 3)\).
The centroid formula for a triangle with vertices \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), and \((x_3, y_3, z_3)\) is given by:
\(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right) = \left(\frac{8}{3}, -1, 3\right)\)
Equating the x-coordinates:
\(\frac{2 + 4 + x_3}{3} = \frac{8}{3}\)
\(6 + x_3 = 8 \implies x_3 = 2\)
Equating the y-coordinates:
\(\frac{-6 + (-2) + y_3}{3} = -1\)
\(-8 + y_3 = -3 \implies y_3 = 5\)
Equating the z-coordinates:
\(\frac{4 + 3 + z_3}{3} = 3\)
\(7 + z_3 = 9 \implies z_3 = 2\)
Therefore, the third vertex is \((2, 5, 2)\).
In simple words: The centroid is simply the average of the x, y, and z coordinates of the three vertices. We equate each coordinate average to the given centroid values to solve for the missing vertex.

Exam Tip: Always verify your answer by plugging all three vertices back into the centroid formula to check if it matches the given centroid.