Get detailed, step-by-step solutions for the NCERT Class 11 Maths Chapter 11 Miscellaneous Exercise
Learn to apply 3D formulas to find the fourth vertex of a parallelogram using the midpoint formula (Q.1). Calculate the lengths of medians of a triangle (Q.2). Use the centroid formula to find unknown coordinates when the centroid is the origin (Q.3). Determine the locus of a point where the sum of the squares of distances from two fixed points is constant (resulting in a sphere, Q.4).
1. Finding the Fourth Vertex of a Parallelogram
Problem: Three vertices of a parallelogram $ABCD$ are $A(3, – 1, 2)$, $B (1, 2, – 4)$ and $C (– 1, 1, 2)$. Find the coordinates of the fourth vertex $D$.
Solution:
In a parallelogram, the diagonals bisect each other. This means the midpoint of $AC$ must be the same as the midpoint of $BD$.
Let $D$ have coordinates $(x, y, z)$.
Step 1: Find the Midpoint of $AC$
$$M_{AC} = \left(\frac{3 + (-1)}{2}, \frac{-1 + 1}{2}, \frac{2 + 2}{2}\right) = \left(\frac{2}{2}, \frac{0}{2}, \frac{4}{2}\right) = (1, 0, 2)$$
Step 2: Find the Midpoint of $BD$
$$M_{BD} = \left(\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{-4 + z}{2}\right)$$
Step 3: Equate the Midpoints
Setting $M_{AC} = M_{BD}$:
- $x$-coordinate: $\frac{1 + x}{2} = 1 \implies 1 + x = 2 \implies \mathbf{x = 1}$
- $y$-coordinate: $\frac{2 + y}{2} = 0 \implies 2 + y = 0 \implies \mathbf{y = -2}$
- $z$-coordinate: $\frac{-4 + z}{2} = 2 \implies -4 + z = 4 \implies \mathbf{z = 8}$
The coordinates of the fourth vertex $D$ are $(1, -2, 8)$.
2. Lengths of the Medians of a Triangle
Problem: Find the lengths of the medians of the triangle with vertices $A (0, 0, 6)$, $B (0, 4, 0)$ and $C (6, 0, 0)$.
Solution:
A median connects a vertex to the midpoint of the opposite side. Let $D, E, F$ be the midpoints of $BC, AC, AB$, respectively. We need to find the lengths of the medians $AD$, $BE$, and $CF$.
Step 1: Find the Midpoints
- $D$ (midpoint of $BC$): $\left(\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}\right) = (3, 2, 0)$
- $E$ (midpoint of $AC$): $\left(\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}\right) = (3, 0, 3)$
- $F$ (midpoint of $AB$): $\left(\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2}\right) = (0, 2, 3)$
Step 2: Find the Lengths of the Medians using the Distance Formula
- Length of Median $AD$ (from $A(0, 0, 6)$ to $D(3, 2, 0)$):$$AD = \sqrt{(3 – 0)^2 + (2 – 0)^2 + (0 – 6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = \mathbf{7}$$
- Length of Median $BE$ (from $B(0, 4, 0)$ to $E(3, 0, 3)$):$$BE = \sqrt{(3 – 0)^2 + (0 – 4)^2 + (3 – 0)^2} = \sqrt{9 + 16 + 9} = \sqrt{34}$$
- Length of Median $CF$ (from $C(6, 0, 0)$ to $F(0, 2, 3)$):$$CF = \sqrt{(0 – 6)^2 + (2 – 0)^2 + (3 – 0)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = \mathbf{7}$$
The lengths of the medians are $7, \sqrt{34},$ and $7$ units.
3. Centroid of a Triangle
Problem: If the origin is the centroid of the triangle $PQR$ with vertices $P (2a, 2, 6)$, $Q (– 4, 3b, –10)$ and $R(8, 14, 2c)$, then find the values of $a, b$ and $c$.
Solution:
The centroid $G(x_g, y_g, z_g)$ of a triangle with vertices $P(x_1, y_1, z_1)$, $Q(x_2, y_2, z_2)$, and $R(x_3, y_3, z_3)$ is given by:
$$x_g = \frac{x_1 + x_2 + x_3}{3}, \quad y_g = \frac{y_1 + y_2 + y_3}{3}, \quad z_g = \frac{z_1 + z_2 + z_3}{3}$$
Given that the centroid $G$ is the origin $(0, 0, 0)$.
Step 1: Equate the $x$-coordinate to 0
$$0 = \frac{2a + (-4) + 8}{3}$$
$$0 = 2a + 4 \implies 2a = -4 \implies \mathbf{a = -2}$$
Step 2: Equate the $y$-coordinate to 0
$$0 = \frac{2 + 3b + 14}{3}$$
$$0 = 3b + 16 \implies 3b = -16 \implies \mathbf{b = -\frac{16}{3}}$$
Step 3: Equate the $z$-coordinate to 0
$$0 = \frac{6 + (-10) + 2c}{3}$$
$$0 = -4 + 2c \implies 2c = 4 \implies \mathbf{c = 2}$$
The values are $a = -2$, $b = -\frac{16}{3}$, and $c = 2$.
4. Locus of Points with Constant Sum of Squares of Distances
Problem: If $A$ and $B$ be the points $(3, 4, 5)$ and $(-1, 3, -7)$, respectively, find the equation of the set of points $P$ such that $PA^2 + PB^2 = k^2$, where $k$ is a constant.
Solution:
Let $P$ be the point $(x, y, z)$. We use the distance formula for $PA^2$ and $PB^2$.
Step 1: Calculate $PA^2$
$$PA^2 = (x – 3)^2 + (y – 4)^2 + (z – 5)^2$$
$$PA^2 = (x^2 – 6x + 9) + (y^2 – 8y + 16) + (z^2 – 10z + 25)$$
Step 2: Calculate $PB^2$
$$PB^2 = (x – (-1))^2 + (y – 3)^2 + (z – (-7))^2$$
$$PB^2 = (x + 1)^2 + (y – 3)^2 + (z + 7)^2$$
$$PB^2 = (x^2 + 2x + 1) + (y^2 – 6y + 9) + (z^2 + 14z + 49)$$
Step 3: Substitute into $PA^2 + PB^2 = k^2$
$$(x^2 – 6x + 9 + y^2 – 8y + 16 + z^2 – 10z + 25) + (x^2 + 2x + 1 + y^2 – 6y + 9 + z^2 + 14z + 49) = k^2$$
Step 4: Combine like terms
- $x, y, z$ squared terms: $2x^2 + 2y^2 + 2z^2$
- $x, y, z$ linear terms: $(-6x + 2x) + (-8y – 6y) + (-10z + 14z) = -4x – 14y + 4z$
- Constant terms: $(9 + 16 + 25) + (1 + 9 + 49) = 50 + 59 = 109$
The equation becomes:
$$2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z + 109 = k^2$$
Step 5: Rearrange to the final form
$$2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z = k^2 – 109$$
Divide by 2:
$$\mathbf{x^2 + y^2 + z^2 – 2x – 7y + 2z = \frac{k^2 – 109}{2}}$$
The locus of $P$ is a sphere.