Q1. 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.

Solution

Let the coordinates of the fourth vertex \(D\) of the parallelogram be \((x, y, z)\). In a parallelogram, the diagonals bisect each other. Hence, the midpoint of diagonal \(AC\) is the same as the midpoint of diagonal \(BD\).

The given vertices are \(A(3, -1, 2)\), \(B(1, 2, -4)\) and \(C(-1, 1, 2)\). First, we find the midpoint of diagonal \(AC\).

\[ \begin{aligned} E_x&=\frac{3+(-1)}{2}=1\\ E_y&=\frac{-1+1}{2}=0\\ E_z&=\frac{2+2}{2}=2\\ E&=(1, 0, 2) \end{aligned} \]

Now, using the midpoint formula for points \(B(1, 2, -4)\) and \(D(x, y, z)\), and equating it to the midpoint \(E(1, 0, 2)\), we obtain

\[ \begin{aligned} \frac{1+x}{2}&=1\\ x&=1\\ \frac{2+y}{2}&=0\\ y&=-2\\ \frac{-4+z}{2}&=2\\ z-4&=4\\ z&=8 \end{aligned} \]

Therefore, the coordinates of the fourth vertex of the parallelogram are \((1, -2, 8)\).

Q2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Solution

The vertices of the given triangle are \(A(0, 0, 6)\), \(B(0, 4, 0)\) and \(C(6, 0, 0)\). To find the lengths of the medians, we first determine the midpoints of the opposite sides and then calculate the distances from the corresponding vertices.

The median from vertex \(A\) meets side \(BC\) at its midpoint. The midpoint of \(BC\) is obtained as

\[ \begin{aligned} x&=\frac{6+0}{2}=3\\ y&=\frac{0+4}{2}=2\\ z&=\frac{0+0}{2}=0\\ (x,y,z)&=(3, 2, 0) \end{aligned} \]

The length of the median from \(A\) to \(BC\) is

\[ \begin{aligned} &\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}\\ &=\sqrt{9+4+36}\\ &=\sqrt{49}\\ &=7 \end{aligned} \]

Next, the midpoint of side \(AB\) is calculated to find the median from vertex \(C\)

\[ \begin{aligned} x&=\frac{0+0}{2}=0\\ y&=\frac{0+4}{2}=2\\ z&=\frac{6+0}{2}=3\\ (x,y,z)&=(0, 2, 3) \end{aligned} \]

The length of the median from \(C\) to \(AB\) is

\[ \begin{aligned} &\sqrt{(0-6)^2+(2-0)^2+(3-0)^2}\\ &=\sqrt{36+4+9}\\ &=\sqrt{49}\\ &=7 \end{aligned} \]

Finally, the midpoint of side \(AC\) is obtained to find the median from vertex \(B\)

\[ \begin{aligned} x&=\frac{0+6}{2}=3\\ y&=\frac{0+0}{2}=0\\ z&=\frac{6+0}{2}=3\\ (x,y,z)&=(3, 0, 3) \end{aligned} \]

The length of the median from \(B\) to \(AC\) is

\[ \begin{aligned} &\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}\\ &=\sqrt{9+16+9}\\ &=\sqrt{34} \end{aligned} \]

Hence, the lengths of the medians of the given triangle are \(7\), \(7\) and \(\sqrt{34}\).

Q3. 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

Since the origin is the centroid of triangle \(PQR\), the coordinates of the centroid are \(G(0, 0, 0)\). The vertices of the triangle are given as \(P(2a, 2, 6)\), \(Q(-4, 3b, -10)\) and \(R(8, 14, 2c)\). Using the formula for the centroid of a triangle, we equate the coordinates of the centroid to zero.

\[ \begin{aligned} G(x)&=\frac{2a+(-4)+8}{3}\\ 0&=\frac{2a+4}{3}\\ 2a&=-4\\ a&=-2 \end{aligned} \]

Now, equating the y-coordinate of the centroid to zero, we obtain

\[ \begin{aligned} G(y)&=\frac{2+3b+14}{3}\\ 0&=\frac{3b+16}{3}\\ 3b&=-16\\ b&=-\frac{16}{3} \end{aligned} \]

Finally, equating the z-coordinate of the centroid to zero, we get

\[ \begin{aligned} G(z)&=\frac{6+(-10)+2c}{3}\\ 0&=\frac{2c-4}{3}\\ 2c&=4\\ c&=2 \end{aligned} \]

Hence, the values of the constants are \(a=-2\), \(b=-\dfrac{16}{3}\) and \(c=2\).

Q4. 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 PA2 + PB2 = k2, where k is a constant.

Solution

Let the coordinates of point \(P\) be \((x, y, z)\). The given fixed points are \(A(3, 4, 5)\) and \(B(-1, 3, -7)\). Using the distance formula, we first express the squares of the distances \(PA\) and \(PB\).

\[ \begin{aligned} PA^{2}&=(x-3)^2+(y-4)^2+(z-5)^2\\ PB^{2}&=(x+1)^2+(y-3)^2+(z+7)^2 \end{aligned} \]

According to the given condition, the sum of the squares of these distances is equal to \(k^2\). Hence,

\[ \begin{aligned} &PA^{2}+PB^{2}\\&=(x-3)^2+(x+1)^2+(y-4)^2+(y-3)^2+(z-5)^2+(z+7)^2 \end{aligned} \]

Expanding and combining like terms, we get

\[ \begin{aligned} &(x-3)^2+(x+1)^2\\&=x^2-6x+9+x^2+2x+1\\ &=2x^2-4x+10\\\\ &(y-4)^2+(y-3)^2\\&=y^2-8y+16+y^2-6y+9\\ &=2y^2-14y+25\\\\ &(z-5)^2+(z+7)^2\\&=z^2-10z+25+z^2+14z+49\\ &=2z^2+4z+74 \end{aligned} \]

Substituting these results, we obtain

\[ \begin{aligned} PA^{2}+PB^{2} &=2x^2+2y^2+2z^2-4x-14y+4z+109\\ &=k^2 \end{aligned} \]

Dividing throughout by 2 and rearranging, the required equation of the set of points is

\[ \begin{aligned} x^2+y^2+z^2-2x-7y+2z=\frac{k^2-109}{2} \end{aligned} \]