Q.Find the sum of the vectors $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$.

The key idea here is that the tangents of half-angles in a triangle are directly linked to the sides through the inradius and semiperimeter. Specifically, for any triangle, $\tan\frac{A}{2} = \frac{r}{s-a}$, and similarly for the other angles. This gives us a clean algebraic handle on the side lengths.

Given the ratio $\tan\frac{A}{2} : \tan\frac{B}{2} : \tan\frac{C}{2} = 1 : 2 : 3$, we can set:

for some positive constant $k$.

1. Use the half-angle formula. We know $\tan\frac{A}{2} = \frac{r}{s-a}$, so:

Hence:

2. Find the semiperimeter $s$. Adding the three equations:

So:

Thus:

3. Express the sides in terms of $r$ and $k$. From $s-a = \frac{r}{k}$, we get $a = s - \frac{r}{k} = \frac{11r}{6k} - \frac{r}{k} = \frac{5r}{6k}$. From $s-b = \frac{r}{2k}$, we get $b = s - \frac{r}{2k} = \frac{11r}{6k} - \frac{r}{2k} = \frac{11r}{6k} - \frac{3r}{6k} = \frac{8r}{6k} = \frac{4r}{3k}$. …

With $a=\sqrt3+1,\ b=\sqrt3-1,\ C=60^\circ$: $a-b=2,\ a+b=2\sqrt3,\ \dfrac{C}{2}=30^\circ$.

Napier's analogy:

$\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}=\frac{2}{2\sqrt3}\cdot\sqrt3=1.$ …

The key idea is that when three vectors sum to zero, they form a triangle when placed head-to-tail. The magnitudes are the side lengths, and the angle between two vectors is not the interior angle of that triangle — it’s the supplement. But we can avoid geometry entirely by using dot products: squaring the sum gives a direct equation linking the magnitudes and the cosine of the required angle.

1. Set up the dot product equation. Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, we have $\vec{c} = -(\vec{a}+\vec{b})$. Square both sides (take the dot product of each side with itself):

2. Substitute the given magnitudes. $|\vec{a}|=3$, $|\vec{b}|=5$, $|\vec{c}|=7$:

3. Solve for the dot product.

4. Find the cosine of the angle. The dot product formula: $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

5. Identify the angle. …

Concept and Intuition: The Power of the Medial Triangle

Imagine you have a large triangle, say $\triangle ABC$. Now, if you find the exact middle point of each of its sides and connect these three midpoints, you form a new, smaller triangle inside the original one. This inner triangle is known as the medial triangle.

The medial triangle isn't just a random shape; it has a profound and elegant relationship with its parent triangle. Think of it as a miniature, perfectly scaled-down version of the original triangle, but rotated 180 degrees. The key insight here, which forms the backbone of our solution, is that each side of the medial triangle is exactly half the length of the corresponding parallel side of the original triangle.

This property is incredibly powerful because it allows us to work directly with the given coordinates of the midpoints (which are the vertices of the medial triangle) to find information about the larger, original triangle, without needing to first calculate the coordinates of the original triangle's vertices. This saves a lot of time and reduces the chances of calculation errors.

Let's apply this concept to solve our problem.

Step-by-Step Solution:

1. Identify the Medial Triangle:

   We are given the midpoints of the sides $AB, BC, CA$ as $D=(3, 0, 0)$, $E=(0, 4, 0)$, and $F=(0, 0, 5)$ respectively. These three points, $D, E, F$, are the vertices of the medial triangle $\triangle DEF$.

2. Recall the Medial Triangle Theorem:

   This fundamental theorem in geometry describes the relationship between a triangle and its medial triangle.

   Medial Triangle Theorem: If $D, E, F$ are the midpoints of sides $AB, BC, CA$ respectively in $\triangle ABC$, then:

   1. The segment $DE$ is parallel to side $AC$ and its length is $DE = \frac{1}{2} AC$.
   2. The segment $EF$ is parallel to side $AB$ and its length is $EF = \frac{1}{2} AB$.
   3. The segment $FD$ is parallel to side $BC$ and its length is $FD = \frac{1}{2} BC$.

   From this, we can express the side lengths of the original triangle in terms of the medial triangle:

   $AB = 2EF$

   $BC = 2FD$

   $CA = 2DE$

3. Calculate the Squared Lengths of the Medial Triangle's Sides:

   We'll use the 3D distance formula, $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$, to find the squared lengths of the sides of $\triangle DEF$.

   • Side $DE$: Connecting $D(3,0,0)$ and $E(0,4,0)$.

     $DE^2 = (0-3)^2 + (4-0)^2 + (0-0)^2$

     $DE^2 = (-3)^2 + 4^2 + 0^2$

     $DE^2 = 9 + 16 + 0 = 25$.

   • Side $EF$: Connecting $E(0,4,0)$ and $F(0,0,5)$.

     $EF^2 = (0-0)^2 + (0-4)^2 + (5-0)^2$

     $EF^2 = 0^2 + (-4)^2 + 5^2$

     $EF^2 = 0 + 16 + 25 = 41$.

   • Side $FD$: Connecting $F(0,0,5)$ and $D(3,0,0)$. …

The core idea: when a vector is expressed as a linear combination of three given vectors, the coefficients are unique (since the three vectors are linearly independent). Equating components gives a system of linear equations.

1. Write the given equation in component form. The vector on the left is $\vec{i} - 2\vec{j} + 5\vec{k}$. On the right:

   • $\alpha(\vec{i} + \vec{j} + \vec{k}) = \alpha\vec{i} + \alpha\vec{j} + \alpha\vec{k}$
   • $\beta(\vec{i} + 2\vec{j} + 3\vec{k}) = \beta\vec{i} + 2\beta\vec{j} + 3\beta\vec{k}$
   • $\gamma(2\vec{i} - \vec{j} + \vec{k}) = 2\gamma\vec{i} - \gamma\vec{j} + \gamma\vec{k}$

   Adding these, the coefficient of $\vec{i}$ is $\alpha + \beta + 2\gamma$, of $\vec{j}$ is $\alpha + 2\beta - \gamma$, and of $\vec{k}$ is $\alpha + 3\beta + \gamma$.

2. Equate coefficients with the left side:

3. Solve the system. Subtract the first equation from the second:

Subtract the first from the third:

We are given the exradii $r_1, r_2, r_3$ of a triangle with sides $a, b, c$. The exradius opposite side $a$ is $r_1$, opposite $b$ is $r_2$, and opposite $c$ is $r_3$. The standard formulas are:

where $\Delta$ is the area of the triangle and $s = \frac{a+b+c}{2}$ is the semi-perimeter.

The problem asks for $a + 2b + 3c$. We don't know $a, b, c$ individually, but we can find $s$ and $\Delta$ from the exradii, and then express the sides in terms of $s$ and $\Delta$.

1. Find $s$ using the reciprocal sum of exradii. A well-known identity is:

where $r$ is the inradius. But more directly, we can use:

So:

Plug in $r_1=12$, $r_2=6$, $r_3=4$:

Hence $\frac{s}{\Delta} = \frac{1}{2}$, so $s = \frac{\Delta}{2}$.

2. Find $\Delta$ using the product of exradii. Another identity: $r_1 r_2 r_3 = \Delta^2 s$. Let's verify:

But by Heron's formula, $\Delta^2 = s(s-a)(s-b)(s-c)$, so $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.

Therefore:

So $r_1 r_2 r_3 = \Delta s$.

Compute: $12 \times 6 \times 4 = 288$, so $\Delta s = 288$.

3. Solve for $s$ and $\Delta$.

   From step 1: $s = \frac{\Delta}{2}$ → $\Delta = 2s$.

   Substitute into $\Delta s = 288$: $(2s) \cdot s = 2s^2 = 288$ → $s^2 = 144$ → $s = 12$ (positive).

   Then $\Delta = 2 \times 12 = 24$.

4. Find the sides $a, b, c$. …