Q.Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

The problem gives two conditions on $\vec{c}$: a cross product $\vec{b} \times \vec{c} = \vec{a}$ and a dot product $\vec{b} \cdot \vec{c} = 5$. To find $\vec{c} \times \vec{a}$, we can avoid solving for $\vec{c}$ component-wise by using the vector triple product identity. The identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$ will let us relate $\vec{c} \times \vec{a}$ to known quantities.

1. Compute $\vec{b} \times \vec{a}$ using the given cross product.

   Since $\vec{b} \times \vec{c} = \vec{a}$, take cross product with $\vec{b}$ on the left:

   $\vec{b} \times (\vec{b} \times \vec{c}) = \vec{b} \times \vec{a}$.

   Apply the triple product identity:

   $\vec{b} \times (\vec{b} \times \vec{c}) = (\vec{b} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{b})\vec{c}$.

   We know $\vec{b} \cdot \vec{c} = 5$, and $\vec{b} \cdot \vec{b} = 1^2 + 3^2 + (-2)^2 = 1 + 9 + 4 = 14$.

   So $\vec{b} \times \vec{a} = 5\vec{b} - 14\vec{c}$.

2. Find $\vec{b} \times \vec{a}$ directly.

   $\vec{a} = 3\vec{i} - 2\vec{j} + 5\vec{k}$, $\vec{b} = \vec{i} + 3\vec{j} - 2\vec{k}$.

   Compute the cross product:

   $\vec{b} \times \vec{a} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & -2 \\ 3 & -2 & 5 \end{vmatrix}$

   $= \vec{i}(3\cdot5 - (-2)(-2)) - \vec{j}(1\cdot5 - (-2)(3)) + \vec{k}(1\cdot(-2) - 3\cdot3)$

   $= \vec{i}(15 - 4) - \vec{j}(5 + 6) + \vec{k}(-2 - 9)$

   $= 11\vec{i} - 11\vec{j} - 11\vec{k} = 11(\vec{i} - \vec{j} - \vec{k})$.

3. Solve for $\vec{c}$.

   From step 1: $11(\vec{i} - \vec{j} - \vec{k}) = 5\vec{b} - 14\vec{c}$.

   So $14\vec{c} = 5\vec{b} - 11(\vec{i} - \vec{j} - \vec{k})$.

   Compute $5\vec{b} = 5\vec{i} + 15\vec{j} - 10\vec{k}$.

   Then $14\vec{c} = (5\vec{i} + 15\vec{j} - 10\vec{k}) - (11\vec{i} - 11\vec{j} - 11\vec{k})$

   $= (5-11)\vec{i} + (15+11)\vec{j} + (-10+11)\vec{k}$ …

We are given a triangle with side lengths $AB = 2$, $BC = 3$, and $CA = 4$. The expression $\overline{AB} \cdot \overline{BC} + \overline{BC} \cdot \overline{CA} + \overline{CA} \cdot \overline{AB}$ uses the notation $\overline{AB}$ to mean the vector from $A$ to $B$, and the dot is the usual dot product. So we need the sum of dot products of consecutive side vectors taken around the triangle.

Why this approach works:

The dot product of two vectors depends on their lengths and the angle between them. For vectors along the sides of a triangle, the angle between, say, $\overline{AB}$ and $\overline{BC}$ is not the interior angle at $B$ — careful: $\overline{AB}$ points from $A$ to $B$, and $\overline{BC}$ points from $B$ to $C$. The angle between these two vectors is actually the supplement of the interior angle at $B$. This sign reversal is crucial. Using the law of cosines, we can express each dot product in terms of the side lengths, and then sum them.

Let’s work through it step by step.

1. Label the vectors and interior angles.

   Let $\vec{a} = \overline{BC}$, $\vec{b} = \overline{CA}$, $\vec{c} = \overline{AB}$.

   Their lengths are $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{c}| = 2$.

   The interior angles at vertices $A, B, C$ are opposite sides $BC, CA, AB$ respectively. So:

   • Angle at $A$ is opposite side $BC = 3$.
   • Angle at $B$ is opposite side $CA = 4$.
   • Angle at $C$ is opposite side $AB = 2$.

2. Find the dot product $\overline{AB} \cdot \overline{BC}$.

   $\overline{AB}$ goes from $A$ to $B$; $\overline{BC}$ goes from $B$ to $C$.

   The angle between these two vectors is the external angle at $B$, which is $180^\circ - \angle B$.

   So

3. Similarly, find $\overline{BC} \cdot \overline{CA}$. $\overline{BC}$ goes from $B$ to $C$; $\overline{CA}$ goes from $C$ to $A$. The angle between them is $180^\circ - \angle C$. So

4. And $\overline{CA} \cdot \overline{AB}$. $\overline{CA}$ goes from $C$ to $A$; $\overline{AB}$ goes from $A$ to $B$. The angle between them is $180^\circ - \angle A$. So

5. Sum them up:

6. Use the law of cosines to find the cosines. For a triangle with sides $a=BC=3$, $b=CA=4$, $c=AB=2$: $$$$

The exradii satisfy $r_1=\dfrac{\Delta}{s-a},\ r_2=\dfrac{\Delta}{s-b},\ r_3=\dfrac{\Delta}{s-c}$, so

Compute the left side:

Also $r_1 r_2 r_3=\dfrac{\Delta^3}{(s-a)(s-b)(s-c)}$ and by Heron $\Delta^2=s(s-a)(s-b)(s-c)$, hence $(s-a)(s-b)(s-c)=\dfrac{\Delta^2}{s}$. Therefore …

We start with two vectors:

The expression to evaluate is $(\vec{a} + 2\vec{b}) \times (3\vec{a} - \vec{b})$.

Concept and intuition

The cross product is bilinear — it distributes over addition and respects scalar multiplication. That means we can expand the product just like multiplying two binomials, but we must keep the order because $\vec{u}\times\vec{v} = -\,\vec{v}\times\vec{u}$. This property will let us reduce the whole expression to a simple multiple of $\vec{a}\times\vec{b}$, which we then compute directly.

Step-by-step solution

1. Expand using bilinearity

2. Pull out scalars (scalars factor out of a cross product):

3. Use $\vec{u}\times\vec{u}=0$ and $\vec{b}\times\vec{a} = -\,\vec{a}\times\vec{b}$:

So the whole expression simplifies to $-7\,(\vec{a}\times\vec{b})$.

4. Compute $\vec{a}\times\vec{b}$ using the determinant formula:

Expand:

The key idea: when three vectors form a triangle and one equals the sum of the other two, the triangle’s sides are those two vectors placed head-to-tail. The area of the triangle is then half the magnitude of the cross product of those two vectors — because the cross product gives the area of the parallelogram they span.

We are told $\vec{p} = \vec{q} + \vec{r}$, so $\vec{q}$ and $\vec{r}$ are two sides of the triangle meeting at a vertex, and $\vec{p}$ is the third side (the resultant). The area of $\triangle ABC$ is therefore $\frac{1}{2} |\vec{q} \times \vec{r}|$.

Let’s work through it.

1. Write the given vectors clearly

   $\vec{q} = d\hat{i} + 3\hat{j} + 4\hat{k}$

   $\vec{r} = 3\hat{i} + \hat{j} - 2\hat{k}$

   $\vec{p} = a\hat{i} + b\hat{j} + c\hat{k}$

   And $\vec{p} = \vec{q} + \vec{r}$.

2. Find $\vec{q} \times \vec{r}$

Compute component by component:

$\hat{i}$-component: $(3)(-2) - (4)(1) = -6 - 4 = -10$

$\hat{j}$-component: $-(d)(-2) - (4)(3) = -(-2d - 12) = 2d + 12$? Wait carefully — the $\hat{j}$ coefficient is $-\left( d\cdot(-2) - 4\cdot 3 \right) = -(-2d - 12) = 2d + 12$

$\hat{k}$-component: $(d)(1) - (3)(3) = d - 9$

So

3. Area condition

   Area of triangle $= \frac{1}{2} |\vec{q} \times \vec{r}| = 5\sqrt{6}$

   So $|\vec{q} \times \vec{r}| = 10\sqrt{6}$.

   Compute the magnitude squared:

Set equal to $(10\sqrt{6})^2 = 600$:

Divide by 5:

So $d = 5$ or $d = -11$. …

Let $P=(x,y)$, with $A=(1,2)$, $B=(2,1)$, $C=(-1,-1)$.

Area of quadrilateral $PABC$ (shoelace on $P,A,B,C$):

Area of triangle $PAB$:

Condition $\text{area}(PABC)=2\,\text{area}(PAB)$:

Squaring both sides:

$(3x-2y-4)^2=4(x+y-3)^2.$ …

The key idea here is that the areas of the incircle and excircles of a triangle are not independent — they are linked through the triangle’s semi-perimeter $s$, its sides, and its area $\Delta$. Once you express each area in terms of $\Delta$ and $s$, a clean relation emerges that lets you find $A$ directly from $A_1, A_2, A_3$ without ever needing the triangle’s sides.

Recall the standard formulas:

• In-circle radius $r = \frac{\Delta}{s}$, so $A = \pi r^2 = \pi \frac{\Delta^2}{s^2}$.
• Ex-circle radii: $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, $r_3 = \frac{\Delta}{s-c}$, so $A_1 = \pi \frac{\Delta^2}{(s-a)^2}$, $A_2 = \pi \frac{\Delta^2}{(s-b)^2}$, $A_3 = \pi \frac{\Delta^2}{(s-c)^2}$.

The trick is to notice that $s$, $s-a$, $s-b$, $s-c$ are all positive numbers whose sum is $s + (s-a) + (s-b) + (s-c) = 4s - (a+b+c) = 4s - 2s = 2s$. But more usefully, we can relate the reciprocals of the radii.

1. Set up the relation between reciprocals. Since $r = \frac{\Delta}{s}$ and $r_1 = \frac{\Delta}{s-a}$, we have $\frac{1}{r} = \frac{s}{\Delta}$ and $\frac{1}{r_1} = \frac{s-a}{\Delta}$. Adding the three ex-circle reciprocals:

So we have the elegant identity:

2. Convert areas to radii. Since $A = \pi r^2$, $A_1 = \pi r_1^2$, etc., we have $r = \sqrt{A/\pi}$, $r_1 = \sqrt{A_1/\pi}$, and so on. Substituting into the reciprocal identity:

The $\sqrt{\pi}$ cancels from every term, giving:

3. Plug in the given numbers. …

The key here is to connect the inradius $r$ with the area $\Delta$ and the semiperimeter $s$, and then express each $\cot \frac{A}{2}$ in terms of $s$, $a$, $b$, $c$, and $r$. Once you see that $\cot \frac{A}{2} = \frac{s-a}{r}$, the product becomes a clean cancellation.

1. Recall the fundamental relation between $r$, $\Delta$, and $s$.

   For any triangle, the area $\Delta$ is given by $\Delta = r s$, where $s$ is the semiperimeter. This is because the incenter is the intersection of angle bisectors, and the area can be split into three triangles each with height $r$ and base a side.

2. Express $\cot \frac{A}{2}$ in terms of $s$, $a$, and $r$.

   From the geometry of the incircle, the distance from vertex $A$ to the points of tangency on sides $AB$ and $AC$ is $s-a$. In the right triangle formed by the incenter and vertex $A$, we have

Similarly,

3. Substitute into the given expression. The expression becomes

4. Use Heron’s formula to simplify. …