Q.(a) Show that the area of a parallelogram whose diagonals are represented by $\vec{a}$ and $\vec{b}$ is given by $\frac{1}{2}|\vec{a} \times \vec{b}|$. Also find the area of a parallelogram whose diagonals are $2\hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + 3\hat{j} - \hat{k}$.

Parallelogram Diagonal Vectors – From Scratch

A parallelogram is a slanted rectangle: opposite sides are equal and parallel. Draw both diagonals — each runs from one corner to the opposite corner. The question is: how do we describe these diagonals using the two side vectors that start from the same corner?

The Setup

Take a parallelogram with vertices $A$, $B$, $C$, $D$ in order, with $A$ at the origin. From $A$, two vectors emerge:

• $\vec{a}$ goes from $A$ to $B$ (one side)
• $\vec{b}$ goes from $A$ to $D$ (the other side)

Because opposite sides are equal, $B$ to $C$ is also $\vec{b}$ and $D$ to $C$ is also $\vec{a}$, so the fourth vertex $C$ sits at $\vec{a} + \vec{b}$. The two diagonals run from $A$ to $C$ and from $B$ to $D$.

The Diagonal from the Common Vertex

From $A$ to $C$ you go $\vec{a}$ then $\vec{b}$, ending at the opposite corner:

That's the vector sum of the two sides — walk along one side then the other and you land on the opposite corner.

The Other Diagonal

$B$ is at $\vec{a}$ and $D$ is at $\vec{b}$. To go from $B$ to $D$, you travel from $\vec{a}$ to $\vec{b}$:

The reverse, from $D$ to $B$, is $\vec{a} - \vec{b}$. Both are correct; they just differ in direction.

The Precise Statement

Why This Matters

• Vector addition — the diagonal from the common vertex is the sum of the sides. This is the parallelogram law of vector addition.
• Finding midpoints — the diagonals bisect each other; both midpoints are the same point, $\frac{\vec{a} + \vec{b}}{2}$.
• Physics — the resultant of two forces acting at a point is the diagonal of the parallelogram formed by the force vectors.

A Quick Check

Take $\vec{a} = (3, 0)$ (horizontal) and $\vec{b} = (1, 2)$ (slanted). Then:

• Diagonal from the common vertex: $(3,0) + (1,2) = (4,2)$ …

Line Perpendicular to Two Lines

In 3D geometry a very common task is this: two lines are given, and you must find the direction of a third line that is perpendicular to both of them. This shows up when finding a common perpendicular, the shortest distance between two lines, or the normal to a plane containing two directions.

The Core Idea

A line in space is fixed by two things: a point it passes through, and its direction vector. So "find a line perpendicular to two lines" really means "find a vector perpendicular to both of the two given direction vectors."

If the two given lines have direction vectors

then a vector perpendicular to both is their cross product:

Why the Cross Product?

The defining property of $\vec{b_1}\times\vec{b_2}$ is that it is perpendicular to each factor:

So it points in exactly the direction we need — along both perpendicularity conditions at once. This is why one cross product replaces solving a pair of dot-product equations by hand.

Using It

Suppose a line must be perpendicular to $\vec{b_1} = \hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b_2}=\hat{i}-\hat{j}+\hat{k}$.

So the required line has direction ratios $\langle 5, 2, -3\rangle$. Through a point $A(x_0,y_0,z_0)$ its equation is …