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

The key idea is to split $\vec{b}$ into two parts: one parallel to $\vec{a}$ (call it $\vec{d}$) and one perpendicular to $\vec{a}$ (call it $\vec{c}$). This is exactly the standard decomposition of a vector into components along and orthogonal to a given direction. Once we have $\vec{c}$ and $\vec{d}$, the quantity $(\vec{c} \times \vec{d})^2$ is simply $|\vec{c}|^2 |\vec{d}|^2$ because $\vec{c}$ and $\vec{d}$ are perpendicular (since $\vec{c} \perp \vec{a}$ and $\vec{d} \parallel \vec{a}$, so $\vec{c} \perp \vec{d}$). The cross product of two perpendicular vectors has magnitude equal to the product of their lengths.

1. Find $\vec{d}$, the component of $\vec{b}$ parallel to $\vec{a}$. The projection formula gives

Compute the dot product:

The magnitude squared of $\vec{a}$ is

Hence

2. Find $\vec{c}$, the component perpendicular to $\vec{a}$. Since $\vec{b} = \vec{c} + \vec{d}$, we have

This gives

3. Check perpendicularity.

   $\vec{c} \cdot \vec{a} = \frac{2}{3}(1) + \left(-\frac{4}{3}\right)(1) + \frac{2}{3}(1) = \frac{2}{3} - \frac{4}{3} + \frac{2}{3} = 0$, so indeed $\vec{c} \perp \vec{a}$.

4. Compute $|\vec{c}|^2$ and $|\vec{d}|^2$.

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Concept & Intuition

The internal bisector of the angle between two vectors points in the direction of their sum when the vectors have equal length. But here the vectors have different magnitudes, so we must first convert them to unit vectors. The bisector direction is then the sum of those unit vectors. Once we have that direction, we scale it to the required magnitude $\sqrt{2}$ and compare with the options.

1. Find unit vectors along each given vector

   Let $\vec{a} = 2\vec{i} - 2\vec{j} + \vec{k}$ and $\vec{b} = \vec{i} + 2\vec{j} + 2\vec{k}$.

   Their magnitudes:

   $|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$

   $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$

   Both have magnitude 3, so the unit vectors are:

   $\hat{a} = \frac{1}{3}(2\vec{i} - 2\vec{j} + \vec{k})$

   $\hat{b} = \frac{1}{3}(\vec{i} + 2\vec{j} + 2\vec{k})$

2. Direction of the internal bisector

   The internal bisector of the angle between $\vec{a}$ and $\vec{b}$ is along $\hat{a} + \hat{b}$ (since both are unit vectors, their sum points exactly halfway).

   Compute:

   $\hat{a} + \hat{b} = \frac{1}{3}\big[(2+1)\vec{i} + (-2+2)\vec{j} + (1+2)\vec{k}\big] = \frac{1}{3}(3\vec{i} + 0\vec{j} + 3\vec{k}) = \vec{i} + \vec{k}$

3. Scale to magnitude $\sqrt{2}$

   The vector $\vec{i} + \vec{k}$ has magnitude $\sqrt{1^2+0^2+1^2} = \sqrt{2}$.

   So it already has exactly the required magnitude $\sqrt{2}$.

   Therefore the required vector is $\vec{i} + \vec{k}$.

4. Check the options …