Q.Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$.

We are given two vectors $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$, and a unit vector $\vec{a}$ such that

We need $\left| \vec{a} \times (\vec{b} \times \vec{c}) \right|$.

Concept and intuition:

The expression $\cos(\vec{a}, \vec{b} \times \vec{c})$ means the cosine of the angle between $\vec{a}$ and the vector $\vec{d} = \vec{b} \times \vec{c}$. If we know that cosine, we know the sine of that same angle (since $\sin^2\theta + \cos^2\theta = 1$). Then the magnitude of the cross product $\vec{a} \times \vec{d}$ is $|\vec{a}| |\vec{d}| \sin\theta$. Since $\vec{a}$ is a unit vector, this simplifies to $|\vec{d}| \sin\theta$. So we just need $|\vec{d}|$ and $\sin\theta$.

1. Compute $\vec{d} = \vec{b} \times \vec{c}$.

So $\vec{d} = 3(-\hat{i} + \hat{j} + \hat{k})$.

2. Find $|\vec{d}|$.

3. Find $\sin\theta$ where $\theta$ is the angle between $\vec{a}$ and $\vec{d}$. Given $\cos\theta = \frac{\sqrt{2}}{\sqrt{3}}$, we have

Hence $\sin\theta = \frac{1}{\sqrt{3}}$ (positive, since angle between vectors is between $0$ and $\pi$, and sine is nonnegative there).

4. Compute $\left| \vec{a} \times \vec{d} \right|$. Since $\vec{a}$ is a unit vector,

We are given three vectors $\vec{a}, \vec{b}, \vec{c}$ each of magnitude $\sqrt{2}$, and the angle between any two is $\pi/3$.

We define $\vec{x} = \vec{a} \times (\vec{b} \times \vec{c})$ and $\vec{y} = \vec{b} \times (\vec{c} \times \vec{a})$.

We need to compare $|\vec{x}|$ and $|\vec{y}|$.

Concept and intuition:

The vector triple product identity is

This lets us rewrite $\vec{x}$ and $\vec{y}$ in terms of dot products. Since all magnitudes and pairwise angles are equal, the dot products are all the same. That symmetry suggests $\vec{x}$ and $\vec{y}$ will have the same magnitude.

Let’s work it out step by step.

1. Compute the common dot product. For any two vectors, say $\vec{a}$ and $\vec{b}$,

So $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 1$.

2. Express $\vec{x}$ using the identity.

Substituting the dot products:

3. Express $\vec{y}$ similarly.

Again, each dot product is 1:

4. Find $|\vec{x}|$.

We have $|\vec{b}|^2 = |\vec{c}|^2 = (\sqrt{2})^2 = 2$, and $\vec{b}\cdot\vec{c}=1$. So

5. Find $|\vec{y}|$.

so $|\vec{y}| = \sqrt{2}$ as well.

Thus $|\vec{x}| = |\vec{y}|$.