Q.If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$, then $\vec{a}$ and $\vec{b}$ are: (A) collinear vectors which are not parallel (B) parallel vectors (C) perpendicular vectors (D) unit vectors

To determine the relationship between vectors, we typically check for parallelism/collinearity and perpendicularity.

Concept: Orthogonal And Parallel Vectors

1. Check for Parallelism/Collinearity: Two vectors $\vec{a}$ and $\vec{b}$ are parallel if $\vec{a} = k\vec{b}$ for some scalar $k$.

   Comparing components of $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$:

   $\frac{2}{1} = 2$, $\frac{-1}{1} = -1$, $\frac{1}{-1} = -1$.

   Since the ratios are not equal ($2 \neq -1$), $\vec{a}$ is not a scalar multiple of $\vec{b}$, so they are not parallel or collinear.

2. Check for Perpendicularity: Two vectors are perpendicular if their dot product is zero. …

When we are given two vectors and asked to describe their relationship, we typically check for a few key properties: whether they are unit vectors, parallel (or collinear), or perpendicular. Each of these properties has a specific mathematical condition that we can test.

Concept and Intuition

1. Unit Vectors: A vector is a unit vector if its magnitude (length) is exactly 1. Unit vectors are important because they represent direction without any associated "strength" or scale.

   • How to check: Calculate the magnitude of each vector using the formula $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. If $|\vec{v}| = 1$, it's a unit vector.

2. Parallel Vectors (and Collinear Vectors): Two non-zero vectors are parallel if they point in the same direction or in exactly opposite directions. Geometrically, if you place their initial points at the same origin, they would lie along the same line. Algebraically, one vector is a scalar multiple of the other. Collinear vectors are essentially parallel vectors; the term "collinear" emphasizes that they lie on the same line. In the context of multiple-choice options, "parallel" usually implies collinearity.

   • How to check: See if $\vec{a} = k\vec{b}$ for some scalar $k$. This means the ratio of their corresponding components must be equal: $\frac{a_x}{b_x} = \frac{a_y}{b_y} = \frac{a_z}{b_z} = k$. Alternatively, their cross product $\vec{a} \times \vec{b}$ will be the zero vector $\vec{0}$.

3. Perpendicular Vectors (Orthogonal Vectors): Two non-zero vectors are perpendicular if the angle between them is $90^\circ$. Geometrically, they form a right angle.

   • How to check: Their dot product must be zero. This comes directly from the definition of the dot product: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$. If $\theta = 90^\circ$, then $\cos\theta = 0$, so $\vec{a} \cdot \vec{b} = 0$.

   For two vectors $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ and $\vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$:

   $\vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z$

Let's apply these checks to the given vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$.

Step-by-Step Solution

1. Check if they are unit vectors.

   First, we calculate the magnitude of each vector.

   For $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$:

   $|\vec{a}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

   Since $\sqrt{6} \neq 1$, $\vec{a}$ is not a unit vector.

   For $\vec{b} = \hat{i} + \hat{j} - \hat{k}$:

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

   Since $\sqrt{3} \neq 1$, $\vec{b}$ is not a unit vector.

   Therefore, option (D) is incorrect. …