if \(2 \mathrm{i}+\mathrm{j}, \underline{\mathrm{i}}-3 \mathrm{j}, 3 \mathrm{i}+2 \mathrm{j} \& \underline{\mathrm{i}}+\lambda \mathrm{j}\) are position vectors of points \(\mathrm{A}\), \(\mathrm{B}, \mathrm{C}, \mathrm{D}\) respectively and \(\mathrm{AB} \| \mathrm{CD}\). Then \(\lambda\) is (a) 6 (b) 8 (c) \(-8\) (d) -6

Imagine standing on a giant grid, with each point on the grid representing a location in space. Now, think of a position vector as the arrow that points directly from the origin of the grid, where you are standing, to any point on this grid. It tells us exactly where that point is relative to where you're standing.

Each position vector has components that correspond to how far you need to go along the grid lines to reach its point. In two dimensions, we describe this with 'i' and 'j', the unit vectors along the x and y-axis respectively. So, if you have a position vector like \(2i + j\), it means you'd move 2 units along the x-axis and 1 unit along the y-axis to reach the point this vector points to. This is essential in physics and engineering to describe locations or in our case, to figure out how points relate to each other in space, like for understanding vector parallelism.

When we talk about moving from one point to another, vector subtraction comes into play. It's like having two position vectors, say one pointing to your friend and another pointing to a nearby tree. If you're curious about the direction from your friend to the tree, you can use vector subtraction.

Subtracting one position vector from another gives you a vector that points directly from one to the other. In our exercise, to find vector AB, we subtract the position vector of point A from that of point B. Mathematically, it's akin to subtracting each component: \((i - 3j) - (2i + j)\) simply becomes \(-i - 4j\) after aligning and subtracting each corresponding 'i' and 'j' component. Vector subtraction thus helps us understand the relationship and the relative position between two points.

Imagine stretching or shrinking an arrow without changing its direction. That's what scalar multiplication does to vectors. It takes a number, known as a scalar, and multiplies it by the vector, scaling its magnitude by that number. If the scalar is positive, the vector’s direction remains unchanged. If it's negative, the vector flips direction.

In the context of our problem, the concept of scalar multiplication is used to express vector parallelism. Two vectors are parallel if one is a stretched or shrunk version of the other, or mathematically, one is a scalar multiple of the other. This very idea enables us to write \(AB = k \times CD\). We are basically saying one vector can be turned into the other by simply scaling it up or down. By comparing the components, we can find the scalar 'k' that aligns one vector perfectly with the other, confirming their parallelism or helping us solve for unknowns like λ in the exercise.