Vectors - The Basics — Question 3

Question 3

Find all real $k$ for which $\vec v=\langle k,2k-1,k+2\rangle$ has magnitude $3$.

Question 3 – Solution

Identify the initial and terminal information first. Build the vector by subtracting coordinates, then use the magnitude formula or normalize only after the components are correct.

Square the magnitude equation: $k^2+(2k-1)^2+(k+2)^2=9$.

Expanding gives $6k^2+5=9$, so $3k^2=2$.

Therefore $\boxed{k=\pm\sqrt{2/3}}$; substitution verifies both values.

The result follows from the defining vector formulas used above, and each component, magnitude, or scalar condition has been checked against the information in the question.

A vector separates two ideas: its components describe displacement, while its magnitude and unit vector describe size and direction. Always check signs and confirm that a unit vector has magnitude $1$.