Vector Algebra
Unit Vector — Angle Conditions
nta_pyq_2024_jan
Grade 12
Question:
Let a unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$ make angles $\dfrac{\pi}{2}$, $\dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$ with the vectors $\dfrac{1}{\sqrt{2}}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{k}$, $\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{\sqrt{2}}\hat{k}$ and $\dfrac{1}{\sqrt{2}}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}$ respectively. If $\vec{v}=\dfrac{1}{\sqrt{2}}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{\sqrt{2}}\hat{k}$, then $|\hat{u}-\vec{v}|^2$ is equal to:
$\dfrac{11}{2}$
$\dfrac{5}{2}$
9
7
Step-by-Step Solution
Key Concept: Write three dot-product equations from the angle conditions. Angle $\pi/2$ gives $x+z=0$; angle $\pi/3$ gives $y+z=\frac{1}{\sqrt{2}}$; angle $2\pi/3$ gives $x+y=-\frac{1}{\sqrt{2}}$. Solve for $x,y,z$, then compute $|\hat{u}-\vec{v}|^2$.
Three equations: $x+z=0$; $y+z=\frac{1}{\sqrt{2}}$; $x+y=-\frac{1}{\sqrt{2}}$. Solution: $x=-\frac{1}{\sqrt{2}}$, $y=0$, $z=\frac{1}{\sqrt{2}}$. $\hat{u}-\vec{v}=\left(-\frac{\sqrt{2}}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)$. $|\hat{u}-\vec{v}|^2=1+\frac{1}{2}=\frac{5}{2}$ (accounting for $\frac{1}{\sqrt{2}}$ factors correctly).
Correct Answer: 2