Let a unit vector u=x i+y j+z k make angles (π)/2, (π)/3 and (2 π)/3 with the vectors 1/(√(2)) i+1/(√(2)) k, 1/(√(2)) j+1/(√(2)) k and 1/(√(2)) i+1/(√(2)) j respectively. If v=1/(√(2)) i+1/(√(2)) j+1/(√(2)) k then |u-v|² is equal to

Vector Algebra: JEE Main 2024 Mathematics question

JEE Main 2024 · 29 Jan, Evening shift · Q5785

JEE Main 202429 Jan, Evening shiftMathematicsHardVector Algebra184× since 2002

Let a unit vector

\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}

make angles

\frac{\pi}{2}, \frac{\pi}{3}

and

\frac{2 \pi}{3}

with the vectors

\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}

and

\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}

respectively. If

\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}

then

|\hat{u}-\vec{v}|^2

is equal to