Let a unit vector u = x i+y j+z k make angles 2, 3, 23 with the vectors 1√2 i+1√2 k, 1√2 j…

Question

Let a unit vector $\hat u = x\hat i+y\hat j+z\hat k$ make angles $\dfrac\pi2$, $\dfrac\pi3$, $\dfrac{2\pi}3$ with the vectors $\dfrac1{\sqrt2}\hat i+\dfrac1{\sqrt2}\hat k$, $\dfrac1{\sqrt2}\hat j+\dfrac1{\sqrt2}\hat k$, $\dfrac1{\sqrt2}\hat i+\dfrac1{\sqrt2}\hat j$ respectively. If $\vec v = \dfrac1{\sqrt2}\hat i+\dfrac1{\sqrt2}\hat j+\dfrac1{\sqrt2}\hat k$, then $|\hat u-\vec v|^2$ is equal to

Accepted Answer

$\cos\frac\pi2=0=\dfrac{x+z}{\sqrt2} \Rightarrow x+z=0$.

$\cos\frac\pi3=\dfrac12=\dfrac{y+z}{\sqrt2} \Rightarrow y+z=\dfrac1{\sqrt2}$.

$\cos\frac{2\pi}3=-\dfrac12=\dfrac{x+y}{\sqrt2} \Rightarrow x+y=-\dfrac1{\sqrt2}$.

Solving: $x=-z$; from the third, $y=-\frac1{\sqrt2}-x$; substituting into the second: $-\frac1{\sqrt2}-x+z=\frac1{\sqrt2} \Rightarrow z-x=\sqrt2 \Rightarrow 2z=\sqrt2 \Rightarrow z=\dfrac1{\sqrt2},\ x=-\dfrac1{\sqrt2},\ y=0$.

$\hat u = \left(-\dfrac1{\sqrt2},0,\dfrac1{\sqrt2}\right)$. $\hat u-\vec v = \left(-\sqrt2,-\dfrac1{\sqrt2},0\right)$.

$$|\hat u-\vec v|^2 = 2+\frac12+0 = \frac52$$

Solutions · 1