Question

Find a unit vector in the same direction as the vector A = 4i−2j+ 4k, and another unit vector in the same direction as B = −4i + 3k. Show that the vector sum of these unit vectors bisects the angle between A and B. Hint: Sketch the rhombus having the two unit vectors as adjacent sides.

Answer 1

Answer with Step-by-step explanation:

We are given that

A=4i-2j+4k

B=-4i+3k

$\mid A\mid=\sqrt{4^2+(-2)^2+4^2}=6$

$mid B\mid=\sqrt{3^2+(-4)^2}=5$

$\hat{A}=\frac{A}{\mid A\mid}$

$\hat{A}=\frac{4i-2j+4k}{6}=\frac{2}{3}i-\frac{1}{3}j+\frac{2}{3}k$

$\hat{B}=\frac{-4i+3k}{5}=\frac{-4}{5}i+\frac{3}{5}k$

Sum of unit vectors=$\hat{A}+\hat{B}$

Sum of unit vectors=$\frac{2}{3}i-\frac{1}{3}j+\frac{2}{3}k+\frac{-4i+3k}{5}=\frac{-4}{5}i+\frac{3}{5}k$

Sum of unit vectors=$\frac{-2}{15}i-\frac{1}{3}j+\frac{19}{15}k$

$\mid \hat{A}+\hat{B}\mid=\sqrt{(\frac{-2}{15})^2+(\frac{1}{3})^2+(\frac{19}{15})^2}$

$\mid \hat{A}+\hat{B}\mid=1.32$

$\theta_1=Cos^{-1}(\frac{A\cdot B)}{\mid A\mid \mid B\mid}$

$\theta_1=cos^{-1}(\frac{-4}{30})=97.6^{\circ}$

$\theta_2=cos^{-1}(\frac{(Sum\;of\;unt\;vectors\cdot A)}{\mid sum\mid \mid A\mid }$

$\theta_2=cos^{-1}(\frac{78}{15\cdot 2\cdot 1.32\cdot 6})=49^{\circ}$

$\frac{1}{2}\theta_1=\frac{1}{2}(97.6)=48.8\sim 49^{\circ}=\theta_2$

Hence, proved.

Answer 2

A unit vector in the same direction as a vector $v$ can be obtained by multiplying $v$ by the reciprocal of its norm. So $\vec A$ and $\vec B$ have corresponding unit vectors $\vec a$ and $\vec b$,

$\vec a=\dfrac{\vec A}{\|\vec A\|}=\dfrac{4\,\vec\imath-2\,\vec\jmath+4\,\vec k}{\sqrt{4^2+(-2)^2+4^2}}=\dfrac23\,\vec\imath-\dfrac13\,\vec\jmath+\dfrac23\,\vec k$

$\vec b=\dfrac{-4\,\vec\imath+3\,\vec k}{\sqrt{(-4)^2+3^2}}=-\dfrac45\,\vec\imath+\dfrac35\,\vec k$

They have vector sum

$\vec a+\vec b=-\dfrac2{15}\,\vec\imath-\dfrac13\,\vec\jmath+\dfrac{19}{15}\,\vec k$

Find the angle between $\vec A$ and $\vec B$ using the dot product formula:

$\vec A\cdot\vec B=\|\vec A\|\|\vec B\|\cos\theta$

$-4=30\cos\theta$

$\cos\theta=-\dfrac2{15}$

$\theta\approx97.66^\circ$

Then find the angle between either $\vec A$ or $\vec B$ and $\vec a+\vec b$:

$\vec A\cdot(\vec a+\vec b)=\|\vec A\|\|\vec a+\vec b\|\cos\varphi$

$\dfrac{26}5=2\sqrt{\dfrac{78}5}\cos\varphi$

$\cos\varphi=\sqrt{\dfrac{13}{30}}$

$\varphi\approx48.831^\circ$

(and this is half of $\theta$)