91Ó°ÊÓ

If two vectors are$\stackrel{→}{a}$and$\stackrel{→}{b}$,and the angle between them is$\mathrm{θ}$, using the triangle rule of vector addition, the resultant vector has a magnitude$R=\sqrt{a^2+b^2+2ab\cos \mathrm{θ}}$.

Given data:

The first vector is ${\stackrel{→}{v}}_1$.

The second vector is ${\stackrel{→}{v}}_2$.

Assumptions:

Let $\mathrm{θ}$be the angle between the vectors for part (a), and $\mathrm{ϕ}$be the angle between the vectors for part (b).

It is given that the magnitude of the resultant vector is $v_{\mathrm{R}}=v_1+v_2$.

You know that the resultant is

$\begin{array}{c}{v}_{\mathrm{R}}=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{θ}}\\ v_1+v_2=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{θ}}\\ {\left(v_1+v_2\right)}^2=v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{θ}\\ v_1^2+v_2^2+2v_1{v}_2=v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{θ}\end{array}$

Comparing both sides of the above equation,

$\begin{array}{c}\cos \mathrm{θ}=1\\ \mathrm{θ}={\cos }^{-1}\left(1\right)\\ \mathrm{θ}=0°\end{array}$

Hence, ${\stackrel{→}{v}}_1$and ${\stackrel{→}{v}}_2$are in the same direction when $v_{\mathrm{R}}=v_1+v_2$.

The given condition is $v_{\mathrm{R}}^2=v_1^2+v_2^2$.

You know that the resultant is

$\begin{array}{c}{v}_{\mathrm{R}}=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{Ï•}}\\ {v_{\mathrm{R}}}^2=v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{Ï•}\\ v_1^2+v_2^2=v_1^2+v_2^2+2v_1{v}_2\cos \mathrm{Ï•}\end{array}$

Comparing both sides of the above equation,

role="math" localid="1644912448369" $\begin{array}{c}\cos \mathrm{ϕ}=0\\ \mathrm{ϕ}={\cos }^{-1}\left(0\right)\\ \mathrm{ϕ}=90°\end{array}$

Hence, ${\stackrel{→}{v}}_1$and ${\stackrel{→}{v}}_2$are perpendicular when $v_{\mathrm{R}}^2=v_1^2+v_2^2$.

The given condition is $v_1+v_2=v_1-v_2$.

Now, simplifying the given condition,

$\begin{array}{c}{v}_1+v_2-\left(v_1-v_2\right)=0\\ v_1+v_2-v_1+v_2=0\\ 2v_2=0\\ v_2=0\end{array}$

Hence, the magnitude of ${\stackrel{→}{v}}_2$is zero when $v_1+v_2=v_1-v_2$.