Question

Find a vector v whose magnitude is 3 and whose component in the i direction is equal to the component in the j direction.

Short answer

Vector v is $\frac{3}{\sqrt{2}}\stackrel{\rightharpoonup }{i}+\frac{3}{\sqrt{2}}\stackrel{\rightharpoonup }{j}$

Step-by-step solution

Step. 1 Given 

|v| = 3 and,

angle of the vector from both $\stackrel{\rightharpoonup }{i}and\stackrel{\rightharpoonup }{j}$directions is same suppose that angle be $\theta$.

Step. 2 Vector representation

Now, As we know the general form of a vector is,

$\stackrel{\rightharpoonup }{v}=r\left(\cos \left(\theta \right)\stackrel{\rightharpoonup }{i}+\sin \left(\theta \right)\stackrel{\rightharpoonup }{j}\right)$

where, r is the magnitude of the vector i.e., distance from the origin and $\theta$is the angle of that vector from x-axis or $\stackrel{\rightharpoonup }{i}$direction.

Now, as magnitude in the question is given itself as 3. So,

role="math" localid="1646738176258" $r=3$.

So,

$\stackrel{\rightharpoonup }{v}=3\left(\cos \left(\theta \right)\stackrel{\rightharpoonup }{i}+\sin \left(\theta \right)\stackrel{\rightharpoonup }{j}\right)$

Step. 3 Final solution

Now, we have left with only $\theta$,

So, as it is given that it makes equal angles with and we know these directions are mutually perpendicular thatswhy

$\theta =45°or\frac{\pi }{4},$

Finally,

$$\begin{gathered} \stackrel{\rightharpoonup }{v}=3\left(\cos \left(\frac{\mathrm{\pi }}{4}\right)\stackrel{\rightharpoonup }{i}+\sin \left(\frac{\mathrm{\pi }}{4}\right)\stackrel{\rightharpoonup }{j}\right), \\ \stackrel{\rightharpoonup }{v}=3(\frac{\stackrel{\rightharpoonup }{i}}{\sqrt{2}}+\frac{\stackrel{\rightharpoonup }{j}}{\sqrt{2}}), \\ \stackrel{\rightharpoonup }{v}=\frac{3}{\sqrt{2}}\stackrel{\rightharpoonup }{i}+\frac{3}{\sqrt{2}}\stackrel{\rightharpoonup }{j}. \end{gathered}$$