1. Use the condition: equal angles with the coordinate axes

If $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ makes equal angles with the coordinate axes, then its direction cosines are equal: $\cos\alpha=\cos\beta=\cos\gamma.$ Hence, $a_1=a_2=a_3=t \quad (t>0).$ So, $\vec a=t(\hat i+\hat j+\hat k).$

2. Use the given projection of $\vec a$ on $3\hat i+4\hat j$

Let $\vec v=3\hat i+4\hat j.$ Its magnitude is $|\vec v|=\sqrt{3^2+4^2}=5.$

The scalar projection of $\vec a$ on $\vec v$ is $\frac{\vec a\cdot \vec v}{|\vec v|}=7.$ Now, $\vec a\cdot \vec v=t(1,1,1)\cdot(3,4,0)=t(3+4)=7t.$ Thus, $\frac{7t}{5}=7 \implies t=5.$ Therefore, $\vec a=5(\hat i+\hat j+\hat k)=(5,5,5).$

3. Use the condition on $\vec b$

$\vec b$ is obtained by rotating $\vec a$ by $90^\circ$, and $\vec a,\vec b$ and the x-axis are coplanar.

So:

• $\vec a\perp \vec b$,
• $|\vec b|=|\vec a|$ (rotation preserves magnitude),
• $\vec b$ lies in the plane containing $\vec a$ and the x-axis.

Let the x-axis direction vector be $\hat i=(1,0,0).$ Since $\vec b$ lies in the plane of $\vec a$ and $\hat i$, write $\vec b=\alpha \vec a+\beta \hat i.$ That is, $\vec b=\alpha(5,5,5)+\beta(1,0,0)=(5\alpha+\beta,5\alpha,5\alpha).$

4. Apply perpendicularity: $\vec a\cdot\vec b=0$

$(5,5,5)\cdot(5\alpha+\beta,5\alpha,5\alpha)=0$ $5(5\alpha+\beta)+25\alpha+25\alpha=0$ $75\alpha+5\beta=0$ $15\alpha+\beta=0 \implies \beta=-15\alpha.$

Hence, $\vec b=(-10\alpha,5\alpha,5\alpha)=5\alpha(-2,1,1).$

5. Apply equal magnitude: $|\vec b|=|\vec a|$

First, $|\vec a|=\sqrt{5^2+5^2+5^2}=5\sqrt3.$ Also, $|\vec b|=|5\alpha|\sqrt{(-2)^2+1^2+1^2}=5|\alpha|\sqrt6.$ So, $5|\alpha|\sqrt6=5\sqrt3$ $|\alpha|=\frac{1}{\sqrt2}.$ Thus, $\vec b=\pm \frac{5}{\sqrt2}(-2,1,1).$

6. Find the projection of $\vec b$ on $3\hat i+4\hat j$

Using $\vec v=(3,4,0)$,