a) In any triangle PQR, if p\, ^2 R2+r\, ^2 P2= q2 , prove that the sides are in A.P. [3] b) If a=4 i 3 j+2 k and b=3 i 2 j+4 k are two vectors, find the projection b on a [2]

a) Proof that the sides are in A.P.

In $△PQR$ let the sides opposite $P,Q,R$ be $p,q,r$ and $s=\frac{p+q+r}{2}$ be the semiperimeter.

$${\sin }^2\frac{A}{2}=\frac{(s-b)(s-c)}{bc}\text{(for any angle }A\text{ opposite side }a)$$

Hence

$${\sin }^2\frac{R}{2}=\frac{(s-p)(s-q)}{pq},{\sin }^2\frac{P}{2}=\frac{(s-q)(s-r)}{qr}$$

Substituting in the given relation

$$p{\sin }^2\frac{R}{2}+r{\sin }^2\frac{P}{2}=\frac{q}{2}$$

$$\begin{array}{rl}p\frac{(s-p)(s-q)}{pq}+r\frac{(s-q)(s-r)}{qr}& =\frac{(s-p)(s-q)}{q}+\frac{(s-q)(s-r)}{q}\\ & =\frac{(s-q)[(s-p)+(s-r)]}{q}\\ & =\frac{(s-q)[2s-(p+r)]}{q}\end{array}$$

Since $2s=p+q+r$,

$$2s-(p+r)=q$$

Thus

$$\frac{(s-q)q}{q}=s-q=\frac{q}{2}⟹s=\frac{3q}{2}$$

But $s=\frac{p+q+r}{2}$; therefore

$$\frac{p+q+r}{2}=\frac{3q}{2}⟹p+r=2q$$

Hence the three sides satisfy

$$p,q,r\text{with}p+r=2q$$

which is the condition for an arithmetic progression. ∎

b) Projection of $\vec{b}$ on $\vec{a}$

The projection of $\vec{b}$ on $\vec{a}$ is

$${\mathrm{proj}}_{\vec{a}}\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}{|}^2}\vec{a}$$

$$\begin{array}{rl}\vec{a}\cdot \vec{b}& =(4)(3)+(-3)(-2)+(2)(4)=12+6+8=26\\ |\vec{a}{|}^2& =4^2+(-3{)}^2+2^2=16+9+4=29\end{array}$$

Scalar factor

$$\frac{\vec{a}\cdot \vec{b}}{|\vec{a}{|}^2}=\frac{26}{29}$$

Therefore

$${\mathrm{proj}}_{\vec{a}}\vec{b}=\frac{26}{29}(4\vec{i}-3\vec{j}+2\vec{k})=\boxed{\frac{104}{29}\vec{i}-\frac{78}{29}\vec{j}+\frac{52}{29}\vec{k}}$$