Vector Algebra Part 3

Concept:Lines $L_1$ and $L_2$ intersect at point $R$.We find $R$ by equating their vector equations.Then we locate points $P$ on $L_1$ and $Q$ on $L_2$ using given distances $|\overrightarrow{PR}|$ and $|\overrightarrow{PQ}|$.Finally compute $27(QR)^2$.Explanation:Equate the position vectors of $L_1$ and $L_2$:$L_1: (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}+4\hat{k})$$L_2: (4\hat{i}+\hat{j}) + \mu(5\hat{i}+2\hat{j}+\hat{k})$Componentwise:$1+2\lambda = 4+5\mu \implies 2\lambda-5\mu=3$ ... (1)$2+3\lambda = 1+2\mu \implies 3\lambda-2\mu=-1$ ... (2)$3+4\lambda = \mu \implies \mu-4\lambda=3$ ... (3)Solve (1) and (2): Multiply (1) by 3 and (2) by 2, subtract:$6\lambda-15\mu - (6\lambda-4\mu) = 9 - (-2) \implies -11\mu=11 \implies \mu=-1$From (1): $2\lambda-5(-1)=3 \implies 2\lambda=-2 \implies \lambda=-1$Check (3): $(-1)-4(-1)=3$, consistent.Find $R$ using $\lambda=-1$ in $L_1$:$\overrightarrow{OR}= (\hat{i}+2\hat{j}+3\hat{k}) - (2\hat{i}+3\hat{j}+4\hat{k}) = -\hat{i}-\hat{j}-\hat{k}$Point $P$ on $L_1$: $\overrightarrow{OP}= (1+2\lambda)\hat{i}+(2+3\lambda)\hat{j}+(3+4\lambda)\hat{k}$Given $|\overrightarrow{PR}|=\sqrt{29} \implies |\overrightarrow{PR}|^2=29$$\overrightarrow{PR}= \overrightarrow{OR}-\overrightarrow{OP}= (-2-2\lambda)\hat{i}+(-3-3\lambda)\hat{j}+(-4-4\lambda)\hat{k}$$|\overrightarrow{PR}|^2 = 4(1+\lambda)^2+9(1+\lambda)^2+16(1+\lambda)^2 = 29(1+\lambda)^2=29$$\implies (1+\lambda)^2=1 \implies \lambda=0 \text{ or } -2$$P$ lies in first octant, so all coordinates positive. $\lambda=0$ gives $\overrightarrow{OP}=\hat{i}+2\hat{j}+3\hat{k}$ (positive). $\lambda=-2$ gives negative coordinates, discard.Point $Q$ on $L_2$: $\overrightarrow{OQ}= (4+5\mu)\hat{i}+(1+2\mu)\hat{j}+\mu\hat{k}$Given $|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}} \implies |\overrightarrow{PQ}|^2=\frac{47}{3}$$\overrightarrow{PQ}= \overrightarrow{OQ}-\overrightarrow{OP}= (3+5\mu)\hat{i}+(-1+2\mu)\hat{j}+(\mu-3)\hat{k}$Square and expand:$(3+5\mu)^2+(-1+2\mu)^2+(\mu-3)^2 = \frac{47}{3}$$9+30\mu+25\mu^2 +1-4\mu+4\mu^2 +\mu^2-6\mu+9 = 30\mu^2+20\mu+19 = \frac{47}{3}$Multiply by 3: $90\mu^2+60\mu+57=47 \implies 90\mu^2+60\mu+10=0$Divide by 10: $9\mu^2+6\mu+1=0 \implies (3\mu+1)^2=0 \implies \mu=-\frac{1}{3}$Thus $\overrightarrow{OQ}= \left(4-\frac{5}{3}\right)\hat{i}+\left(1-\frac{2}{3}\right)\hat{j}-\frac{1}{3}\hat{k}= \frac{7}{3}\hat{i}+\frac{1}{3}\hat{j}-\frac{1}{3}\hat{k}$Now find $\overrightarrow{QR}= \overrightarrow{OR}-\overrightarrow{OQ}= \left(-1-\frac{7}{3}\right)\hat{i}+\left(-1-\frac{1}{3}\right)\hat{j}+\left(-1+\frac{1}{3}\right)\hat{k}= -\frac{10}{3}\hat{i}-\frac{4}{3}\hat{j}-\frac{2}{3}\hat{k}$$|\overrightarrow{QR}|^2 = \left(\frac{10}{3}\right)^2+\left(\frac{4}{3}\right)^2+\left(\frac{2}{3}\right)^2 = \frac{100+16+4}{9} = \frac{120}{9} = \frac{40}{3}$Then $27(QR)^2 = 27 \times \frac{40}{3} = 360$Answer:360 (Option A)