Let a line passing through the point (4,1,0) intersect the line L_1:(x-1)/(2)=(y-2)/(3)=(z…

Question

Let a line passing through the point $(4,1,0)$ intersect the line ${L}_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha ,\beta ,\gamma )$ and the line ${L}_{2}:x-6=y=-z+4$ at the point $B(a,b,c)$. Then $\begin{vmatrix} 1 \quad  0 \quad  1 \ \alpha  \quad  \beta  \quad  \gamma \ a \quad  b \quad  c \end{vmatrix}$ is equal to

Accepted Answer

$$\begin{aligned}
& A=(\alpha, \beta, 4)=(2 \lambda+1,3 \lambda+2,4 \lambda+3) \
& \text { and } B=(a, b, c)=(\mu+6, \mu,-\mu+4) \
& \because \frac{\mu+2}{2 \lambda-3}=\frac{\mu-1}{3 \lambda+1}=\frac{-\mu+4}{4 \lambda+3} \
& \therefore \quad \lambda \mu+8 \lambda+4 \mu=1  \quad \ldots(1)\
& \text { and } 7 \lambda \mu-16 \lambda+4 \mu=7 \quad \ldots(2)
\end{aligned}$$

from equation (1) and (2)

$$\lambda=-1 \text { and } \mu=3$$

Or $\lambda=-\frac{1}{3}$ and $\mu=1$

By taking, $\lambda=-1$ and $\mu=3$, we get

$$\begin{aligned}
& \therefore \quad A(\alpha, \beta, 4)=(-1,-1,-1) \
& \text { and } B(a, b, c)=(9,3,1)
\end{aligned}$$

$$\therefore\left|\[\begin{array}{ccc} 1 & 0 & 1 \ -1 & -1 & -1 \ 9 & 3 & 1 \end{array}$$\right|=8\]

Solutions · 1