If the equation of the plane containing the line x+2 y+3 z-4=0=2 x+y-z+5 and perpendicular…

Question

If the equation of the plane containing the line $x+2 y+3 z-4=0=2 x+y-z+5$ and perpendicular to the plane $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$ is $a x+b y+c z=4$, then $(a-b+c)$ i...

Accepted Answer

Equation of plane $\mathrm{P}$ containing the given lines is

$$
\begin{aligned}
& (x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0 \\
& \Rightarrow(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0
\end{aligned}
$$

Now, plane $\mathrm{P}$ is perpendicular to plane $\mathrm{P}^{\prime}$

$$
\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})
$$

So, normal to plane $\mathrm{P}^{\prime}$ is

$$
\begin{aligned}
& \vec{n}=(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}-2 \hat{j}+3 \hat{k}) \\
& \Rightarrow \vec{n}=5 \hat{i}-2 \hat{j}-3 \hat{k}
\end{aligned}
$$

$\therefore \mathrm{P}$ and $\mathrm{P}^{\prime}$ are perpendicular

$$
\begin{aligned}
& \therefore 5(1+2 \lambda)-2(2+\lambda)-3(3-\lambda)=0 \\
& \Rightarrow 5+10 \lambda-4-2 \lambda-9+3 \lambda=0 \\
& \Rightarrow 11 \lambda=8 \Rightarrow \lambda=\frac{8}{11}
\end{aligned}
$$

$$
\[\begin{gathered} \therefore P:\left(1+\frac{16}{11}\right) x+\left(2+\frac{8}{11}\right) y+\left(3-\frac{8}{11}\right) z+\left(5 \times \frac{8}{11}-4\right)
=0 \end{gathered}$$
\]

i.e., $27 x+30 y+25 z=4$

which is same as $a x+b y+c z=4$

$$
\begin{aligned}
& \therefore a=27, b=30 \text { and } c=25 \\
& \Rightarrow a-b+c=27-30+25=22
\end{aligned}
$$

Solutions · 1