Let P_1 be the plane 3x-y-7z=11 and P_2 be the plane passing through the points (2,-1,0),(…

Question

Let ${P}_{1}$ be the plane $3x-y-7z=11$ and ${P}_{2}$ be the plane passing through the points $(2,-1,0),(2,0,-1)$, and $(5,1,1)$. If the foot of the perpendicular drawn from the point $(7,4,-1)$ on the line of intersection of the planes ${P}_{1}$ and ${P}_{2}$ is $(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +\gamma $ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 11

Accepted Answer

Equation of plane ${P}_{2}$ passing through $(2,-1,0),(2,0$, $-1)$ and $(5,1,1)$ is

$\begin{gathered} |\begin{array}{ccc}x-5 \quad  y-1 \quad  z-1 \ 3 \quad  2 \quad  1 \ 3 \quad  1 \quad  2\end{array}|=0 \ \Rightarrow (x-5)(4-1)-(y-1)(6-3)+(z-1)(3-6)=0 \ \Rightarrow 3x-15-3y+3-3z+3=0 \ \Rightarrow 3x-3y-3z-9=0 \ \Rightarrow x-y-z=3.......(i) \end{gathered}$

Now, direction ratios of line of intersection of ${P}_{1}$ and ${P}_{2}$ is

$\begin{gathered} |\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  -1 \quad  -1 \ 3 \quad  -1 \quad  -7\end{array}| \ =\hat{i}(7-1)-\hat{j}(-7+3)+\hat{k}(-1+3) \ =6\hat{i}+4\hat{j}+2\hat{k} \end{gathered}$

At $z=0,x-y=3$
[from (i)]

$3x-y=11$

On solving, we get

$x=4y=1$

So, equation of line is

$\frac{x-4}{6}=\frac{y-1}{4}=\frac{z-2}{6}=k$

$\begin{gathered} 	herefore (\alpha ,\beta ,\gamma )=(6k+4,4k+1,2k) \ \Rightarrow (6)(\alpha -7)+4(\beta -4)+2(\gamma +1)=0 \ \Rightarrow 6(6k+4-7)+4(4k+1-4)+2(2k+1)=0 \ \Rightarrow 36k-18+16k-12+4k+4=0 \ \Rightarrow 56k=26\Rightarrow k=\frac{1}{2} \ \alpha =7,\beta =3\gamma =1 \ 	herefore \alpha +\beta +\gamma =7+3+1=11 \end{gathered}$

Solutions · 1