Question

Two points $A(7,0,6)$ and $B(3,4,2)$ lie on a plane $P_1$ and plane $P_2:2x-5y=15$ is perpendicular to the plane $P_1$ and point $(2,\alpha ,\beta )$ also lie on plane $P_1$. Then find the value of $(2\alpha -3\beta )$

• A. $12$
• B. $17$
• C. $5$
• D. $7$

Answer 1

The correct option is A $12$

Cartesian equation of plane passing through a given point and normal to a given vector

$(x-x_o)a+(y-y_o)b+(z-z_o)c=0$

So, $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{a}=x_o\hat{i}+y_o\hat{j}+z_o\hat{k}$

$\overrightarrow{n}=a\hat{i}+b\hat{j}+c\hat{k}$

Putting in $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

We get $(x-x_o)a+(y-y_o)b+(z-z_o)c=0$

Conversion of equation is normal form (vector form)

The equation $ax+by+cz+D=0$ is converted in normal by

$ax+by+cz=-D$

or $-ax-by-cz=D$

Dividing by $\sqrt{a^2+b^2+c^2}$

$\frac{-ax-by-cz}{\sqrt{a^2+b^2+c^2}}=\frac{D}{\sqrt{a^2+b^2+c^2}}$

We get , $lx+my+nz=d$

Normal vector of plane where direction ratios are $2,-5,0$ from $2x-5y=15$ and $4,-4,4$ from $A(3,4,2)$ & $B(7,0,6)$

$\therefore$ normal vector of plane $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -5& 0\\ 4& -4& 4\end{array}\right|=-4(5\hat{i}+2\hat{j}-3\hat{k}$

Equation of plane is 5 (x-7) + 2y - 3 (z-6) = 0

$\Rightarrow 5x+2y-3z=0\Rightarrow 2\alpha -3\beta =17-(5\times 2)=7$

Correct answer is D