Question

The equation of the plane passing through the points $A(2,5,-3),B(-2,-3,5)$ and $C(5,3,-3)$ is

• A. $\overline{r}.(4\hat{i}-3\hat{j}+2\hat{k})=9$
• B. $\overline{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7$
• C. $\overline{r}.(3\hat{i}+2\hat{j}+4\hat{k})=11$
• D. $\overline{r}.(2\hat{i}+5\hat{j}-\hat{k})=7$

Answer 1

Answer: (A)

$\overline{r}.(4\hat{i}-3\hat{j}+2\hat{k})=9$

Consider the problem,

As we know that equation of line passing through $(x_1,y_1,z_1)$

$a(x-x_1)+b(y-y_1)+z(z-z_1)=0$

where $a,b,c$ are direction ratios

Now, equation of plane passing through $(2,5,-3)$

$a(x-2)+b(y-5)+c(z+3)=0$ ---- $1$

Since it also passes through $(-2,-3,5)$

$a(-2-2)+b(-3-5)+c(5+3)=0$

$-4a-8b+8c=0$

$a+2b-2c=0$ ----- $\left(i\right)$

And it also passes through $(5,3,-3)$

Therefore,

$a(5-2)+b(3-5)+c(-3+3)=0$

$3a-2b=0$ ----- $\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$ by cross multiplication we get

$\frac{a}{0+2}=\frac{b}{-6-0}=\frac{c}{-2-6}$

$\frac{a}{2}=\frac{b}{-6}=\frac{c}{-8}$

$\frac{a}{-2}=\frac{b}{6}=\frac{c}{8}=k$ Consider

Therefore,

$a=-2k,b=6k,c=8k$

Now, from equation $1$

$a(x-2)+b(y-5)+c(z+3)=0$

$-2k(x-2)+6k(y-5)+8k(z+3)=0$

$-2(x-2)+6(y-5)+8(z+3)=0$

$-2x+4+6y-30+8z+24=0$

$x-3y-4z+1=0$

Let $\overrightarrow{r}$ be the position vector. which is $x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{r}(\hat{i}-3\hat{j}-4\hat{k})=-1$